Ch.6 Exam Prep Continuous Distributions nan - Business Stats Contemporary Decision 10e | Test Bank by Ken Black by Ken Black. DOCX document preview.
Ch06: Chapter 6, Continuous Distributions
True/False
1. A uniform continuous distribution is also referred to as a rectangular distribution.
Response: See section 6.1, The Uniform Distribution
Difficulty: Easy
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
2. The height of the rectangle depicting a uniform distribution is the probability of each outcome and it is the same for all of the possible outcomes
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution
3. The area of the rectangle depicting a uniform distribution is always equal to the mean of the distribution.
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution
4. Many human characteristics such as height and weight and many variables such as household insurance and cost per square foot of rental space are normally distributed.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
5. Normal distribution is a symmetrical distribution with its tails extending to infinity on either side of the mean.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
6. Since a normal distribution curve extends from minus infinity to plus infinity, the area under the curve is infinity.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
7. A z-score is the number of standard deviations that a value of a random variable is above or below the mean.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
8. A normal distribution with a mean of zero and a standard deviation of 1 is called a null distribution.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
9. A standard normal distribution has a mean of one and a standard deviation of three.
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
10. The standard normal distribution is also called a finite distribution because its mean is zero and standard deviation one, always.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
11. In a standard normal distribution, if the area under curve to the right of a z-value is 0.10, then the area to the left of the same z-value is -0.10.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
12. The area under the standard normal distribution between -1 and 1 is twice the area between 0 and 1.
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
13. The area under the standard normal distribution between 0 and 2 is twice the area between 0 and 1.
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
14. The normal approximation for binomial distribution can be used when n=10 and p=1/5.
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
15. Binomial distributions in which the sample sizes are large may be approximated by a Poisson distribution.
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
16. A correction for continuity must be made when approximating the binomial distribution problems using a normal distribution.
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
17. If the probability of a binomial distribution was considering values that were greater than 14, then the correction for continuity would start at 13.5.
Ans: False
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
18. If the probability of a binomial distribution was considering values that were less than 14, then the correction for continuity would start at 13.5.
Ans: True
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
19. Correcting for continuity is necessary when a researcher is using a discrete distribution to estimate probabilities from a continuous distribution.
Ans: False
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
20. If arrivals at a bank followed a Poisson distribution, then the time between arrivals would follow a binomial distribution.
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
21. For an exponential distribution, the mean is always equal to its variance.
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
22. For an exponential distribution, the mean and the median are equal.
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
23. The exponential distribution is a continuous distribution that is closely related to the Poisson distribution.
Ans: True
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
24. The mean of an exponential distribution is equal to 1/λ.
Ans: True
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
25. The exponential distribution is especially useful in queuing problems to analyze the interarrival times.
Ans: True
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
Multiple Choice
26. Suppose the number of parking spots at urban grocery stores is uniformly distributed over the interval 90 to 140, inclusively (90 ≤ x ≤ 140), then the height of this distribution, f(x), is __________________.
a) 1/90
b) 1/50
c) 1/140
d) 1/200
e) 1/500
Response: See section 6.1, The Uniform Distribution
Difficulty: Easy
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
27. If the number of parking spots at urban grocery stores is uniformly distributed over the interval 90 to 140, inclusively (90 ≤ x ≤ 140), then the mean of this distribution is __________________.
a) 115
b) 230
c) 45
d) 70
e) unknown
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
28. If the number of parking spots at urban grocery stores is uniformly distributed over the interval 90 to 140, inclusively (90 ≤ x ≤ 140), then the standard deviation of this distribution is __________________.
a) 4.16
b) 50
c) 14.4
d) 7.07
e) 28.2
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
29. If the number of parking spots at urban grocery stores is uniformly distributed over the interval 90 to 140, inclusively (90 ≤ x ≤ 140), then P(x = exactly 100) is __________________.
a) 0.750
b) 0.000
c) 0.333
d) 0.500
e) 0.900
Response: See section 6.1, The Uniform Distribution
Difficulty: Hard
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
30. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then the probability, P(9 ≤ x ≤ 11) is __________________.
a) 0.250
b) 0.500
c) 0.333
d) 0.750
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
31. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then the probability, P(10.0 ≤ x ≤ 11.5), is __________________.
a) 0.250
b) 0.333
c) 0.375
d) 0.500
e) 0.750
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
32. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then the probability, P(13 ≤ x ≤ 15), is __________________.
