**INSTANT DOWNLOAD WITH ANSWERS**

#### Business Statistics A Decision Making Approach 9th Edition by David F. Groebner – Test Bank

*Business Statistics, 9e ***(Groebner/Shannon/Fry)**

**Chapter 6Ā Ā Introduction to Continuous Probability Distributions**

**Ā **

1) The normal distribution is one of the most frequently used discrete probability distributions.

Answer:Ā FALSE

Diff: 1

Keywords:Ā normal distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

2) Typically, a continuous random variable is one whose value is determined by measurement instead of counting.

Answer:Ā TRUE

Diff: 1

Keywords:Ā continuous, random, variable

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

3) The number of defects manufactured by workers in a small engine plant is an example of a discrete random variable.

Answer:Ā TRUE

Diff: 1

Keywords:Ā discrete, random, variable

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

4) One example of a difference between discrete random variables and continuous random variables is that in a discrete distribution P(x > 2) = P(x ā„ 3) while in a continuous distribution P(x > 2) is treated the same as P(x ā„ 2).

Answer:Ā TRUE

Diff: 2

Keywords:Ā discrete, continuous, random, variable

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

5) The probability distribution for a continuous random variable is represented by a probability density function that defines a curve.

Answer:Ā TRUE

Diff: 1

Keywords:Ā probability, density, continuous, variable

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

6) When graphed, the probability distribution for a discrete random variable looks like a histogram.

Answer:Ā TRUE

Diff: 2

Keywords:Ā discrete, variable, distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

7) For a continuous distribution the total area under the curve is equal to 100.

Answer:Ā FALSE

Diff: 1

Keywords:Ā normal, distribution, continuous, variable

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

8) A continuous random variable approaches normality as the level of skewness increases.

Answer:Ā FALSE

Diff: 2

Keywords:Ā continuous, variable, random

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

9) If the mean, median and mode are all equal for a continuous random variable, then the random variable is normally distributed.

Answer:Ā FALSE

Diff: 2

Keywords:Ā continuous, random, variable, normal

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 3

10) When a single die is rolled, each of the six sides are equally likely. This is an example of a uniform distribution.

Answer:Ā TRUE

Diff: 1

Keywords:Ā uniform, random, variable

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 1

11) All symmetric distributions can be assumed normally distributed.

Answer:Ā FALSE

Diff: 1

Keywords:Ā normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

12) The parameters of a normal distribution are the mean and the standard deviation.

Answer:Ā TRUE

Diff: 1

Keywords:Ā normal, distribution, mean, standard deviation

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

13) The actual weight of 2-pound sacks of salted peanuts is found to be normally distributed with a mean equal to 2.04 pounds and a standard deviation of 0.25 pounds. Given this information, the probability of a sack weighing more than 2.40 pounds is 0.4251.

Answer:Ā FALSE

Diff: 2

Keywords:Ā standardized normal, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

14) The standard normal distribution table provides probabilities for the area between the z-value and the population mean.

Answer:Ā TRUE

Diff: 2

Keywords:Ā z-value, standard normal distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

15) The standard normal distribution has a mean of 0 and a standard deviation of 1.0.

Answer:Ā TRUE

Diff: 1

Keywords:Ā standard normal

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

16) The time it takes a parent to assemble a children’s bicycle has been shown to be normally distributed with a mean equal to 295 minutes with a standard deviation equal to 45 minutes. Given this information, the probability that it will take a randomly selected parent more than 220 minutes is about 0.0475.

Answer:Ā FALSE

Diff: 2

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

17) The State Department of Forests has determined that annual tree growth in a particular forest area is normally distributed with a mean equal to 17 inches and a standard deviation equal to 6 inches. Based on this information, it is possible for a randomly selected tree not to have grown any during a year.

Answer:Ā TRUE

Diff: 2

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

18) The State Department of Forests has determined that annual tree growth in a particular forest area is normally distributed with a mean equal to 17 inches and a standard deviation equal to 6 inches. If 2 trees are randomly chosen, the probability that both trees will have grown more than 20 inches during the year is approximately .037.

Answer:Ā FALSE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

19) Watersports Rental at Flathead Lake rents jet skis and power boats for day use. Each piece of equipment has a clock that records the time that it was actually in use while rented. The company has observed over time that the distribution of time used is normally distributed with a mean of 3.6 hours and a standard deviation equal to 1.2 hours. Watersports management has decided to give a rebate to customers who use the equipment for less than 2.0 hours. Based on this information, the probability that a customer will get the rebate is 0.4082.

Answer:Ā FALSE

Diff: 2

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

20) Watersports Rental at Flathead Lake rents jet skis and power boats for day use. Each piece of equipment has a clock that records the time that it was actually in use while rented. The company has observed over time that the distribution of time used is normally distributed with a mean of 3.6 hours and a standard deviation equal to 1.2 hours. Watersports management has decided to give a rebate to customers who use the equipment for only a short amount of time. They wish to grant a rebate to no more than 10 percent of all customers. Based on the information provided, the amount of time that should be set as the cut-off between getting the rebate and not getting the rebate is approximately 2.06 hours.

Answer:Ā TRUE

Diff: 2

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

21) The Varden Packaging Company has a contract to fill 50 gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 49.92 gallons.

Answer:Ā FALSE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

22) The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. In order to meet this requirement, Varden should set the mean fill to approximately 50.225 gallons.

Answer:Ā TRUE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

23) The Varden Packaging Company has a contract to fill 50-gallon barrels with gasoline for use by the U.S. Army. The machine that Varden uses has an adjustable device that allows the average fill per barrel to be adjusted as desired. However, the actual distribution of fill volume from the machine is known to be normally distributed with a standard deviation equal to 0.5 gallons. The contract that Varden has with the military calls for no more than 2 percent of all barrels to contain less than 49.2 gallons of gasoline. Suppose Varden managers are unwilling to set the mean fill at any level higher than 50 gallons. Given that, in order to meet the requirements, they will need to increase the standard deviation of fill volume.

Answer:Ā FALSE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

24) A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. They want to classify the largest fillets as extra large and charge a higher price for them. If they want the largest 15 percent of the fillets to be classified as extra large, the minimum weight for an extra large fillet should be 1.91 pounds.

Answer:Ā TRUE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

25) A seafood shop sells salmon fillets where the weight of each fillet is normally distributed with a mean of 1.6 pounds and a standard deviation of 0.3 pounds. Based on this information we can conclude that 90 percent of the fillets weight more than 1.0 pound.

Answer:Ā FALSE

Diff: 2

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

26) The vehicle speeds on a city street have been determined to be normally distributed with a mean of 33.2 mph and a variance of 16. Based on this information, the probability that if three randomly selected vehicles are monitored and that two of the three will exceed the 35 mph speed limit is slightly greater than 0.18.

Answer:Ā FALSE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

27) Any normal distribution can be converted to a standard normal distribution.

Answer:Ā TRUE

Diff: 2

Keywords:Ā standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

28) For a normal distribution, the probability of a value being between a positive z-value and its population mean is the same as that of a value being between a negative z-value and its population mean.

Answer:Ā TRUE

Diff: 3

Keywords:Ā z-value, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

29) Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that it will take exactly 5 minutes is 0.20.

Answer:Ā FALSE

Diff: 2

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

30) One of the basic differences between a uniform probability distribution and a normal probability distribution is that the uniform is symmetrical but the normal is skewed depending on the value of the standard deviation.

Answer:Ā FALSE

Diff: 1

Keywords: Ā uniform, normal, distribution, skewed

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

31) Suppose the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that a customer is served in less than 3 minutes is 0.

Answer:Ā TRUE

Diff: 1

Keywords:Ā uniform distribution

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

32) If the time it takes for a customer to be served at a fast-food chain business is thought to be uniformly distributed between 3 and 8 minutes, then the probability that the time it takes for a randomly selected customer to be served will be less than 5 minutes is 0.40.

Answer:Ā TRUE

Diff: 2

Keywords:Ā uniform, probability, distribution

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

33) If a uniform distribution and normal distribution both have the same mean and the same range, the normal distribution will have a larger standard deviation than the uniform distribution

Answer:Ā FALSE

Diff: 2

Keywords:Ā uniform distribution, normal distribution

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

34) It has been determined the weight of bricks made by the Dillenger Stone Company is uniformly distributed between 1 and 1.5 pounds. Based on this information, the probability that two randomly selected bricks will each weigh more than 1.3 pounds is 0.16.

Answer:Ā TRUE

Diff: 3

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

35) The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. Based on this, the probability that a part selected will not be dry is approximately 0.33.

Answer:Ā TRUE

Diff: 2

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

36) The amount of drying time for the paint applied to a plastic component part is thought to be uniformly distributed between 30 and 60 minutes. Currently, the automated process selects the part from the drying bin after the part has been there for 50 minutes. The probability that none of three parts picked are still wet when they are selected is approximately 0.04.

Answer:Ā FALSE

Diff: 3

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

37) An assembly process takes between 20 and 40 minutes to complete with the distribution of time thought to be uniformly distributed. Based on this, the percentage of assemblies that require less than 25 minutes is 0.05.

Answer:Ā FALSE

Diff: 2

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

38) Service time for customers at a drive-through coffee shop has been shown to be uniformly distributed between 2 and 10 minutes. Customers will complain when service time exceeds 7.5 minutes. Based on this information, the probability of getting a complaint based on service time is 0.3125.

Answer:Ā TRUE

Diff: 2

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

39) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail in the first 20 hours is approximately 0.095.

Answer:Ā TRUE

Diff: 3

Keywords:Ā exponential distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

40) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will fail between 20 and 100 hours is approximately 0.30.

Answer:Ā TRUE

Diff: 3

Keywords:Ā exponential distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

41) An electronics repair shop has determined that the time between failures for a particular electronic component part is exponentially distributed with a mean time between failures of 200 hours. Based on this information, the probability that a part will not fail in the first 200 hours is 0.50.

Answer:Ā FALSE

Diff: 2

Keywords:Ā exponential distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

42) A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. Based on this information, the mean number of cars arriving per minute is about 0.83.

Answer:Ā TRUE

Diff: 2

Keywords:Ā exponential distribution, lambda

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

43) A study of cars arriving at a parking structure at the local airport shows that the time between arrivals is 1.2 minutes and is exponentially distributed. The probability that more than 2 minutes will elapse between the arrivals of cars is about 0.81.

Answer:Ā FALSE

Diff: 3

Keywords:Ā exponential distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

44) Which of the following probability distributions can be used to describe the distribution for a continuous random variable?

- A) Normal distribution
- B) Binomial distribution
- C) Poisson distribution
- D) Hypergeometric

Answer:Ā A

Diff: 1

Keywords:Ā normal, distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

45) Which of the following is not a characteristic of the normal distribution?

- A) Symmetric
- B) Mean = median = mode
- C) Bell-shaped
- D) Equal probabilities at all values of x

Answer:Ā D

Diff: 2

Keywords:Ā normal, uniform, distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

46) Which of the following probability distributions could be used to describe the distribution for a continuous random variable?

