Modern Business Statistics with Microsoft Office Excel 4th Edition by David R. Anderson - Test Bank

Modern Business Statistics with Microsoft Office Excel 4th Edition by David R. Anderson – Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

CHAPTER 6—CONTINUOUS PROBABILITY DISTRIBUTIONS

 

MULTIPLE CHOICE

 

1. If arrivals follow a Poisson probability distribution, the time between successive arrivals must follow

a.	a Poisson probability distribution
b.	a normal probability distribution
c.	a uniform probability distribution
d.	an exponential probability distribution

 

 

ANS:  D                    PTS:   1

 

2. Whenever the probability is proportional to the length of the interval in which the random variable can assume a value, the random variable is

a.	uniformly distributed
b.	normally distributed
c.	exponentially distributed
d.	Poisson distributed

 

 

ANS:  A                    PTS:   1

 

3. There is a lower limit but no upper limit for a random variable that follows the

a.	uniform probability distribution
b.	normal probability distribution
c.	exponential probability distribution
d.	binomial probability distribution

 

 

ANS:  C                    PTS:   1

 

4. The form of the continuous uniform probability distribution is

a.	triangular
b.	rectangular
c.	bell-shaped
d.	a series of vertical lines

 

 

ANS:  B                    PTS:   1

 

5. The mean, median, and mode have the same value for which of the following probability distributions?

a.	uniform
b.	normal
c.	exponential
d.	Poisson

 

 

ANS:  B                    PTS:   1

 

6. The probability distribution that can be described by just one parameter is the

a.	uniform
b.	normal
c.	exponential
d.	binomial

 

 

ANS:  C                    PTS:   1

 

7. A continuous random variable may assume

a.	all values in an interval or collection of intervals
b.	only integer values in an interval or collection of intervals
c.	only fractional values in an interval or collection of intervals
d.	all the positive integer values in an interval

 

 

ANS:  A                    PTS:   1

 

8. For a continuous random variable x, the probability density function f(x) represents

a.	the probability at a given value of x
b.	the area under the curve at x
c.	Both the probability at a given value of x and the area under the curve at x are correct answers.
d.	the height of the function at x

 

 

ANS:  D                    PTS:   1

 

9. The function that defines the probability distribution of any continuous random variable is a

a.	normal function
b.	uniform function
c.	Both the normal function and the uniform function are correct.
d.	probability density function

 

 

ANS:  D                    PTS:   1

 

10. For any continuous random variable, the probability that the random variable takes on exactly a specific value is

a.	1.00
b.	0.50
c.	any value between 0 to 1
d.	zero

 

 

ANS:  D                    PTS:   1

 

11. The uniform probability distribution is used with

a.	a continuous random variable
b.	a discrete random variable
c.	a normally distributed random variable
d.	any random variable

 

 

ANS:  A                    PTS:   1

 

12. A uniform probability distribution is a continuous probability distribution where the probability that the random variable assumes a value in any interval of equal length is

a.	different for each interval
b.	the same for each interval
c.	Either a or b could be correct depending on the magnitude of the standard deviation.
d.	None of the alternative answers is correct.

 

 

ANS:  B                    PTS:   1

 

13. For a uniform probability density function, the height of the function

a.	can not be larger than one
b.	is the same for each value of x
c.	is different for various values of x
d.	decreases as x increases

 

 

ANS:  B                    PTS:   1

 

14. A continuous random variable is uniformly distributed between a and b. The probability density function between a and b is

a.	zero
b.	(a – b)
c.	(b – a)
d.	1/(b – a)

 

 

ANS:  D                    PTS:   1

 

15. The probability density function for a uniform distribution ranging between 2 and 6 is

a.	4
b.	undefined
c.	any positive value
d.	0.25

 

 

ANS:  D                    PTS:   1

 

16. The random variable x is known to be uniformly distributed between 70 and 90. The probability of x having a value between 80 to 95 is

a.	0.75
b.	0.5
c.	0.05
d.	1

 

 

ANS:  B                    PTS:   1

 

17. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability density function has what value in the interval between 6 and 10?