a) 0.250
b) 0.500
c) 0.375
d) 0.000
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
33. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then P(x < 7) is __________________.
a) 0.500
b) 0.000
c) 0.375
d) 0.250
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
34. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then P(x ≤ 11) is __________________.
a) 0.750
b) 0.000
c) 0.333
d) 0.500
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Hard
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
35. If x is uniformly distributed over the interval 8 to 12, inclusively (8 ≤ x ≤ 12), then P(x ≥ 10) is __________________.
a) 0.750
b) 0.000
c) 0.333
d) 0.500
e) 0.900
Response: See section 6.1, The Uniform Distribution
Difficulty: Hard
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
36. If x, the time (in minutes) to complete an oil change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the height of this distribution, f(x), is __________________.
a) 1/10
b) 1/20
c) 1/30
d) 12/50
e) 1/60
Response: See section 6.1, The Uniform Distribution
Difficulty: Easy
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
37. If x, the time (in minutes) to complete an oil change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the mean of this distribution is __________________.
a) 50
b) 25
c) 10
d) 15
e) 5
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
38. If x, the time (in minutes) to complete an oil change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the standard deviation of this distribution is __________________.
a) unknown
b) 8.33
c) 0.833
d) 2.89
e) 1.89
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
39. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job is completed in 25 to 28 minutes, inclusively, i.e., P(25 ≤ x ≤ 28) is __________________.
a) 0.250
b) 0.500
c) 0.300
d) 0.750
e) 81.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
40. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job is completed in 21.75 to 24.25 minutes, inclusively, i.e., P(21.75 ≤ x ≤ 24.25) is __________________.
a) 0.250
b) 0.333
c) 0.375
d) 0.000
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
41. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job is completed in 33 to 35 minutes, inclusively, i.e., P(33 ≤ x ≤ 35) is __________________.
a) 0.5080
b) 0.000
c) 0.375
d) 0.200
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
42. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job is completed in less than 17 minutes, i.e., P(x < 17) is __________________.
a) 0.500
b) 0.300
c) 0.000
d) 0.250
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Medium
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
43. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job is completed in less than or equal to 22 minutes, i.e., P(x ≤ 22) is __________________.
a) 0.200
b) 0.300
c) 0.000
d) 0.250
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Hard
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
44. If x, the time (in minutes) to complete an change job at certain auto service station, is uniformly distributed over the interval 20 to 30, inclusively (20 ≤ x ≤ 30), then the probability that an oil change job will be completed 24 minutes or more, i.e., P(x ≥ 24) is __________________.
a) 0.100
b) 0.000
c) 0.333
d) 0.600
e) 1.000
Response: See section 6.1, The Uniform Distribution
Difficulty: Hard
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
45. The normal distribution is an example of _______.
a) a discrete distribution
b) a continuous distribution
c) a bimodal distribution
d) an exponential distribution
e) a binomial distribution
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
46. The total area underneath any normal curve is equal to _______.
a) the mean
b) one
c) the variance
d) the coefficient of variation
e) the standard deviation
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
47. The area to the left of the mean in any normal distribution is equal to _______.
a) the mean
b) 1
c) the variance
d) 0.5
e) -0.5
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
48. A standard normal distribution has the following characteristics:
a) the mean and the variance are both equal to 1
b) the mean and the variance are both equal to 0
c) the mean is equal to the variance
d) the mean is equal to 0 and the variance is equal to 1
e) the mean is equal to the standard deviation
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
49. Let z be a normal random variable with mean 0 and standard deviation 1. What is P(z < 1.3)?
a) 0.4032
b) 0.9032
c) 0.0968
d) 0.3485
e) 0. 5485
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
50. Let z be a normal random variable with mean 0 and standard deviation 1. What is P(1.3 < z < 2.3)?
a) 0.4032
b) 0.9032
c) 0.4893
d) 0.0861
e) 0.0086
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
51. Let z be a normal random variable with mean 0 and standard deviation 1. What is P(z > 2.4)?
a) 0.4918
b) 0.9918
c) 0.0082
d) 0.4793
e) 0.0820
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
52. Let z be a normal random variable with mean 0 and standard deviation 1. What is P(z < -2.1)?
a) 0.4821
b) -0.4821
c) 0.9821
d) 0.0179
e) -0.0179
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
53. Let z be a normal random variable with mean 0 and standard deviation 1. What is P(z > -1.1)?
a) 0.3643
b) 0.8643
c) 0.1357
d) -0.1357
e) -0.8643
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
54. Let z be a normal random variable with mean 0 and standard deviation 1. Which of these most closely reflects P(-2.25 < z < -1.1)?