- A) Exponential distribution
- B) Normal distribution
- C) Uniform distribution
- D) All of the above

Answer:Ā D

Diff: 1

Keywords:Ā normal, exponential, uniform, continuous, distribution

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

47) Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more?

- A) 0.3869
- B) 0.1131
- C) 0.7100
- D) 0.8869

Answer:Ā B

Diff: 2

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

48) Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $0.35 and a standard deviation of $0.33. Based on this information, what is the probability that a randomly selected stock will be lower by $0.40 or more?

- A) 2.27
- B) 0.4884
- C) 0.0116
- D) 0.9884

Answer:Ā C

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

49) Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. Based on this information, what is the probability of a student reading at more than 1400 words per minute after finishing the course?

- A) 0.0202
- B) 0.5207
- C) 0.4798
- D) 0.9798

Answer:Ā A

Diff: 2

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

50) Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. If two students were selected at random, what is the probability that they would both read at less than 400 words per minute?

- A) 0.4938
- B) 0.0062
- C) 0.00004
- D) 0.2438

Answer:Ā C

Diff: 3

Keywords:Ā normal, probability, z-value, multiplication

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

51) The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a variance equal to 1,456. Based on this information, what are the chances that the revenue on the first show will exceed $800?

- A) 0.1255
- B) Essentially zero
- C) 0.3745
- D) 0.9999

Answer:Ā B

Diff: 2

Keywords:Ā normal, z-value, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

52) The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a standard deviation equal to $80. Based on this information, what are the chances that the revenue on the first show will be between $300 and $500?

- A) About 0.3062
- B) Approximately 0.6534
- C) 0.1736
- D) Approximately 0.4798

Answer:Ā B

Diff: 2

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

53) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. Based on this, what is the probability that a call will last longer than 13 minutes?

- A) About 0.0125
- B) Approximately 0.4875
- C) About 0.5125
- D) About 0.9875

Answer:Ā A

Diff: 2

Keywords:Ā normal, z-score, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

54) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. The manager has decided to have a signal system attached to the phone so that after a certain period of time, a sound will occur on her employees’ phone if she exceeds the time limit. The manager wants to set the time limit at a level such that it will sound on only 8 percent of all calls. The time limit should be:

- A) 10.35 minutes.
- B) approximately 5.19 minutes.
- C) about 14.58 minutes.
- D) about 11.23 minutes.

Answer:Ā D

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

55) The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. What is the probability that three randomly monitored calls will each be completed in 4 minutes or less?

- A) 0.4756
- B) Approximately 0.1076
- C) About 0.00001
- D) Can’t be determined without more information.

Answer:Ā C

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

56) The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, with a mean weight of 2 pounds, what must the standard deviation be? Assume that the box weights are normally distributed.

- A) Approximately 0.05 pounds
- B) -0.133 pounds
- C) 1.144 pounds
- D) None of the above

Answer:Ā A

Diff: 3

Keywords:Ā normal, probability, z-value, standard deviation

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

57) The makers of Sweet-Things candy sell their candy by the box. Based on company policy, the mean target weight of all boxes is 2.0 pounds. To make sure that they are not putting too much in the boxes, the manager wants no more than 3 percent of all boxes to contain more than 2.10 pounds of candy. In order to do this, what should the mean fill weight be set to if the fill standard deviation is 0.13 pounds? Assume that the box weights are normally distributed.

- A) Just over 2 pounds
- B) Approximately 2.33 pounds
- C) Nearly 1.27 pounds
- D) Approximately 1.86 pounds

Answer:Ā D

Diff: 3

Keywords:Ā normal, mean, z-value, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

58) A major U.S. automaker has determined that the city mileage for one of its new SUV models is normally distributed with a mean equal to 15.2 mpg. A report issued by the company indicated that 22 percent of the SUV model vehicles will get more than 17 mpg in the city. Given this information, what is the city mileage standard deviation for this SUV model?

- A) 0.77 mpg
- B) Approximately 2.34 mpg
- C) 1.8 mpg
- D) Approximately 3.1 mpg

Answer:Ā B

Diff: 3

Keywords:Ā normal, z-value, standard deviation

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

59) A recent study showed that the length of time that juries deliberate on a verdict for civil trials is normally distributed with a mean equal to 12.56 hours with a standard deviation of 6.7 hours. Given this information, what is the probability that a deliberation will last between 10 and 15 hours?

- A) Approximately 0.29
- B) Nearly 0.75
- C) About 0.48
- D) About 0.68

Answer:Ā A

Diff: 2

Keywords:Ā normal, z-value, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

60) Suppose that it is believed that investor returns on equity investments at a particular brokerage house are normally distributed with a mean of 9 percent and a standard deviation equal to 3.2 percent. What percent of investors at this brokerage house earned at least 5 percent?

- A) 89.44 percent
- B) 10.56 percent
- C) 39.44 percent
- D) 100 percent

Answer:Ā A

Diff: 2

Keywords:Ā normal, z-value, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

61) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. As a promotion, the company plans to hold a drawing to give away one free vacation to Hawaii for a customer who uses between 400 and 402 minutes during a particular month. Based on the information provided, what proportion of the company’s customers would be eligible for the drawing?

- A) Approximately 0.1026
- B) About 0.004
- C) Approximately 0.2013
- D) About 0.02

Answer:Ā B

Diff: 2

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

62) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of changing its fee structure so that anyone who uses the phone less than 250 minutes during a given month will pay a reduced monthly fee. Based on the available information, what percentage of current customers would be eligible for the reduced fee?

- A) About 36.4 percent
- B) Approximately 52 percent
- C) About 86.6 percent
- D) About 13.6 percent

Answer:Ā D

Diff: 2

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

63) A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of charging a lower rate for customers who use the phone less than a specified amount. If it wishes to give the rate reduction to no more than 12 percent of its customers, what should the cut-off be?

- A) About 237 minutes
- B) About 654 minutes
- C) About 390 minutes
- D) About 325 minutes

Answer:Ā A

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

64) In a standard normal distribution, the probability that *z *is greater than 0 is:

- A) 0.5
- B) equal to 1
- C) at least 0.5
- D) 1.96

Answer:Ā A

Diff: 1

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome: Ā 2

65) In a standard normal distribution, the probability P(-1.00< z < 1.20) is the same as:

- A) P(1< z < 1.20) – P(0 < z < 1.00).
- B) P(1< z < 1.20) – 2*P(0 < z < 1.00).
- C) 2 ā P(1< z < 1.20) – P(0 < z <1.00).
- D) P(1 < z < 1.20) + 2 ā P(0 < z <1.00).

Answer:Ā D

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

66) A professor noted that the grades of his students were normally distributed with a mean of 75.07 and a standard deviation of 11.65. If only 10 percent of the students received grades of A, what is the minimum score needed to receive an A?

- A) 80.00
- B) 85.00
- C) 90.00
- D) 95.00

Answer:Ā C

Diff: 3

Keywords:Ā normal, probability, z-value

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

67) A store sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:

- A) uniform distribution.
- B) Poisson distribution.
- C) continuous distribution.
- D) relative frequency distribution.

Answer:Ā A

Diff: 1

Keywords:Ā uniform distribution

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 1

68) Which of the following probability distributions would most likely be used to describe the time between failures for electronic components?

- A) Binomial distribution
- B) Exponential distribution
- C) Uniform distribution
- D) Normal distribution

Answer:Ā B

Diff: 1

Keywords:Ā exponential distribution, continuous

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

69) It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes. Based on this information, what is the probability that a customer will spend more than 9 minutes in the record store?

- A) 0.33
- B) 0.1111
- C) 0.67
- D) 0.25

Answer:Ā A

Diff: 2

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

70) It is assumed that the time customers spend in a record store is uniformly distributed between 3 and 12 minutes. Based on this information, what is the probability that a customer will be exactly 7.50 minutes in the record store?

- A) 0.1250
- B) 0.05
- C) Essentially zero
- D) 0.111

Answer:Ā C

Diff: 1

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

71) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. Based on this information, what is the probability that an employee will have less than 20 minutes of unused sick time?

- A) 0.002
- B) 0.966
- C) 0.063
- D) 0.042

Answer:Ā D

Diff: 2

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

72) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. Based on this information, what is the probability that three randomly chosen employees have over 400 unused sick minutes at the end of the year?

- A) 0.1667
- B) 0.0046
- C) 0.5001
- D) 0.0300

Answer:Ā B

Diff: 3

Keywords:Ā uniform, probability, multiplication

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

73) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. The company has decided to give a cash payment to any employee that returns over 400 minutes of sick leave at the end of the year. What percentage of employees could expect a cash payment?

- A) 16.67 percent
- B) 0.1667 percent
- C) Just over 43 percent
- D) 80 percent

Answer:Ā A

Diff: 2

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

74) Employees at a large computer company earn sick leave in one-minute increments depending on how many hours per month they work. They can then use the sick leave time any time throughout the year. Any unused time goes into a sick bank account that they or other employees can use in the case of emergencies. The human resources department has determined that the amount of unused sick time for individual employees is uniformly distributed between 0 and 480 minutes. The company has decided to give a cash payment to any employee that returns over a specified amount of sick leave minutes. Assuming that the company wishes no more than 5 percent of all employees to get a cash payment, what should the required number of minutes be?

- A) 24 minutes
- B) 419 minutes
- C) 456 minutes
- D) 470 minutes

Answer:Ā C

Diff: 2

Keywords:Ā uniform, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

75) It is thought that the time between customer arrivals at a fast food business is exponentially distributed with Ī» equal to 5 customers per hour. Given this information, what is the mean time between arrivals?

- A) 12 minutes
- B) 5 minutes
- C) 5 hours
- D) 2 minutes

Answer:Ā A

Diff: 1

Keywords:Ā exponential, expected value

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

76) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. Based on this information, what is the probability that a randomly selected part will fail in less than 10 hours?

- A) About 0.82
- B) About 0.20
- C) About 0.33
- D) About 0.18

Answer:Ā D

Diff: 2

Keywords:Ā exponential, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

77) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. What is the probability that a component will be functioning after 60 hours?

- A) Approximately 0.30
- B) About 0.70
- C) About 0.21
- D) About 0.49

Answer:Ā A

Diff: 2

Keywords:Ā exponential, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

78) It is assumed that the time failures for an electronic component are exponentially distributed with a mean of 50 hours between consecutive failures. If one extra component is installed as a backup, what is the probability of at least one of the two components working for at least 60 hours?

- A) About 0.51
- B) About 0.09
- C) About 0.06
- D) About 0.70

Answer:Ā A

Diff: 3

Keywords:Ā exponential, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

79) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state’s turnpike is exponentially distributed with Ī» = 4 cars per minute. Based on this, the average time between arrivals is:

- A) 15 seconds.
- B) 12 seconds.
- C) 25 seconds.
- D) 4 minutes.

Answer:Ā A

Diff: 2

Keywords:Ā exponential, lambda

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

80) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state’s turnpike is exponentially distributed with Ī» = 4 cars per minute. Based on this information, the standard deviation for the time between arrivals is:

- A) 25 seconds.
- B) 3.87 seconds.
- C) 15 seconds.
- D) 2 minutes.