a.	0.25
b.	4.00
c.	5.00
d.	zero

 

 

ANS:  A                    PTS:   1

 

18. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product between 7 to 9 minutes is

a.	zero
b.	0.50
c.	0.20
d.	1

 

 

ANS:  B                    PTS:   1

 

19. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product in less than 6 minutes is

a.	zero
b.	0.50
c.	0.15
d.	1

 

 

ANS:  A                    PTS:   1

 

20. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product in 7 minutes or more is

a.	0.25
b.	0.75
c.	zero
d.	1

 

 

ANS:  B                    PTS:   1

 

21. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The expected assembly time (in minutes) is

a.	16
b.	2
c.	8
d.	4

 

 

ANS:  C                    PTS:   1

 

22. The assembly time for a product is uniformly distributed between 6 to 10 minutes. The standard deviation of assembly time (in minutes) is approximately

a.	0.3333
b.	0.1334
c.	16
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

23. A normal probability distribution

a.	is a continuous probability distribution
b.	is a discrete probability distribution
c.	can be either continuous or discrete
d.	always has a standard deviation of 1

 

 

ANS:  A                    PTS:   1

 

24. Which of the following is not a characteristic of the normal probability distribution?

a.	The mean, median, and the mode are equal
b.	The mean of the distribution can be negative, zero, or positive
c.	The distribution is symmetrical
d.	The standard deviation must be 1

 

 

ANS:  D                    PTS:   1

 

25. Which of the following is not a characteristic of the normal probability distribution?

a.	The graph of the curve is the shape of a rectangle
b.	The total area under the curve is always equal to 1.
c.	99.72% of the time the random variable assumes a value within plus or minus three standard deviations of its mean
d.	The mean is equal to the median, which is also equal to the mode.

 

 

ANS:  A                    PTS:   1

 

26. The highest point of a normal curve occurs at

a.	one standard deviation to the right of the mean
b.	two standard deviations to the right of the mean
c.	approximately three standard deviations to the right of the mean
d.	the mean

 

 

ANS:  D                    PTS:   1

 

27. If the mean of a normal distribution is negative,

a.	the standard deviation must also be negative
b.	the variance must also be negative
c.	a mistake has been made in the computations, because the mean of a normal distribution can not be negative
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

28. Larger values of the standard deviation result in a normal curve that is

a.	shifted to the right
b.	shifted to the left
c.	narrower and more peaked
d.	wider and flatter

 

 

ANS:  D                    PTS:   1

 

29. A standard normal distribution is a normal distribution with

a.	a mean of 1 and a standard deviation of 0
b.	a mean of 0 and a standard deviation of 1
c.	any mean and a standard deviation of 1
d.	any mean and any standard deviation

 

 

ANS:  B                    PTS:   1

 

30. In a standard normal distribution, the range of values of z is from

a.	minus infinity to infinity
b.	-1 to 1
c.	0 to 1
d.	-3.09 to 3.09

 

 

ANS:  A                    PTS:   1

 

31. For a standard normal distribution, a negative value of z indicates

a.	a mistake has been made in computations, because z is always positive
b.	the area corresponding to the z is negative
c.	the z is to the left of the mean
d.	the z is to the right of the mean

 

 

ANS:  C                    PTS:   1

 

32. For the standard normal probability distribution, the area to the left of the mean is

a.	-0.5
b.	0.5
c.	any value between 0 to 1
d.	1

 

 

ANS:  B                    PTS:   1

 

33. The mean of a standard normal probability distribution

a.	is always equal to 1
b.	can be any value as long as it is positive
c.	can be any value
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

34. The standard deviation of a standard normal distribution

a.	is always equal to zero
b.	is always equal to one
c.	can be any positive value
d.	can be any value

 

 

ANS:  B                    PTS:   1

 

35. For a standard normal distribution, the probability of z £ 0 is

a.	zero
b.	-0.5
c.	0.5
d.	one

 

 

ANS:  C                    PTS:   1

 

36. Z is a standard normal random variable. The P(1.20 £ z £ 1.85) equals

a.	0.4678
b.	0.3849
c.	0.8527
d.	0.0829

 

 

ANS:  D                    PTS:   1

 

37. Z is a standard normal random variable. The P(1.05 £ z £ 2.13) equals

a.	0.8365
b.	0.1303
c.	0.4834
d.	None of the alternative answers is correct.