a) 0.3643
b) 0.8643
c) 0.1235
d) 0.4878
e) 0.5000
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
55. Let z be a normal random variable with mean 0 and standard deviation 1. The 50th percentile of z is most closely reflected by ____________.
a) 0.6700
b) -1.254
c) 0.0000
d) 1.2800
e) 0.5000
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
56. Let z be a normal random variable with mean 0 and standard deviation 1. The 75th percentile of z is most closely reflected by ____________.
a) 0.6700
b) -1.254
c) 0.0000
d) 1.2800
e) 0.5000
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
57. Let z be a normal random variable with mean 0 and standard deviation 1. The 90th percentile of z is most closely reflected by ____________.
a) 1.645
b) -1.254
c) 1.960
d) 1.280
e) 1.650
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
58. A z score is the number of __________ that a value is from the mean.
a) variances
b) standard deviations
c) units
d) miles
e) minutes
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
59. Within a range of z scores from -1 to +1, you can expect to find _______ per cent of the values in a normal distribution.
a) 95
b) 99
c) 68
d) 34
e) 100
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
60. Within a range of z scores from -2 to +2, you can expect to find _______ per cent of the values in a normal distribution.
a) 95
b) 99
c) 68
d) 34
e) 100
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
61.Suppose the total time to fill a routine prescription at a local pharmacy averages 35 minutes starting from the time the physician places the order to the time it is dispensed. Assume the standard deviation is 11 minutes. The z-score for x = 46 is ________.
a) 1.0
b) -1.0
c) 11
d) -11
e) .10
Response: See section 6.2, Normal Distribution
Difficulty: Easy
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
62. Suppose the total time to fill a routine prescription at a local pharmacy averages 35 minutes starting from the time the physician places the order to the time it is dispensed. Assume the standard deviation is 11 minutes. A z score was calculated for a number, and the z score is 3.4. What is x?
a) 37.4
b) 72.4
c) 0.00
d) 68.0
e) 2.0.8
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
63. Suppose the total time to fill a routine prescription at a local pharmacy averages 35 minutes starting from the time the physician places the order to the time it is dispensed. Assume the standard deviation is 11 minutes and the z score is -1.3. What is x?
a) 20.7
b) 0.0
c) -14.3
d) 14.3
e) -20.7
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
64. Suppose the total time to fill a routine prescription at a local pharmacy averages 35 minutes starting from the time the physician places the order to the time it is dispensed. Assume the standard deviation is 11 minutes and the z score is 0. What is x?
a) -35.0
b) 0.0
c) 70.0
d) 35.0
e) -1.0
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve
65. The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. Which most closely reflects the probability that a randomly selected bulb would last longer than 1150 hours?
a) 0.4987
b) 0.9987
c) 0.0013
d) 0.5013
e) 0.5513
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
66. The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. Which most closely reflects the probability that a randomly selected bulb would last fewer than 1100 hours?
a) 0.4772
b) 0.9772
c) 0.0228
d) 0.5228
e) 0.5513
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
67. The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. The life of this bulb is normally distributed. Which most closely reflects the probability that a randomly selected bulb would last fewer than 940 hours?
a) 0.3849
b) 0.8849
c) 0.1151
d) 0.6151
e) 0.6563
Response: See section 6.2, Normal Distribution
Difficulty: Medium
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
68. Suppose you are working with a data set that is normally distributed with a mean of 400 and a standard deviation of 20. Determine the most likely value of x such that 60% of the values are greater than x.
a) 404.5
b) 395.5
c) 405.0
d) 395.0
e) 415.0
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
69. Sure Stone Tire Company has established that the useful life of a particular brand of its automobile tires is normally distributed with a mean of 40,000 miles and a standard deviation of 5000 miles. What is the probability that a randomly selected tire of this brand has a life of at most 30,000 miles?
a) 0.5000
b) 0.4772
c) 0.0228
d) 0.9772
e) 1.0000
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
70. Sure Stone Tire Company has established that the useful life of a particular brand of its automobile tires is normally distributed with a mean of 40,000 miles and a standard deviation of 5000 miles. What is the probability that a randomly selected tire of this brand has a life of at least 50,000 miles?