Answer:Ā C

Diff: 1

Keywords:Ā exponential, lambda, mean, standard deviation

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

81) The transportation manager for the State of New Jersey has determined that the time between arrivals at a toll booth on the state’s turnpike is exponentially distributed with Ī» = 4 cars per minute. Based on this information, what is the probability that the time between any two cars arriving will exceed 11 seconds?

- A) Approximately 1.0
- B) Approximately 0.48
- C) About 0.52
- D) About 0.75

Answer:Ā B

Diff: 2

Keywords:Ā exponential, lambda, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

82) The time between calls to an emergency 911-call center is exponentially distributed with a mean time between calls of 645 seconds. Based on this information, what is the probability that the time between the next two calls is between 200 and 400 seconds?

- A) Approximately 0.47
- B) About 0.199
- C) About 0.747
- D) About 0.801

Answer:Ā B

Diff: 3

Keywords:Ā exponential, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

83) For a standardized normal distribution, calculate P(z < 1.5).

- A) 0.9332
- B) 0.0668
- C) 0.333
- D) 0.667

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

84) For a standardized normal distribution, calculate P(z ā„ 0.85).

- A) 0.8033
- B) 0.1977
- C) 0.2340
- D) 0.7660

Answer:Ā B

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

85) For a standardized normal distribution, calculate P(-1.28 < z < 1.75).

- A) 0.3997
- B) 0.4599
- C) 0.1404
- D) 0.8596

Answer:Ā D

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

86) For a standardized normal distribution, calculate P(0.00 < z < 2.33).

- A) 0.7181
- B) 0.5099
- C) 0.4901
- D) 0.2819

Answer:Ā C

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

87) For a standardized normal distribution, calculate P(-1.00 < z < 1.00).

- A) 0.6826
- B) 0.6667
- C) 0.4572
- D) 0.5521

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

88) For a standardized normal distribution, calculate P(1.78 < z < 2.34).

- A) 0.0124
- B) 0.0341
- C) 0.0412
- D) 0.0279

Answer:Ā D

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

89) For a standardized normal distribution, determine a value, say z0, so that P(0 < z < z0) = 0.4772.

- A) 2.00
- B) 2.33
- C) 1.85
- D) 1.66

Answer: Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

90) For a standardized normal distribution, determine a value, say z0, so that P(-z0Ā ā¤ z < 0) = 0.45.

- A) 1.84
- B) 1.645
- C) 1.96
- D) 1.33

Answer:Ā B

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

91) For a standardized normal distribution, determine a value, say z0, so that P(-z0Ā ā¤Ā z ā¤Ā z0) = 0.95.

- A) 2.14
- B) 1.65
- C) 1.96
- D) 1.24

Answer:Ā C

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

92) For a standardized normal distribution, determine a value, say z0, so that P(z > z0) = 0.025.

- A) 1.96
- B) 1.65
- C) 1.24
- D) 2.14

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

93) For a standardized normal distribution, determine a value, say z0, so that P(z ā¤ z0) = 0.01.

- A) -2.33
- B) -1.96
- C) 2.33
- D) 1.96

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

94) Consider a random variable, z, that has a standardized normal distribution. Determine P(0 < z < 1.96).

- A) 0.1250
- B) 0.5250
- C) 0.3250
- D) 0.4750

Answer:Ā D

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

95) Consider a random variable, z, that has a standardized normal distribution. Determine P(z > 1.645).

- A) 0.05
- B) 0.01
- C) 0.03
- D) 0.45

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

96) Consider a random variable, z, that has a standardized normal distribution. Determine P (1.28 < z < 2.33).

- A) 0.0126
- B) 0.3997
- C) 0.0904
- D) 0.4901

Answer:Ā C

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

97) Consider a random variable, z, that has a standardized normal distribution. Determine (-2 ā¤ z ā¤ 3).

- A) 0.12414
- B) 0.97587
- C) 0.47722
- D) 0.49865

Answer:Ā B

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

98) Consider a random variable, z, that has a standardized normal distribution. Determine P(z > -1).

- A) 0.8413
- B) 0.1251
- C) 0.1512
- D) 0.2124

Answer:Ā A

Diff: 1

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

99) A random variable, *x*, has a normal distribution with *Ī¼* = 13.6 and *Ļ* = 2.90. Determine a value, *x*0, so that *P*(*x* > *x*0) = 0.05.

- A) 14.46
- B) 15.33
- C) 18.37
- D) 12.45

Answer:Ā C

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

100) A random variable, *x*, has a normal distribution with *Ī¼* = 13.6 and *Ļ* = 2.90. Determine a value, *x*0, so that *P*(*x* ā¤ *x*0) = 0.975.

- A) 16.678
- B) 19.284
- C) 23.360
- D) 14.475

Answer:Ā B

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

101) A random variable, *x*, has a normal distribution with *Ī¼* = 13.6 and *Ļ* = 2.90. Determine a value, *x*0, so that *P*(*Ī¼* – *x*0Ā ā¤ *x* ā¤ *Ī¼* + *x*0) = 0.95.

- A) 7.916
- B) 4.535
- C) 3.178
- D) 9.425

Answer:Ā A

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

102) For the normal distribution with parameters *Ī¼* = 5, *Ļ*= 2; calculate *P*(0 < *x* < 8).

- A) 0.8023
- B) 0.4152
- C) 0.9270
- D) 0.8845

Answer:Ā C

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

103) For the normal distribution with parameters *Ī¼* = 5, *Ļ* = 4; calculate P(0 < *x* < 8).

- A) 0.8841
- B) 0.8812
- C) 0.4215
- D) 0.6678

Answer:Ā D

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

104) For the normal distribution with parameters *Ī¼* = 3, *Ļ* = 2; calculate *P*(0 < *x* < 8).

- A) 0.3124
- B) 0.9270
- C) 0.8123
- D) 0.6723

Answer:Ā B

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

105) For the normal distribution with parameters *Ī¼* = 4, *Ļ* = 3; calculate *P*(*x* > 1).

- A) 0.8413
- B) 0.4562
- C) 0.7812
- D) 0.4152

Answer:Ā A

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

106) For the normal distribution with parameters *Ī¼* = 0, *Ļ* = 3; calculate *P*(*x* > 1).

- A) 0.5812
- B) 0.1214
- C) 0.3707
- D) 0.4412

Answer:Ā C

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

107) A randomly selected value from a normal distribution is found to be 2.1 standard deviations above its mean. What is the probability that a randomly selected value from the distribution will be greater than 2.1 standard deviations above the mean?

- A) 0.0179
- B) 0.0512
- C) 0.0231
- D) 0.0024

Answer:Ā A

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

108) A randomly selected value from a normal distribution is found to be 2.1 standard deviations above its mean. What is the probability that a randomly selected value from the distribution will be less than 2.1 standard deviations from the mean?

- A) 0.9488
- B) 0.9821
- C) 0.9976
- D) 0.9712

Answer:Ā B

Diff: 2

Keywords: Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

109) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, what value will be exceeded 10% of the time?

- A) 31.40
- B) 28.60
- C) 66.23
- D) 14.56

Answer:Ā A

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

110) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, what value will be exceeded 85% of the time?

- A) 16.2
- B) 17.9
- C) 19.8
- D) 14.2

Answer:Ā C

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

111) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, determine two values of which the smallest has 25% of the values below it and the largest has 25% of the values above it.

- A) 18.85 and 27.94
- B) 19.31 and 21.12
- C) 16.23 and 18.82
- D) 21.65 and 28.35

Answer:Ā D

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

112) A random variable is normally distributed with a mean of 25 and a standard deviation of 5. If an observation is randomly selected from the distribution, what value will 15% of the observations be below?

- A) 19.8
- B) 16.2
- C) 18.7
- D) 17.2

Answer:Ā A

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

113) If a continuous random variable is said to be exponentially distributed, what would be the easiest way to reduce the standard deviation?

Answer:Ā The standard deviation of an exponential distribution is equal to the mean (1/Ī») so the easiest way to reduce the standard deviation is to reduce the mean. This would be done by increasing Ī».

Diff: 1

Keywords:Ā exponential distribution, standard deviation, mean

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

114) What is the difference between a normal distribution and the standard normal distribution?

Answer:Ā A normal distribution is a bell-shaped distribution defined by two parameters, Ī¼ and Ļ. The values for the mean and standard deviation reflect the population data and may be any values. Thus, a normal distribution may be centered at any value and may have any spread, but it will have the common bell shape. The standard normal distribution is a specific normal distribution with mean = 0 and standard deviation equal to 1.0. The horizontal axis of the standard normal distribution represents z-values. Any normal distribution can be converted to the standard normal by converting the random variable values to z-values.

Diff: 2

Keywords:Ā normal distribution, standard normal distribution, mean, standard deviation

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 1

115) The weight of sacks of potatoes is normally distributed with a mean of 20 pounds and a standard deviation of 2 pounds. The weight of sacks of onions is also normally distributed with a mean of 20 pounds and a standard deviation of 0.50 pounds. Based on this information, which product will yield the highest probability of getting a very heavy sack?

Answer:Ā Since both products have the same mean and are both normally distributed, the one with the largest standard deviation will provide the higher probability of a heavy sack. Since potatoes have a standard deviation of 2 pounds compared to 0.50 pounds for onions, you would be more apt to see a very heavy sack of potatoes than onions.

Diff: 1

Keywords:Ā normal distribution, mean, standard deviation, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

116) A class takes an exam where the average time to complete the exam is normally distributed with a time of 40 minutes and standard deviation of 9 minutes. If the class lasts 1 hour, what percent of the students will have turned in the exam after 60 minutes?

Answer:Ā We are looking for the area under the curve to the left of 60 minutes, which is above the mean so we need the area between 40 and 60 and then 0.5 will be added.

*z *= Ā = Ā = 2.22.Ā Looking up 2.22 in the standard normal table we find 0.4868, which is the area between 40 and 60. So P(time ā¤ 60) = .5 + .4868 = .9868, which means 98.7 percent of class finishes the test before time is up.

Diff: 2

Keywords:Ā normal distribution, standard normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

117) The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard deviation of $3.25. Based on this information, what is the probability that a specific fare will exceed $15.00?

Answer:Ā Since this is a normal distribution problem, the first step is to convert the normal distribution to a standard normal distribution. We do this by converting the $15.00 value to a z-value using:

*z *= Ā = Ā = 0.7692 ā 0.77. Then we go to the standard normal table and locate a z-value = 0.77. The probability corresponding to z = 0.77 is 0.2794. The table in the text always gives the probability between the z-value and the mean. Since we want *P*(*x* > $15.00), we need to subtract from 0.50, giving 0.5000 – 0.2794 = 0.2206.

Diff: 2

Keywords:Ā normal distribution, standard normal, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

118) The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard deviation of $3.25. Suppose a driver has four consecutive fares that are less than $6.00. What is the probability of this happening?

Answer:Ā Since this is a normal distribution problem, the first step is to convert the normal distribution to a standard normal distribution. We do this by converting the $6.00 value to a z-value using:

*z *= Ā = Ā = -2.00.Ā Then we go to the standard normal table and locate a z-value = -2.00. The probability corresponding to z = -2.00 is 0.4772. The table in the text always gives the probability between the z-value and the mean. Since we want P(x ā¤ $6.00), we need to subtract from 0.50, giving 0.5000 – 0.4772 = 0.0228. This is the probability of one fare being less than $6.00. To get the probability of four consecutive fares being less than $6.00, we can use the multiplication rule for independent events discussed in Chapter 4.

This gives: 0.0228 Ć 0.0228 Ć 0.0228 Ć 0.0228 = 0.0000003. Since this is such a low probability, we would not expect such an event to occur. If it did, then it is likely that the fare distribution has changed.

Diff: 2

Keywords:Ā normal distribution, probability

Section:Ā 6-1 The Normal Probability Distribution

Outcome:Ā 2

119) The money spent by people at an amusement park, after paying to get in the gate, is thought to be uniformly distributed between $5.00 and $25.00. Based on this, what is the probability that someone will spend between $8.00 and $12.00?

Answer:Ā The continuous uniform probability distribution has a function: f(x) = Ā where b is the upper extreme of the distribution ($25.00) and a is the lower extreme ($5.00).

Then f(x) = Ā = Ā = 0.05.

Now, p(8.00 ā¤ x ā¤ x ā¤ 12.00) = f(x)(12.00 – 8.00) = 0.05(4.00) = 0.20. Thus, there is a 0.20 probability that someone will spend between $8.00 and $12.00 after getting into the amusement park.

Diff: 2

Keywords:Ā uniform distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

120) In comparing a uniform distribution with a normal distribution where both distributions have the same mean and the same range, explain which distribution will have the larger standard deviation.

Answer:Ā A picture of this situation would show a larger area of the uniform distribution farther away from the mean. In the normal distribution, the majority of the distribution is close to the mean and a much smaller proportion is near the edges. As the empirical rule says, about 68 percent should be within one standard deviation of the mean. In the uniform distribution this percentage will be lower (about 58 percent). Further the standard deviation of the normal distribution can be approximated using the 6-sigma rule of thumb for the range, meaning that the standard deviation is approximately 1/6 of the range. In the uniform distribution the standard deviation is (range)/3.46, where 3.46 is the square root of 12. All of which gives a larger standard deviation to the uniform distribution.

Diff: 3

Keywords:Ā uniform distribution, normal distribution

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 4

121) At the West-Side Drive-Inn, customers arrive at the rate of 10 every 30 minutes. The time between arrivals is exponentially distributed. Given this, what is the mean time between arrivals?

Answer:Ā The parameter for the exponential distribution is lambda, Ī». This was given as 10 per 30 minutes. Then the mean time between arrivals is Ā = = 0.10. The value 0.10 represents the fraction of the 30 minutes that occurs between arrivals. Thus, the mean time between arrivals is 0.10(30 minutes) = 3 minutes. On average, customers arrive every 3 minutes.

Diff: 2

Keywords:Ā exponential distribution, mean

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

122) At the West-Side Drive-Inn, customers arrive at the rate of 10 every 30 minutes. The time between arrivals is exponentially distributed. Based on this information, what is the probability that the time between two customers arriving will exceed 6 minutes?

Answer:Ā To solve this problem, we are looking for P(x > 6). It will be helpful to use the complement approach. Therefore, we wish to find 1 – P(x ā¤ 6). Since we are dealing with an exponential distribution with arrival rate equal to 10 per 30 minutes, Ī» = 10 per 30 minutes = 0.33 per minute. Then we use P(x ā¤ a) = 1 – *e*-Ī»a. To do this we can use either the table of exponential values in the back of the text or software such as Excel. We are looking for:

1 – (1 – *e*-33(6)) = 1 – (1 – *e*2) = 1 – (1 – 0.1353) = 0.1353

Thus, there is approximately a 0.1353 probability that the time between two customers arriving will exceed 6 minutes. Note that if Excel or other software is used, the answer might be slightly different due to rounding differences.

Diff: 2

Keywords:Ā exponential distribution, probability

Section:Ā 6-2 Other Continuous Probability Distributions

Outcome:Ā 5

*Business Statistics, 9e ***(Groebner/Shannon/Fry)**

**Chapter 7Ā Ā Introduction to Sampling Distributions**

**Ā **

1) The sample mean is a parameter.

Answer:Ā FALSE

Diff: 1

Keywords:Ā sample, parameter

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

2) The size of the sampling error that comes from a random sample depends on both the variation in the population and the size of the sample being selected.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling error, random, sample, size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

3) Taking a larger sample size will always result in less sampling error but costs more money and takes more time.

Answer:Ā FALSE

Diff: 1

Keywords:Ā sample, size, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

4) Recently the State Fish and Game planted several thousand tagged fish in a local river. The mean length of these fish, which constitute a population, is 12.6 inches. Yesterday, fishermen caught 100 of these tagged fish. You could expect that the mean length for these fish would be 12.6 inches as well since they come from the population.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sample, mean, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome: Ā 1

5) The actual mean fill volume for all bottles of a soft drink product that were filled on a Tuesday is 11.998 ounces. A sample of 64 bottles was randomly selected and the sample mean fill volume was 12.004 ounces. Based upon this information, the sampling error is .006 ounce.

Answer:Ā TRUE

Diff: 1

Keywords:Ā mean, sample, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

6) Sampling error is the difference between the sample statistic and the population parameter.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling error, population, sample, parameter

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

7) The reason that a population mean and the mean of a random sample selected from that population might be different is that the sample mean is found by dividing by n-1 while the population mean is found by dividing by n.

Answer:Ā FALSE

Diff: 1

Keywords:Ā population, sample, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

8) If the mean age for all students that attend your university is 24.78 years, it would be reasonable to expect that the mean of a sample of students selected from that population would also equal 24.78 years as long at the sampling is done using sound statistical methods.

Answer:Ā FALSE

Diff: 2

Keywords:Ā population, sample, mean, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

9) If a sample is selected using random sampling methods, the primary reason that the sample mean might be different from the corresponding population mean is that the sample might be biased.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sample, random, bias, population, mean

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

10) Suppose it is known that the mean purchase price for all homes sold last year in Blacksburg, Virginia was $203,455. Recently, two studies were done on home sales prices. In the first study, a random sample of 200 homes was selected from the population. In the second study, a random sample of 60 homes was selected. Based on this information, we know that the second study would contain more sampling errors than the first study due to the smaller sample size.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sample size, population, sample, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

11) A smaller sample might provide less sampling error than a larger sample from a given population.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sample, sampling error, population

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

12) A larger sample size reduces the potential for large sampling error.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling error, sample size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

13) A local bank has 1,400 checking account customers. Of these, 1,020 also have savings accounts. A sample of 400 checking account customers was selected from the bank of which 302 also had savings accounts. The sampling error in this situation is .0264.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sampling error, population, sample

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

14) If a population mean is equal to 200, the sample mean for a random sample selected from the population is about as likely to be higher or lower than 200.

Answer:Ā TRUE

Diff: 2

Keywords:Ā population, mean, sample

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

15) An increase in sample size will tend to result in less sampling error.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sample, size, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

16) A simple random sample is selected in a manner such that each possible sample of a given size has an equal chance of being selected.

Answer:Ā TRUE

Diff: 2

Keywords:Ā simple random sample

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

17) Suppose the mean balance of checking accounts at Regions Bank is known to be $4320. A random sample of 10 accounts yields a total of $41,490. This means the sampling error is -$171.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

18) Sampling error occurs when the population parameter and the sample statistic are different.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling error, population, parameter, sample, statistic

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

19) Sampling error can be eliminated if the sampling is done properly.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sampling error, sample

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

20) If it is desired that sampling error be reduced, one step that tends to work is to increase the sample size that is selected from the population.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling error, sample size, population

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

21) It is very unlikely that a nonstatistical sample will ever provide less sampling error than a statistical sample of the same size.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sampling error, statistical sample, sample size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

22) A major automobile manufacturer has developed a new model car that it claims will average 25 mpg on the highway. A random sample of fifty of these cars was tested and they averaged 24 mpg. This means that the claim made by the auto company is incorrect.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sample, population, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

23) Sampling error can be eliminated if the sample size is large enough.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sampling error, sample size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

24) A sampling distribution is the distribution of the individual values that are included in a sample from a population.

Answer:Ā FALSE

Diff: 1

Keywords:Ā sampling distribution, sample, population

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

25) A sampling distribution for a sample mean shows the distribution of the possible values for the sample mean for a given sample size from a population.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

26) Although the concept of sampling distributions is an important concept in statistics, it is very unlikely that a decision maker will actually construct a sampling distribution in any practical business situation.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

27) A sampling distribution for Ā is the distribution of all possible sample means that could be computed from the possible samples of a given sample size.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling distribution, sample mean, mean, sample, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

28) The mean of a sampling distribution would be equal to the mean of the population from which the sampling distribution is constructed.

Answer:Ā TRUE

Diff: 1

Keywords:Ā sampling distribution, mean, population

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

29) The population mean of income for adults in a particular community is known to be $28,600. Given this information, the sampling distribution of Ā values will be less than this depending on the size of the sample used in developing the sampling distribution.

Answer:Ā FALSE

Diff: 2

Keywords:Ā population, mean, sample size, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

30) If a population is normally distributed, then the sampling distribution for the sample mean will always be normally distributed regardless of the sample size.

Answer:Ā TRUE

Diff: 2

Keywords:Ā population, mean, sample, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

31) The population of soft drink cans filled by a particular machine is known to be normally distributed with a mean equal to 12 ounces and a standard deviation equal to .25 ounce. Given this information, the sampling distribution for a random sample of *n* = 25 cans will also be normally distributed with a mean equal to 12 ounces and a standard deviation equal to .05 ounce.

Answer:Ā TRUE

Diff: 2

Keywords:Ā population, mean, standard deviation, sample, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

32) The sampling distribution for Ā is actually the distribution of possible sampling error for samples of a given size selected at random from the population.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sampling distribution, mean, sampling error, sample, population

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

33) A sampling distribution for a sample of *n* = 4 is normally distributed with a standard deviation equal to 5. Based on this information, the population standard deviation, Ļ, is equal to 10 mph.

Answer:Ā TRUE

Diff: 3

Keywords:Ā population, mean, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

34) If a population standard deviation is 100, then the sampling distribution for Ā will have a standard deviation that is less than 100 for all sample sizes greater than 2.

Answer:Ā TRUE

Diff: 2

Keywords:Ā population, standard deviation, sampling distribution, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

35) The Central Limit Theorem is of most use to decision makers when the population is known to be normally distributed.

Answer:Ā FALSE

Diff: 1

Keywords:Ā Central Limit Theorem, normal, population

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

36) One of the things that the Central Limit Theorem tells us is that about half of the sample means will be greater than the population mean and about half will be less.

Answer:Ā TRUE

Diff: 2

Keywords:Ā Central Limit Theorem, normal, population, sample, mean

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

37) When a population is not normally distributed, the Central Limit Theorem states that a sufficiently large sample will result in the sample mean being normally distributed.