a) 0.0228
b) 0.9772
c) 0.5000
d) 0.4772
e) 1.0000
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
71. Sure Stone Tire Company has established that the useful life of a particular brand of its automobile tires is normally distributed with a mean of 40,000 miles and a standard deviation of 5000 miles. Which probability most closely reflects that a randomly selected tire of this brand has a life between 30,000 and 50,000 miles?
a) 0.5000
b) 0.4772
c) 0.9544
d) 0.9772
e) 1.0000
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
72. The net profit from a certain investment is normally distributed with a mean of $2,500 and a standard deviation of $1,000. The probability that the investor will not have a net loss is _____________.
a) 0.4938
b) 0.0062
c) 0.9938
d) 0.5062
e) 0.0000
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
73. The net profit from a certain investment is normally distributed with a mean of $2,500 and a standard deviation of $1,000. The probability that the investor’s net gain will be at least $2,000 is most closely reflected by _____________.
a) 0.0000
b) 0.3413
c) 0.6915
d) 0.0500
e) 0.5000
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
74. Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will be completed within 185 work-days is ______.
a) 0.0668
b) 0.4332
c) 0.5000
d) 0.9332
e) 0.9950
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
75. Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. To be 99% sure that we will not be late in completing the project, we should request a completion time of _______ work-days.
a) 211
b) 207
c) 223
d) 200
e) 250
Response: See section 6.2, Normal Distribution
Difficulty: Hard
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value of x in a normal distribution when given information about the area under the normal curve.
76. Let x be a binomial random variable with n=35 and p=.20. If we use the normal distribution to approximate probabilities for this, we would use a mean of _______.
a) 35
b) 20
c) 70
d) 7
e) 3.5
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Easy
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
77. Let x be a binomial random variable with n=50 and p=.3. If we use the normal distribution to approximate probabilities for this, a correction for continuity should be made. To find the probability of more than 15 successes, we should find _______.
a) P(x>15.5)
b) P(x>15)
c) P(x>14.5)
d) P(x<14.5)
e) P(x < 15)
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
78. Let x be a binomial random variable with n=50 and p=.3. The probability of less than or equal to13 successes, when using the normal approximation for binomial is most closely reflected by ________.
a) -0.6172
b) 0.3086
c) 3.240
c)0 .2324
d) 0.3217
e) -0.23224
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Hard
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
79. Assuming an equal chance of a new baby being a boy or a girl (that is, p= 0.5), we would like to find the probability of 40 or more of the next 100 births at a local hospital will be boys. Using the normal approximation for binomial with a correction for continuity, we should use the z-score _______
a) 0.4
b) -2.1
c) 0.6
d) 2
e) -1.7
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Hard
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
80. The probability that a call to an emergency help line is answered in less than 10 seconds is 0.8. Assume that the calls are independent of each other. Using the normal approximation for binomial with a correction for continuity, the probability that at least 75 of 100 calls are answered within 10 seconds is approximately _______
a) 0.8
b) 0.1313
c) 0.5235
d) 0.9154
e) 0.8687
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Hard
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
81. For the normal approximation to be a good approximation for binomial distribution, n*p must be greater than ___________.
a) 50
b) 5
c) 0.5
d) 0.05
e) 0
Ans: b
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
82. The probability that someone will prefer Coke over Pepsi is 0.56. A researcher asked 90 people which soda pop they preferred, and used the normal approximation to the binomial with a correction for continuity, what would be the mean used in the calculation?
a) 0.56
b) 90
c) 45.0
d) 50.4
e) 106.7
Ans: d
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
83. The probability that someone will prefer Coke over Pepsi is 0.56. A researcher asked 90 people which soda pop they preferred, and used the normal approximation to the binomial with a correction for continuity, what would be the standard deviation used in the calculation?
a) 4.7
b) 22.2
c) 50.4
d) 39.6
e) 22.5
Ans: a
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
84. The probability that someone will prefer Coke over Pepsi is 0.56. A researcher asked 90 people which soda pop they preferred, and used the normal approximation to the binomial with a correction for continuity, what would be the probability at least 50 preferring Coke?
a) 0.615
b) 0.507
c) 0.560
d) 0.900
e) 0.576
Ans: e
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
85. Based on the caterer’s experience, 38% of attendees to events will prefer chicken for the main dish. As the caterer plans for an event attended by 780 individuals, the normal approximation will be used for the binomial with a correction for continuity. In this case, what is the average number that she would expect to prefer chicken, when determining the probability that less than 300 will prefer chicken?