Answer:Ā TRUE

Diff: 2

Keywords:Ā Central Limit Theorem, population, sampling distribution, spread, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

38) If you are sampling from a very large population, a doubling of the sample size will reduce the standard error of the sampling distribution by one-fourth.

Answer:Ā FALSE

Diff: 3

Keywords:Ā sample, population, standard error, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

39) The population of incomes in a particular community is thought to be highly right-skewed with a mean equal to $36,789 and a standard deviation equal to $2,490. Based on this, if a sample of size *n* = 36 is selected, the sampling distribution would have a mean equal to the population mean, but the standard deviation of the sampling distribution will be one-sixth of the population standard deviation.

Answer:Ā TRUE

Diff: 2

Keywords:Ā Central Limit Theorem, population, sampling distribution standard deviation, mean

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

40) The population of incomes in a particular community is thought to be highly right-skewed with a mean equal to $36,789 and a standard deviation equal to $2,490. Based on this, if a sample of size *n* = 36 is selected, the highest sample mean that we would expect to see would be approximately $38,034.

Answer:Ā TRUE

Diff: 3

Keywords:Ā Central Limit Theorem, population, sampling distribution standard deviation, mean

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

41) If a population is not normally distributed, then the sampling distribution for the mean also cannot be normally distributed.

Answer:Ā FALSE

Diff: 2

Keywords:Ā Central Limit Theorem, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

42) The Dilmart Company has 8,000 parts in inventory. The mean dollar value of these parts is $10.79 with a standard deviation equal to $3.34. Suppose the inventory manager selected a random sample of *n* = 64 parts from the inventory and found a sample mean equal to $11.27. The probability of getting a sample mean at least as large as $11.27 is approximately 0.444.

Answer:Ā FALSE

Diff: 2

Keywords:Ā Central Limit Theorem, sample, population, mean

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

43) One of the nation’s biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. Then, the probability that a single turn time selected at random from this population would exceed 20 minutes is approximately 0.2327.

Answer:Ā TRUE

Diff: 2

Keywords:Ā z-value, probability, mean, standard deviation

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

44) One of the nation’s biggest regional airlines has tracked 4,000 landings and take-offs during the past month. Treating these data as the population of interest, the company found that the average time the planes spent on the ground (called the turn time) was 17.23 minutes with a standard deviation of 3.79 minutes. Further, they determined that the distribution of turn times is normally distributed. If a sample of size *n* = 16 turn times was selected at random from the population, the chances of the mean of this sample exceeding 20 minutes is 0.2327.

Answer:Ā FALSE

Diff: 2

Keywords:Ā z-value, probability, mean, standard deviation, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

45) The Fallbrook Distributing Company has a soft drink bottling plant in Plano, Texas. Based on historical records, if its filling machine is working properly, the mean fill volume per can is 12.0 ounces with a standard deviation equal to 0.13 ounce. Further, the distribution of fill amounts is known to be normally distributed. The State of Texas has a department whose job it is to check on such consumer-related processes as soft drink filling. The idea is to protect the consumer. The department arrives at the Fallbrook plant once a month on an unscheduled day. When they arrive, they randomly select *n* = 4 cans and carefully measure the volume in each can. If any of these cans contains less than 11.85 ounces, the plant is shut down until a full inspection of the filling process is performed. Based on this information, the probability that the plant will get shut down if it is operating properly is approximately 0.1251.

Answer:Ā TRUE

Diff: 3

Keywords:Ā z-value, probability, mean, standard deviation

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

46) Regardless of population distribution, the sampling distribution for a random variable X will be approximately normally distributed.

Answer:Ā FALSE

Diff: 3

Keywords:Ā Central Limit Theorem, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

47) A sample proportion can be assumed normally distributed if n ā„ 30.

Answer:Ā FALSE

Diff: 2

Keywords:Ā Central Limit Theorem, sampling distribution, mean, proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

48) Suppose it is known that 93 percent of all parts in an inventory of 18,900 parts are in workable order. If a sample of *n* = 100 parts were selected from the inventory, based on the concept of sampling distributions of proportions, it can be assumed that the sample proportion of workable parts will also be 0.93.

Answer:Ā FALSE

Diff: 1

Keywords:Ā sampling error, proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

49) The size of the standard error of the sample proportion is dependent on the value of the population proportion and the closer the population proportion is to .50, the larger the standard error for a given sample size will be.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sample size, standard error, proportion, population

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

50) Assume that *n* = 18 people are asked a yes/no survey question, and 6 people say “yes” while 12 people say “no.” Based on this information the sample proportion can be assumed normally distributed.

Answer:Ā TRUE

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

51) In order to assume that the sampling distribution for a proportion is approximately normal, the population proportion must be very close to 0.50.

Answer:Ā FALSE

Diff: 1

Keywords:Ā sampling distribution, proportion, normal

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

52) In a particular city, the proportion of cars that would fail an air quality emissions test is thought to be 0.13. Given this, the probability that a random sample of *n* = 200 cars will have a sample proportion between 0.11 and 0.15 is approximately 0.60.

Answer:Ā TRUE

Diff: 2

Keywords:Ā proportion, probability, sample, z-value

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

53) A random sample of 500 customers at a large retail store was selected. These customers were asked whether they had a positive experience the last time they shopped there. Only 50 customers said that they did not have a positive experience. Thus, the population parameter for proportion of customers who did have a positive experience is .90.

Answer:Ā FALSE

Diff: 2

Keywords:Ā sample, population, parameter, proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

54) In a campaign speech, a candidate for governor stated that about 63 percent of the people in the state were in favor of spending additional money on higher education. After the speech, a polling agency surveyed a random sample of 400 people and found 234 people who favored more spending on higher education. Based on the candidate’s statement, the probability of finding 234 or fewer is approximately 0.97.

Answer:Ā FALSE

Diff: 2

Keywords:Ā probability, z-value, sample

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

55) The makers of a particular type of candy have stated that 75 percent of their sacks of candy will contain 6 ounces or more of candy. A consumer group that studies such claims recently selected a random sample of 100 sacks of this candy. Of these, 70 sacks actually contained 6 ounces or more. The probability that 70 or fewer sacks would contain 6 ounces or less is approximately 0.1251.

Answer:Ā TRUE

Diff: 2

Keywords:Ā probability, z-value, sample

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

56) In analyzing the sampling distribution of a proportion, doubling the sample size will cut the standard deviation of the sampling distribution in half.

Answer:Ā FALSE

Diff: 3

Keywords:Ā sampling distribution, proportion, sample size, standard deviation

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

57) If the standard error for the sampling distribution of a proportion is equal to 0.0327 and if the population proportion, *p,* is equal to .80, the sample size must be 150.

Answer:Ā TRUE

Diff: 3

Keywords:Ā proportion, population, sample size

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

58) Regardless of the value of the population proportion, *p*, (with the obvious exceptions of *p* = 0 and *p *= 1) the sampling distribution for the sample proportion, Ā will be approximately normally distributed providing that the sample size is large enough.

Answer:Ā FALSE

Diff: 3

Keywords:Ā Central Limit Theorem, sample size

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

59) When sampling from a population, the sample mean will:

- A) typically exceed the population mean.
- B) likely be different from the population mean.
- C) always be closer to the population mean as the sample size increases.
- D) likely be equal to the population mean if proper sampling techniques are employed.

Answer:Ā B

Diff: 2

Keywords:Ā sample, population, mean, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

60) A measure computed from the entire population is called:

- A) a statistic.
- B) a mean.
- C) a parameter.
- D) a qualitative value.

Answer:Ā C

Diff: 1

Keywords:Ā measure, population, parameter

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

61) The following values represent the population of home mortgage interest rates (in percents) being charged by the banks in a particular city:

6.9 | 7.5 | 6.5 | 7 | 7.3 | 6.8 | 6.5 |

7 | 7 | 7.2 | 7.5 | 7.8 | 6 | 7 |

Given this information, what is the most extreme amount of sampling error possible if a random sample of *n* = 4 banks is surveyed and the mean loan rate is calculated?

- A) -0.55 percent
- B) 0.52 percent
- C) 1.08 percent
- D) Can’t be determined without more information.

Answer:Ā A

Diff: 3

Keywords:Ā sampling error, mean

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

62) The following values represent the population of home mortgage interest rates (in percents) being charged by the banks in a particular city:

6.9 | 7.5 | 6.5 | 7 | 7.3 | 6.8 | 6.5 |

7 | 7 | 7.2 | 7.5 | 7.8 | 6 | 7 |

Given this information, what would the sampling error be if a sample including the seven values in the top row were used to compute the sample mean?

- A) Approximately 6.93
- B) About 0.56
- C) Approximately -0.07
- D) About 0.07

Answer:Ā C

Diff: 2

Keywords:Ā sampling error, mean

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

63) The impact on sampling of increasing the sample size is:

- A) the potential for extreme sampling error is reduced.
- B) the amount of sampling error is always reduced.
- C) the sample mean will always be closer to the population mean.
- D) There is no specific relationship between sample size and sampling error.

Answer:Ā A

Diff: 2

Keywords:Ā sample size, sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

64) Suppose the mean of dogs a pet shop grooms each day is know to be 14.2 dogs. If a sample of *n* = 12 days is chosen and a total of 178 dogs are groomed during those 12 days, then the sampling error is:

- A) 163.8.
- B) about 0.63.
- C) about -0.63.
- D) -163.8.

Answer:Ā B

Diff: 2

Keywords:Ā sampling error, mean

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

65) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows:

2 | 4 | 7 | 3 | 4 |

2 | 4 | 5 | 2 | 3 |

5 | 4 | 6 | 3 | 4 |

2 | 2 | 1 | 4 | 3 |

If a random sample of *n* = 3 homes were selected, what would be the highest possible positive sampling error?

- A) 6.0
- B) 3.0
- C) 0.5
- D) 2.5

Answer:Ā D

Diff: 3

Keywords:Ā sampling error, positive, mean, sample

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

66) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows:

2 | 4 | 7 | 3 | 4 |

2 | 4 | 5 | 2 | 3 |

5 | 4 | 6 | 3 | 4 |

2 | 2 | 1 | 4 | 3 |

Which of the following statements is true when comparing a random sample of size three homes selected from the population to a random sample of size 6 homes selected from the population?

- A) The amount of sampling error that will exist between the sample mean and the population mean will be half for the larger sample.
- B) The most extreme negative sampling error between and Ī¼ is reduced by about 0.167 person.
- C) We can expect that the larger sample will produce more sampling error due to the potential to make coding errors.
- D) The sampling error that will result from the smaller sample will be less than what we would see from the larger sample.

Answer:Ā B

Diff: 3

Keywords:Ā sample, sampling error, sample size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

67) Which of the following statements is false?

- A) Increasing the sample size will always reduce the size of the sampling error when the sample mean is used to estimate the population mean.
- B) Increasing the sample size will reduce the potential for extreme sampling error.
- C) Sampling error can occur when differs from Ī¼ due to the fact that the sample was not a perfect reflection of the population.
- D) There is no way to prevent sampling error short of taking a census of the entire population.

Answer:Ā A

Diff: 2

Keywords:Ā sample size, sampling error, mean, sample, population

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

68) A particular subdivision has 20 homes. The number of people living in each of these homes is listed as follows:

2 | 4 | 7 | 3 | 4 |

2 | 4 | 5 | 2 | 3 |

5 | 4 | 6 | 3 | 4 |

2 | 2 | 1 | 4 | 3 |

If a sample of size *n* = 5 is selected, the largest possible sample mean is:

- A) 7
- B) 6.5
- C) 6
- D) 5.4

Answer:Ā D

Diff: 2

Keywords:Ā sample mean

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

69) If the monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be distributed as a normal distribution with mean equal to $87.00 a month and standard deviation of $36.00, which of the following would be the largest individual customer bill that you might expect to find?

- A) Approximately $811.00
- B) About $195.00
- C) Nearly $123.00
- D) There is no way to determine this without more information.

Answer:Ā B

Diff: 2

Keywords:Ā z-value, normal distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

70) The monthly electrical utility bills of all customers for the Far East Power and Light Company are known to be distributed as a normal distribution with mean equal to $87.00 a month and standard deviation of $36.00. If a statistical sample of *n* = 100 customers is selected at random, what is the probability that the mean bill for those sampled will exceed $75.00?

- A) -0.33
- B) Approximately 0.63
- C) About 1.00
- D) 3.33

Answer:Ā C

Diff: 2

Keywords:Ā Central Limit Theorem, z-value, mean, standard deviation, normal

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

71) The Olsen Agricultural Company has determined that the weight of hay bales is normally distributed with a mean equal to 80 pounds and a standard deviation equal to 8 pounds. Based on this, what is the mean of the sampling distribution for Ā if the sample size is* n* = 64?

- A) 80
- B) 10
- C) Between 72 and 88
- D) 8

Answer:Ā A

Diff: 1

Keywords:Ā mean, sampling distribution, normal

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

72) The Olsen Agricultural Company has determined that the weight of hay bales is normally distributed with a mean equal to 80 pounds and a standard deviation equal to 8 pounds. Based on this, what is the probability that the mean weight of the bales in a sample of *n* = 64 bales will be between 78 and 82 pounds?

- A) 0.4772
- B) 0.0228
- C) 0.6346
- D) 0.9544

Answer:Ā D

Diff: 2

Keywords:Ā Central Limit Theorem, z-value, mean, standard deviation, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

73) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. Based on this information, if the department employee selects a random sample of *n* = 9 containers, what is the probability that the mean volume for the sample will be greater than 1.01 gallons?

- A) 0.3821
- B) 0.1179
- C) 0.6179
- D) 0.2358

Answer:Ā A

Diff: 2

Keywords: Ā Central Limit Theorem, z-value, mean, standard deviation, sample, probability

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

74) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. The department’s test process requires that they select a random sample of *n* = 9 containers. If the sample mean is less than 0.97 gallons, the department will fine the dairy. Based on this information, what is the probability that the dairy will get fined even when its filling process is working correctly?

- A) 0.90
- B) Approximately 0.3159
- C) About 0.1841
- D) Approximately 0.3821

Answer:Ā C

Diff: 2

Keywords:Ā Central Limit Theorem, z-value, mean, standard deviation, sample, type I, alpha, probability

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

75) The State Department of Weights and Measures is responsible for making sure that commercial weighing and measuring devices, such as scales, are accurate so customers and businesses are not cheated. Periodically, employees of the department go to businesses and test their scales. For example, a dairy bottles milk in 1-gallon containers. Suppose that if the filling process is working correctly, the mean volume of all gallon containers is 1.00 gallon with a standard deviation equal to 0.10 gallon. The department’s test process requires that they select a random sample of *n* = 9 containers. If the sample mean is less than 0.97 gallon, the department will fine the dairy. Based on this information, suppose that the dairy wants no more than a 0.05 chance of being fined, which of the following options exist if they can’t alter the filling standard deviation?

- A) They can convince the state to decrease the sample size.
- B) They can change the mean fill level to approximately 1.025 gallons.
- C) They could lower the mean fill level to a level lower than 1 gallon.
- D) There is actually nothing that they can do if they can’t modify the standard deviation.

Answer:Ā B

Diff: 3

Keywords:Ā Central Limit Theorem, z-value, mean, standard deviation, sample, type I, alpha, probability

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

76) Suppose it is known that the income distribution in a particular region is right-skewed and bi-modal. If bank economists are interested in estimating the mean income, which of the following is true?

- A) Provided that the sample size is sufficiently large, the sampling distribution for will be approximately normal with a mean equal to the population mean that they wish to estimate.
- B) The sampling distribution will also be right-skewed for large sample sizes.
- C) The standard deviation of the sampling distribution for will be proportionally larger than the population standard deviation, depending on the size of the sample.
- D) The sampling distribution will be left-skewed.

Answer:Ā A

Diff: 2

Keywords:Ā Central Limit Theorem, z-value, mean, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

77) A company has determined that the mean number of days it takes to collect on its accounts receivable is 36 with a standard deviation of 11 days. The company plans to select a random sample of *n *= 12 accounts and compute the sample mean. Which of the following statements holds true in this situation?

- A) There is no way to determine what the mean of the sampling distribution is without knowing the specific shape of the population.
- B) The sampling distribution will have the same distribution as the population, provided that the population is not normally distributed.
- C) The sampling error will be larger than if they had sampled
*n*= 64 accounts. - D) The sampling distribution may actually be approximately normally distributed depending on what the population distribution is.

Answer:Ā D

Diff: 2

Keywords:Ā Central Limit Theorem, sampling distribution, normal, population

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

78) A golf course in California has determined that the mean time it takes for a foursome to complete an 18 hole round of golf is 4 hours 35 minutes (275 minutes) with a standard deviation of 14 minutes. The time distribution is also thought to be approximately normal. Every month, the head pro at the course randomly selects a sample of 8 foursomes and monitors the time it takes them to play. Suppose the mean time that was observed for the sample last month was 4 hours 44 minutes (284 minutes). What is the probability of seeing a sample mean this high or higher?

- A) Approximately 0.4649
- B) About 0.9649
- C) Approximately 0.0351
- D) About 0.9298

Answer:Ā C

Diff: 2

Keywords:Ā z-value, probability, mean, Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

79) Which of the following statements is not consistent with the Central Limit Theorem?

- A) The Central Limit Theorem applies without regard to the size of the sample.
- B) The Central Limit Theorem applies to non-normal distributions.
- C) The Central Limit Theorem indicates that the sampling distribution will be approximately normal when the sample size is sufficiently large.
- D) The Central Limit Theorem indicates that the mean of the sampling distribution will be equal to the population mean.

Answer:Ā A

Diff: 1

Keywords:Ā Central Limit Theorem, sample size

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

80) The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald’s. The “prime” line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for “prime” fries, they plan to select a random sample of *n* = 100 fries each day. The quality analysts will compute the mean length for the sample. They want to establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99. What should these limits be?

- A) Approximately Ā±0.13 inches
- B) Within the approximate range of 5.87 inches to 6.13 inches
- C) Within the range of about 4.71 inches to 7.29 inches
- D) Approximately Ā±1.29 inches

Answer:Ā B

Diff: 3

Keywords:Ā z-value, mean, standard deviation, Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome: Ā 3

81) The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald’s. The “prime” line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for “prime” fries, they plan to select a random sample of* n* = 100 fries each day. Yesterday, the sample mean was 6.05 inches. What is the probability that the mean would be 6.05 inches or more if they are meeting the quality standards?

- A) 0.2350
- B) 0.3413
- C) 0.9413
- D) 0.1587

Answer:Ā D

Diff: 2

Keywords:Ā z-value, mean, standard deviation, Central Limit Theorem, probability

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

82) The St. Joe Company grows pine trees and the average annual increase in tree diameter is 3.1 inches with a standard deviation of 0.5 inch. A random sample of *n* = 50 trees is collected. What is the probability of the sample mean being less the 2.9 inches?

- A) 0.4977
- B) 0.0023
- C) 0.9977
- D) 0.9954

Answer:Ā B

Diff: 2

Keywords:Ā probability, z-value, mean, standard deviation, Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

83) According to the local real estate board, the average number of days that homes stay on the market before selling is 78.4 with a standard deviation equal to 11 days. A prospective seller selected a random sample of 36 homes from the multiple listing service. Above what value for the sample mean should 95 percent of all possible sample means fall?

- A) About 79.3 days
- B) About 64 days
- C) Approximately 75.4 days
- D) Can’t be determined without knowing whether the population is normally distributed.

Answer:Ā C

Diff: 3

Keywords:Ā probability, z-value, Central Limit Theorem, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

84) A population, with an unknown distribution, has a mean of 80 and a standard deviation of 7. For a sample of 49, the probability that the sample mean will be larger than 82 is:

- A) 0.5228
- B) 0.9772
- C) 0.4772
- D) 0.0228

Answer:Ā D

Diff: 2

Keywords:Ā probability, z-value, Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

85) Which of the following statements is true with respect to the sampling distribution of a proportion?

- A) An increase in the sample size will result in a reduction in the size of the standard deviation.
- B) As long as the sample size is sufficiently large, the sampling distribution will be approximately normal.
- C) The mean of the sampling distribution will equal the population proportion.
- D) All of the above are true.

Answer:Ā D

Diff: 1

Keywords:Ā sampling distribution, proportion, sample size, normal, population

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

86) The Chamber of Commerce in a large Midwestern city has stated that 70 percent of all business owners in the city favor increasing the downtown parking fees. The city council has commissioned a random sample of *n* = 100 business owners. Of these, 63 said that they favor increasing the parking fees. What is the probability of 63 or fewer favoring the idea if the Chamber’s claim is correct?

- A) Approximately 0.0630
- B) About 0.4370
- C) Nearly 0.20
- D) About 0.9370

Answer:Ā A

Diff: 2

Keywords:Ā z-value, proportion, normal, probability

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

87) A claim was recently made on national television that two of every three doctors recommend a particular pain killer. Suppose a random sample of *n* = 300 doctors revealed that 180 said that they would recommend the painkiller. If the TV claim is correct, what is the probability of 180 or fewer in the sample agreeing?

- A) 0.4929
- B) 0.0049
- C) 0.9929
- D) 0.0142

Answer:Ā B

Diff: 2

Keywords:Ā z-value, proportion, normal, probability

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

88) A major textbook publisher has a contract with a printing company. Part of the contract stipulates that no more than 5 percent of the pages should have any type of printing error. Suppose that the company selects a random sample of 400 pages and finds 33 that have an error. If the printer is meeting the standard, what is the probability that a sample would have 33 or more errors?

- A) 0.1245
- B) 0.4986
- C) 0.0014
- D) 0.1250

Answer:Ā C

Diff: 2

Keywords:Ā z-value, proportion, normal, probability

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

89) A major textbook publisher has a contract with a printing company. Part of the contract stipulates that no more than 5 percent of the pages should have any type of printing error. As a quality control measure, the publisher periodically selects a random sample of *n* = 100 pages. Then, depending on the proportion of pages with errors, they either say nothing to the printer or they complain that the quality has slipped. Suppose the publisher wants no more than a .10 chance of mistakenly blaming the printer for poor quality, what should the cut-off proportion be?