a) 296.4
b) 114.0
c) 483.6
d) 13.6
e) 183.8
Ans: a
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
86. Based on the caterer’s experience, 38% of attendees to events will prefer chicken for the main dish. As the caterer plans for an event attended by 780 individuals, the normal approximation will be used for the binomial with a correction for continuity. In this case, what is the standard deviation of the number that she would expect to prefer chicken, when determining the probability that at less than 300 will prefer chicken?
a) 114.0
b) 300.0
c) 183.8
d) 296.4
e) 13.6
Ans: e
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
87. Based on the caterer’s experience, 38% of attendees to events will prefer chicken for the main dish. As the caterer plans for an event attended by 780 individuals, the normal approximation will be used for the binomial with a correction for continuity. In this case, what is the probability that less than 300 will prefer chicken?
a) 0.604
b) 0.380
c) 0.590
d) 0.620
e) 0.618
Ans: c
Response: See section 6.3, Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Medium
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
88. The exponential distribution is an example of _______.
a) a discrete distribution
b) a continuous distribution
c) a bimodal distribution
d) a normal distribution
e) a symmetrical distribution
Response: See section 6.4, Exponential Distribution
Difficulty: Easy
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
89. For an exponential distribution with a lambda () equal to 20, the mean equal to _______.
a) 20
b) .05
c) 4.474
d) 1
e) 2.11
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
90. The average time between phone calls arriving at a call center is 30 seconds. Assuming that the time between calls is exponentially distributed, find the probability that more than a minute elapses between calls.
a) 0.135
b) 0.368
c) 0.865
d) 0.607
e) 0.709
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
91. The average time between phone calls arriving at a call center is 30 seconds. Assuming that the time between calls is exponentially distributed, find the probability that less than two minutes elapse between calls.
a) 0.018
b) 0.064
c) 0.936
d) 0.982
e) 1.000
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
92. At a certain workstation in an assembly line, the time required to assemble a component is exponentially distributed with a mean time of 10 minutes. Find the probability that a component is assembled in 7 minutes or less?
a) 0.349
b) 0.591
c) 0.286
d) 0.714
e) 0.503
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
93. At a certain workstation in an assembly line, the time required to assemble a component is exponentially distributed with a mean time of 10 minutes. Find the probability that a component is assembled in 3 to 7 minutes?
a) 0.5034
b) 0.2592
c) 0.2442
d) 0.2942
e) 0.5084
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
94. Participants in a 2 day biking event, cross the finish line at a rate of 10 bike riders per fifteen minute interval. On average, how much time, in minutes, elapses between bike riders?
a) 1.50
b) .0667
c) .1667
d) 1.00
e) 2.50
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
95. Participants in a 2 day biking event, cross the finish line at a rate of 10 bike riders per fifteen minute interval. The probability that at least 2 minutes will elapse between bike riders is closest to _____________.
a) .0000
b) .0498
c) .2635
d) .1353
e) .4647
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
96. Participants in a 2 day biking event, cross the finish line at a rate of 10 bike riders per fifteen minute interval. The probability that less than 10 minutes will elapse between car arrivals is _____________.
a) .0001
b) .9987
c) .0013
d) .6667
e) .1667
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
97. Inquiries arrive at a record message device according to a Poisson process at a rate of 15 inquiries per minute. The probability that it takes more than 12 seconds for the first inquiry to arrive is approximately _________
a) 0.05
b) 0.75
c) 0.25
d) 0.27
e) 0.73
Response: See section 6.4, Exponential Distribution
Difficulty: Hard
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
98. In a busy gas station, an average of 2.4 cars arrive every 5 minutes. What is the average time between arrivals?
a) 0.42 minutes
b) 2.08 minutes
c) 5.24 minutes
d) 2.40 minutes
e) 5.00 minutes
Ans: b
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
99. In a busy gas station, an average of 2.4 cars arrive every minute. What is the average time between arrivals?
a) 0.42 minutes
b) 2.08 minutes
c) 5.24 minutes
d) 2.40 minutes
e) 5.00 minutes
Ans: a
Response: See section 6.4, Exponential Distribution
Difficulty: Medium
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
100. If x is uniformly distributed over the interval a2 to b2 (x ~ [a2, b2]), the mean value of x is _________.
a) (a + b)/2
b) (b – a)/2
c) (a2 + b2)/2
d) (b2 – a2)/2
e) b2 – a2
Response: See section 6.1 The Uniform Distribution
Difficulty: Medium
AACSB: Reflective thinking
Bloom’s level: Knowledge
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
101. If x is uniformly distributed over the interval a2 to b2 (x ~ [a2, b2]), and a2< a < b < b2, the probability that x is between a and b, P(a ≤ x ≤ b), is _________.