- A) About 0.0279
- B) Approximately 0.0779
- C) About 0.0221
- D) About 0.10

Answer:Ā B

Diff: 3

Keywords:Ā z-value, proportion, normal, probability, type II, beta

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

90) One of the leading dot-com companies has found that the proportion of customers who come into its Web site that actually makes a purchase is 0.045. The company plans to see whether this rate still holds by selecting a random sample of 200 hits on its Web site. Given that the 0.045 rate still applies, what is the standard deviation of the sampling distribution?

- A) Approximately 0.0147
- B) About 0.0002
- C) About 0.0354
- D) Can’t be determined without knowing the mean.

Answer:Ā A

Diff: 2

Keywords:Ā z-value, proportion, normal, standard deviation

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

91) In a recent report, it was stated that the proportion of employees who carpool to their work is 0.14 and that the standard deviation of the sampling proportion is 0.0259. However, the report did not indicate what the sample size was. What was the sample size?

- A) 100
- B) 180
- C) 460
- D) Can’t be determined without more information

Answer:Ā B

Diff: 3

Keywords:Ā proportion, sample size, sampling distribution

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

92) According to an industry report, 26 percent of all households have at least one cell phone. Further, of those that do have a cell phone, the mean monthly bill is $55.90 with a standard deviation equal to $9.60. Recently, a random sample of 400 households was selected. Of these households, 88 indicated that they had cell phones. The mean bill for these 88 households was $57.00. What is the probability of getting 88 or fewer households with cell phones if the numbers provided by the industry report are correct?

- A) Approximately 0.0344
- B) Nearly 0.4656
- C) About 0.1345
- D) Can’t be determined without knowing the standard deviation

Answer:Ā A

Diff: 2

Keywords:Ā proportion, probability, sampling distribution

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

93) A pharmaceutical company claims that only 5 percent of patients experience nausea when they take a particular drug. In a research study, *n* = 100 patients were given this drug and 8 experienced nausea. Assuming that the company’s claim is true, what is the probability of 8 or more patients experiencing nausea?

- A) About 0.9162
- B) About 0.0300
- C) About 0.0838
- D) About 0.4162

Answer:Ā C

Diff: 2

Keywords:Ā probability, sampling distribution, proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

94) A major shipping company has stated that 96 percent of all parcels are delivered on time. To check this, a random sample of *n* = 200 parcels were sampled. Of these, 184 arrived on time. If the company’s claim is correct, what is the probability of 184 or fewer parcels arriving on time?

- A) About 0.0019
- B) Nearly 0.24
- C) Just over 0.98
- D) About 0.4981

Answer:Ā A

Diff: 2

Keywords:Ā probability, sampling distribution, proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

95) A sample of 25 observations is taken to estimate a population proportion *Ļ*. The sampling distribution of sample proportion p is:

- A) not normal since
*n*< 30. - B) approximately normal because is always normally distributed.
- C) approximately normal if
*np ā„*5 and*n*(1 –*p*) ā„ 5. - D) approximately normal if
*p*approaches 0.50.

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution, proportion, Central Limit Theorem

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

96) Hillman Management Services manages apartment complexes in Tulsa, Oklahoma. They currently have 30 units available for rent. The monthly rental prices (in dollars) for this population of 30 units are:

455 | 690 | 450 | 495 | 550 | 780 | 800 | 395 | 500 | 405 |

675 | 550 | 490 | 495 | 700 | 995 | 650 | 550 | 400 | 750 |

600 | 780 | 650 | 905 | 415 | 600 | 600 | 780 | 575 | 750 |

What is the range of possible sampling error if a random sample of size* n* = 6 is selected from the population?

- A) -194.33 to 225.67
- B) -245.23 to 271.86
- C) -184.15 to 215.61
- D) -172.52 to 234.04

Answer:Ā A

Diff: 1

Keywords:Ā sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

97) Hillman Management Services manages apartment complexes in Tulsa, Oklahoma. They currently have 30 units available for rent. The monthly rental prices (in dollars) for this population of 30 units are:

455 | 690 | 450 | 495 | 550 | 780 | 800 | 395 | 500 | 405 |

675 | 550 | 490 | 495 | 700 | 995 | 650 | 550 | 400 | 750 |

600 | 780 | 650 | 905 | 415 | 600 | 600 | 780 | 575 | 750 |

What is the range of possible sampling error if a random sample of size* n* = 10 is selected?

- A) -174.21 to 191.12
- B) -182.59 to 169.91
- C) -164.33 to 178.67
- D) -162.16 to 171.51

Answer:Ā C

Diff: 1

Keywords:Ā sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

98) Princess Cruises recently offered a 16-day voyage from Beijing to Bangkok during the time period from May to August. The announced price, excluding airfare, for a room with an ocean view or a balcony was listed as $3,475. Cruise fares usually are quite variable due to discounting by the cruise line and travel agents. A sample of 20 passengers who purchased this cruise paid the following amounts (in dollars):

3,559 | 3,005 | 3,389 | 3,505 | 3,605 | 3,545 | 3,529 | 3,709 | 3,229 | 3,419 |

3,439 | 3,375 | 3,349 | 3,559 | 3,419 | 3,569 | 3,559 | 3,575 | 3,449 | 3,119 |

Calculate the sample mean cruise fare.

- A) 3715.24
- B) 3445.30
- C) 4581.81
- D) 6314.24

Answer:Ā B

Diff: 1

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

99) Princess Cruises recently offered a 16-day voyage from Beijing to Bangkok during the time period from May to August. The announced price, excluding airfare, for a room with an ocean view or a balcony was listed as $3,475. Cruise fares usually are quite variable due to discounting by the cruise line and travel agents. A sample of 20 passengers who purchased this cruise paid the following amounts (in dollars):

3,559 | 3,005 | 3,389 | 3,505 | 3,605 | 3,545 | 3,529 | 3,709 | 3,229 | 3,419 |

3,439 | 3,375 | 3,349 | 3,559 | 3,419 | 3,569 | 3,559 | 3,575 | 3,449 | 3,119 |

Determine the sampling error for this sample.

- A) -$29.70
- B) -$51.12
- C) -$21.71
- D) -$31.74

Answer:Ā A

Diff: 1

Keywords:Ā sampling error

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

100) A population with a mean of 1,250 and a standard deviation of 400 is known to be highly skewed to the right. If a random sample of 64 items is selected from the population, what is the probability that the sample mean will be less than 1,325?

- A) 0.8981
- B) 0.8141
- C) 0.7141
- D) 0.9332

Answer:Ā D

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

101) Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230. If a random sample of size* n* = 8 is selected, calculate the probability that the sample mean will exceed 2,100.

- A) 0.2141
- B) 0.1871
- C) 0.0712
- D) 0.1093

Answer:Ā D

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

102) A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475.

- A) 0.3251
- B) 0.7124
- C) 0.0475
- D) 0.0712

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

103) A normally distributed population has a mean of 500 and a standard deviation of 60. Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515.

- A) 0.1056
- B) 0.1761
- C) 0.0712
- D) 0.0151

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

104) Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20. The nine randomly sampled values are:

125 | 95 | 66 | 116 | 99 |

91 | 102 | 51 | 110 |

Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values.

- A) 0.1411
- B) 0.1612
- C) 0.1512
- D) 0.2266

Answer:Ā D

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

105) A random sample of 100 items is selected from a population of size 350. What is the probability that the sample mean will exceed 200 if the population mean is 195 and the population standard deviation equals 20? (*Hint*: Use the finite correction factor since the sample size is more than 5% of the population size.)

- A) 0.0415
- B) 0.0016
- C) 0.0241
- D) 0.0171

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

106) SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of *n* = 50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the standards?

- A) 0.0612
- B) 0.0082
- C) 0.0015
- D) 0.0009

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

107) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours?

- A) 0.2412
- B) 0.3830
- C) 0.1712
- D) 0.5121

Answer:Ā B

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

108) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?

- A) 0.9444
- B) 0.5121
- C) 0.7124
- D) 0.1512

Answer:Ā A

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

109) Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours, what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours?

- A) 0.6127
- B) 0.8124
- C) 0.9736
- D) 0.8812

Answer:Ā C

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

110) The branch manager for United Savings and Loan in Seaside, Virginia, has worked with her employees in an effort to reduce the waiting time for customers at the bank. Recently, she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute. However, before making a statement at a managers’ meeting, this branch manager wanted to double-check that the process was working as thought. To make this check, she randomly sampled 25 customers and recorded the time they had to wait. She discovered that mean wait time for this sample of customers was 4.2 minutes. Based on the team’s claims about waiting time, what is the probability that a sample mean for *n* = 25 people would be as large or larger than 4.2 minutes?

- A) 0.0214
- B) 0.0512
- C) 0.0231
- D) 0.0011

Answer:Ā D

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

111) In an article entitled “Fuel Economy Calculations to Be Altered,” James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. Suppose 100 AAA members were to test the 2008 F-150. Determine the probability that the average mpg would be at least 15.7.

- A) 0.0026
- B) 0.0121
- C) 0.0451
- D) 0.0001

Answer:Ā A

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

112) In an article entitled “Fuel Economy Calculations to Be Altered,” James R. Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks. This is the first modification since 1985. It is expected to lower average mileage for city driving in conventional cars from 10% to 20%. AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg). The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg. Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg. The current method of calculating the mpg forecasts that the 2008 F-150 will average 16.8 mpg. Determine the probability that these same 100 AAA members would average more than 16.8 mpg while testing the 2008 F-150.

- A) 0
- B) 0.0155
- C) 0.0412
- D) None of the above

Answer:Ā A

Diff: 3

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 1

113) A population has a proportion equal to 0.30. Calculate the following probabilities with *n* = 100. Find *P*(Ā ā¤ 0.35).

- A) 0.7244
- B) 0.8621
- C) 0.7124
- D) 0.6126

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

114) A population has a proportion equal to 0.30. Calculate the following probabilities with *n *= 100. Find *P(**Ā *> 0.40).

- A) 0.0146
- B) 0.0411
- C) 0.0521
- D) 0.0312

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

115) A population has a proportion equal to 0.30. Calculate the following probabilities with *n *= 100. Find *P*(0.25 < Ā ā¤ 0.40).

- A) 0.8121
- B) 0.7415
- C) 0.8475
- D) 0.5612

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

116) A population has a proportion equal to 0.30. Calculate the following probabilities with* n *= 100. Find *P(*Ā ā„ 0.27).

- A) 0.7422
- B) 0.8141
- C) 0.6125
- D) 0.6841

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

117) If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is *p *= 0.30, what is the probability that the proportion of successes in the sample will be less than or equal to 0.27?

- A) 0.0841
- B) 0.1011
- C) 0.1912
- D) 0.1762

Answer:Ā D

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

118) The proportion of items in a population that possess a specific attribute is known to be 0.70. If a simple random sample of size *n* = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65, what is the sampling error?

- A) -0.03
- B) -0.05
- C) 0.08
- D) 0.01

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

119) Given a population where the proportion of items with a desired attribute isĀ *p* = 0.25, if a sample of 400 is taken, what is the standard deviation of the sampling distribution of ?

- A) 0.0512
- B) 0.0312
- C) 0.0217
- D) 0.0412

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

120) Given a population where the proportion of items with a desired attribute is* p* = 0.25, if a sample of 400 is taken, what is the probability the proportion of successes in the sample will be greater than 0.22?

- A) 0.9162
- B) 0.8812
- C) 0.7141
- D) 0.8412

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

121) Given a population in which the probability of success is *Ā p* =0.20, if a sample of 500 items is taken, then calculate the probability the proportion of successes in the sample will be between 0.18 and 0.23.

- A) 0.7812
- B) 0.8221
- C) 0.9212
- D) 0.6812

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

122) Given a population in which the probability of success is *p* = 0.20, if a sample of 500 items is taken, then calculate the probability the proportion of successes in the sample will be between 0.18 and 0.23 if the sample size is 200.

- A) 0.8911
- B) 0.7121
- C) 0.8712
- D) 0.6165

Answer:Ā D

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

123) United Manufacturing and Supply makes sprinkler valves for use in residential sprinkler systems. United supplies these valves to major companies such as Rain Bird and Nelson, who in turn sell sprinkler products to retailers. United recently entered into a contract to supply 40,000 sprinkler valves. The contract called for at least 97% of the valves to be free of defects. Before shipping the valves, United managers tested 200 randomly selected valves and found 190 defect-free valves in the sample. The managers wish to know the probability of finding 190 or fewer defect-free valves if in fact the population of 40,000 valves is 97% defect-free. The probability is:

- A) 0.0111
- B) 0.0612
- C) 0.0475
- D) 0.0212

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

124) The J R Simplot Company is one of the world’s largest privately held agricultural companies, employing over 10,000 people in the United States, Canada, China, Mexico, and Australia. More information can be found at the company’s Web site: www.Simplot.com. One of its major products is french fries that are sold primarily on the commercial market to customers such as McDonald’s and Burger King. French fries have numerous quality attributes that are important to customers. One of these is called “dark ends,” which are the dark-colored ends that can occur when the fries are cooked. Suppose a major customer will accept no more than 0.06 of the fries having dark ends.

Recently, the customer called the Simplot Company saying that a recent random sample of 300 fries was tested from a shipment and 27 fries had dark ends. Assuming that the population does meet the 0.06 standard, what is the probability of getting a sample of 300 with 27 or more dark ends?

- A) 0.0341
- B) 0.0162
- C) 0.0012
- D) 0.0231

Answer:Ā B

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

125) The National Association of Realtors released a survey indicating that a surprising 43% of first-time home buyers purchased their homes with no-money-down loans during 2005. The fear is that house prices will decline and leave homeowners owing more than their homes are worth. PMI Mortgage Insurance estimated that there existed a 50% risk that prices would decline within two years in major metro areas such as San Diego, Boston, Long Island, New York City, Los Angeles, and San Francisco. A survey taken by realtors in the San Francisco area found that 12 out of the 20 first-time home buyers sampled purchased their home with no-money-down loans. Calculate the probability that at least 12 in a sample of 20 first-time buyers would take out no-money-down loans if San Francisco’s proportion is the same as the nationwide proportion of no-money-down loans.

- A) 0.0618
- B) 0.0124
- C) 0.0512
- D) 0.0441

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

126) According to the most recent Labor Department data, 10.5% of engineers (electrical, mechanical, civil, and industrial) were women. Suppose a random sample of 50 engineers is selected. How likely is it that the random sample of 50 engineers will contain 8 or more women in these positions?

- A) 0.1612
- B) 0.0821
- C) 0.1020
- D) 0.0314

Answer:Ā C

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

127) According to the most recent Labor Department data, 10.5% of engineers (electrical, mechanical, civil, and industrial) were women. Suppose a random sample of 50 engineers is selected. How likely is it that the random sample will contain fewer than 5 women in these positions?

- A) 0.4522
- B) 0.3124
- C) 0.5121
- D) 0.5512

Answer:Ā A

Diff: 2

Keywords:Ā sampling distribution proportion

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 1

128) Indicate why the Central Limit Theorem is so important in the application of statistical analysis.

Answer:Ā The Central Limit Theorem helps decision makers know what the general shape of the sampling distribution is regardless of what the population distribution looks like. This is important because in many cases we won’t know the shape of the population distribution. As long as the sample size is sufficiently large, the sampling distribution of Ā will be approximately normally distributed.

Diff: 2

Keywords:Ā Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

129) Suppose it is known that the ages of all employees working for a very large computer company is normally distributed with a mean of 44.2 and a standard deviation of 5.6 years. Given this information, discuss what the sampling distribution for Ā looks like?

Answer:Ā When the population distribution is known to be normally distributed, the sampling distribution of Ā will also be normally distributed. Further, the mean of the sampling distribution will be equal to the population mean, 44.2, and the standard deviation of the sampling distribution will be .

Diff: 2

Keywords:Ā sampling distribution, mean, sample

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

130) Explain what is meant by the concept of sampling distribution.

Answer:Ā A population measure is called a parameter. It is a fixed value as long as the population does not change. A corresponding value computed from a sample is called a statistic. Depending on which sample we select we will get a different statistic. The difference between the statistic and the parameter due to the sample not being a perfect representative of the population is called sampling error. Each sample will generate a different amount of sampling error. For instance, if we are dealing with the mean, and were to take all possible samples of a given size, the distribution of the resulting sample means is called the sampling distribution. It represents the distribution of potential sampling error. The mean of the sampling distribution will equal the population mean and the standard deviation of the sampling distribution will equal the population standard deviation divided by the square root of the sample size.

Diff: 2

Keywords:Ā sampling distribution

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

131) Explain the impact of the size of the sample on the shape of the sampling distribution.

Answer:Ā First, if the population distribution is unknown or is not normally distributed, the size of the sample is important in helping us know what the shape of the sampling distribution is. Under these conditions, the Central Limit Theorem tells us that the sampling distribution will be approximately normally distributed provided that the sample size is sufficiently large. Second, the size of the sample is key in determining the spread in the sampling distribution. The larger the sample size, the less variability there is in the sampling distribution. When dealing with means, the standard deviation of the sampling distribution is equal to .

Diff: 2

Keywords:Ā sample size, sampling distribution

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

132) How would you respond to a statement that says that by increasing the sample size, the amount of sampling error will be decreased?

Answer:Ā We might respond by saying that in the long run the statement is true. That is, in situations where repeated random samples are selected, larger samples would result in less extreme sampling error on average. However, in a given situation there is no guarantee that a larger sample size will result is less sampling error. By chance, the smaller sample might provide a sample mean that is actually closer to the population mean than would occur from a particular larger sample.

Diff: 2

Keywords:Ā sampling error, sample size

Section:Ā 7-1 Sampling ErrorāWhat It Is and Why It Happens

Outcome:Ā 1

133) Suppose a population is normally distributed with a mean 100 and a standard deviation of 15. When a sample of size *n* = 36 is collected a sampling distribution is created. Explain which is larger: the probability of a value randomly selected from the population being larger than 120, or the probability of a sample mean being larger than 120.

Answer:Ā The probability will be larger for the value from the population than for the sample mean distribution because the population has the larger standard deviation.

Population: *z* = = 20/15 = 1.33, and looking this up in the standard normal table gives us a probability of 0.4082, so P(x > 120) = .5 – .4082 = 0.0918.

Sample mean distribution: z = Ā = 20/2.5 = 8.00, which is too large to look up in the standard.

Normal table, meaning P(*Ā *> 120) is so small it’s practically 0.

Diff: 2

Keywords:Ā sample, population, mean, sampling distribution

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 2

134) The annual income for independent sales representatives in the United States is thought to be highly right-skewed with a mean equal to $144,300 and a standard deviation of $32,450. Given this information, if a sample of 36 independent sales representatives is selected, what is the probability that the mean of the sample will exceed $130,000?

Answer:Ā Even though the population distribution is skewed, the Central Limit Theorem tells us that the sampling distribution will be approximately normally distributed for a sample of *n* = 36. Further, the mean of the sampling distribution should be equal to the population mean

(Ī¼ = $144,300) and the standard deviation should be equal to

= = $5,408.33. Then, we convert the Ā Ā = $130,000 to a standardized z-value using:

z = Ā = Ā = -2.64.

Now, we can go to the standard normal table for z = -2.64, which gives 0.4959. This is the probability of a sample mean between $130,000 and $144,300. To get the probability we are looking for, we add 0.5000 giving 0.9959. Thus, the probability that the sample mean annual income will exceed $130,000 is 0.9959 or almost a sure thing.

Diff: 2

Keywords:Ā probability, mean, sampling distribution, z-value, Central Limit Theorem

Section:Ā 7-2 Sampling Distribution of the Mean

Outcome:Ā 3

135) The proportion of parts in an inventory that are outdated and no longer useful is thought to be 0.10. To check this, a random sample of *n* = 100 parts is selected and 14 are found to be outdated. Based upon this information, what is the probability of 14 or more outdated parts?

Answer:Ā We are interested in finding P(Ā > 0.14). The sampling distribution for a proportion will be approximately normal as long as both *np *and *n*(1 – *p*) are greater than 5.

That applies in this case. The standard deviation for the sampling distribution is given by

. Thus, to find the probability, we standardize the sample proportion as follows:

*z *= Ā = = 1.33.

Then we can go to the standard normal table for z = 1.33. We get 0.4082. Subtracting this from 0.5000, we get 0.0918, which is the probability we are looking for.

Diff: 2

Keywords:Ā probability, proportion, sampling distribution, sample

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4

136) The Good Food chain has a contract to receive eggs from a large egg producer. The eggs come in lots of 4,000 dozen each week. The contract specifies that the rate of broken or defective eggs should not exceed 8 percent. Each time a load comes in, Good Food warehouse employees select a random sample of *n* = 100 eggs and check to see if they are broken or defective. If Good Food wants no more than a 0.05 chance of rejecting the shipment, what should the cut-off be in terms of proportion of broken or defective eggs so that if the proportion is that value or more, the shipment will be rejected?

Answer:Ā Since we are dealing with proportions, and both *np* and *n*(1 – *p)* are greater than 5, the normal distribution can be used to describe the sampling distribution. The standard in the contract calls for a 0.08 defect rate. The cut-off needs to be higher than this so that the chance of exceeding the cut-off is, at most, 0.05. To find what the cut-off should be, we use:

= *p *+ *z* Ā where z is the value from the standard normal table that is associated with 0.45. This z-value is approximately 1.645. Thus, the cut-off that the company should use is found as:

.08 + 1.645 = .1246. Thus, in the sample of 100 if they find more than 12 broken or defective eggs, the shipment should be rejected.

Diff: 3

Keywords:Ā sampling distribution, proportion, probability

Section:Ā 7-3 Sampling Distribution of a Proportion

Outcome:Ā 4