a) 1/(a + b)
b) (a + b)/(a2 + b2)
c) (a2 + b2)/(a + b)
d) 1/(b2 – a2)/
e) b2 – a2
Response: See section 6.1 The Uniform Distribution
Difficulty: Hard
AACSB: Reflective thinking
Bloom’s level: Knowledge
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
102. If x is uniformly distributed over the interval a2 to b2 (x ~ [a2, b2]), and a < a2 < b < b2, the probability that x is between a and b, P(a ≤ x ≤ b), is _________.
a) (b – a)/(b2 – a2)
b) (a + b)/(a2 + b2)
c) (a + b)/( b2 – a2)
d) (b – a2)/(b2 – a2)
e) (b2 – a2)/(a2 + b2)
Response: See section 6.1 The Uniform Distribution
Difficulty: Hard
AACSB: Reflective thinking
Bloom’s level: Knowledge
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
103. If x is uniformly distributed over some interval. The mean value (μ) of x is 2 and its variance (σ2) is 1/3. The probability that x is between 2 and 2.5, P(2 ≤ x ≤ 2.5), is _________.
a) 0.45
b) 0.40
c) 0.35
d) 0.33
e) 0.25
Response: See section 6.1 The Uniform Distribution
Difficulty: Hard
AACSB: Analytic
Bloom’s level: Application
Learning Objective: 6.1: Solve for probabilities in a continuous uniform distribution.
104. If variable x is normally distributed with mean 0 and standard deviation 1, x ~ N(0,1), then the probability that x is exactly 0 is _________.
a) 0.05
b) 0.04
c) 0.02
d) 0.01
e) 0.00
Response: See section 6.2 The Normal Distribution
Difficulty: Medium
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value in x in a normal distribution when given information about the area under the normal curve.
105. Most graduate business schools require applicants to take the GMAT. Scores on this test are approximately normally distributed with a mean of 545 points and a standard deviation of 100 points. What score do you need to be in the top 5% (approximately the scores needed for top schools)?
a) 718
b) 716
c) 714
d) 712
e) 710
Response: See section 6.2 The Normal Distribution
Difficulty: Medium
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.2: Solve for probabilities in a normal distribution using z scores and for the mean, the standard deviation, or a value in x in a normal distribution when given information about the area under the normal curve.
106. A recent market study has determined that the probability that a young adult would be willing to try a new online financial service that your company is offering is 50%. In a random sample of 10 young adults, the probability that exactly 5 will be willing to try this new service is _________, and the approximate probability (using the normal distribution is) _________.
a) 0.251; 0.246
b) 0.246; 0.242
c) 0.246; 0.248
d) 0.249; 0.251
e) 0.248; 0.246
Response: See section 6.3 Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Hard
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
107. A recent market study has determined that the probability that a young adult would be willing to try a new online financial service that your company is offering is 50%. In a random sample of 10 young adults, the approximate probability that at least 2 but no more than 3 will be willing to try this new service is _________.
a) 0.1575
b) 0.1612
c) 0.1775
d) 0.1812
e) 0.1975
Response: See section 6.3 Using the Normal Curve to Approximate Binomial Distribution Problems
Difficulty: Hard
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.3: Solve problems from the discrete binomial distribution using the continuous normal distribution and correcting for continuity.
108. Employees arrive at a cafeteria according to a Poisson process at an average rate of 30 employees per hour. The average waiting time between employees’ arrivals is _________ minutes.
a) 0.03
b) 0.05
c) 0.5
d) 1
e) 2
Response: See section 6.4 Exponential Distribution
Difficulty: Medium
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
109. Employees arrive at a cafeteria according to a Poisson process at an average rate of 30 employees per hour. The probability that after one employee arrives, the next one will arrive at least 3 minutes later is _________.
a) 0.223
b) 0.202
c) 0.183
d) 0.162
e) 0.143
Response: See section 6.4 Exponential Distribution
Difficulty: Hard
AACSB: Reflective thinking
Bloom’s level: Application
Learning Objective: 6.4: Solve for probabilities in an exponential distribution and contrast the exponential distribution to the discrete Poisson distribution.
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