 

 

ANS:  B                    PTS:   1

 

38. Z is a standard normal random variable. The P(1.41 < z < 2.85) equals

a.	0.4772
b.	0.3413
c.	0.8285
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

39. Z is a standard normal random variable. The P(-1.96 £ z £ -1.4) equals

a.	0.8942
b.	0.0558
c.	0.475
d.	0.4192

 

 

ANS:  B                    PTS:   1

 

40. Z is a standard normal random variable. The P (-1.20 £ z £ 1.50) equals

a.	0.0483
b.	0.3849
c.	0.4332
d.	0.8181

 

 

ANS:  D                    PTS:   1

 

41. Z is a standard normal random variable. The P(-1.5 £ z £ 1.09) equals

a.	0.4322
b.	0.3621
c.	0.7953
d.	0.0711

 

 

ANS:  C                    PTS:   1

 

42. Z is a standard normal random variable. The P(z ³ 2.11) equals

a.	0.4821
b.	0.9821
c.	0.5
d.	0.0174

 

 

ANS:  D                    PTS:   1

 

43. Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1112?

a.	0.3888
b.	1.22
c.	2.22
d.	3.22

 

 

ANS:  B                    PTS:   1

 

44. Given that z is a standard normal random variable, what is the value of z if the area to the right of z is 0.1401?

a.	1.08
b.	0.1401
c.	2.16
d.	-1.08

 

 

ANS:  A                    PTS:   1

 

45. Given that z is a standard normal random variable, what is the value of z if the area to the left of z is 0.9382?

a.	1.8
b.	1.54
c.	2.1
d.	1.77

 

 

ANS:  B                    PTS:   1

 

46. Z is a standard normal random variable. What is the value of z if the area between –z and z is 0.754?

a.	0.377
b.	0.123
c.	2.16
d.	1.16

 

 

ANS:  D                    PTS:   1

 

47. Z is a standard normal random variable. What is the value of z if the area to the right of z is 0.9803?

a.	-2.06
b.	0.4803
c.	0.0997
d.	3.06

 

 

ANS:  A                    PTS:   1

 

48. For a standard normal distribution, the probability of obtaining a z value between -2.4 to -2.0 is

a.	0.4000
b.	0.0146
c.	0.0400
d.	0.5000

 

 

ANS:  B                    PTS:   1

 

49. For a standard normal distribution, the probability of obtaining a z value of less than 1.6 is

a.	0.1600
b.	0.0160
c.	0.0016
d.	0.9452

 

 

ANS:  D                    PTS:   1

 

50. For a standard normal distribution, the probability of obtaining a z value between -1.9 to 1.7 is

a.	0.9267
b.	0.4267
c.	1.4267
d.	0.5000

 

 

ANS:  A                    PTS:   1

 

51. X is a normally distributed random variable with a mean of 8 and a standard deviation of 4. The probability that x is between 1.48 and 15.56 is

a.	0.0222
b.	0.4190
c.	0.5222
d.	0.9190

 

 

ANS:  D                    PTS:   1

 

52. X is a normally distributed random variable with a mean of 5 and a variance of 4. The probability that x is greater than 10.52 is

a.	0.0029
b.	0.0838
c.	0.4971
d.	0.9971

 

 

ANS:  A                    PTS:   1

 

53. X is a normally distributed random variable with a mean of 12 and a standard deviation of 3. The probability that x equals 19.62 is

a.	0.000
b.	0.0055
c.	0.4945
d.	0.9945

 

 

ANS:  A                    PTS:   1

 

54. X is a normally distributed random variable with a mean of 22 and a standard deviation of 5. The probability that x is less than 9.7 is

a.	0.000
b.	0.4931
c.	0.0069
d.	0.9931

 

 

ANS:  C                    PTS:   1

 

55. The ages of students at a university are normally distributed with a mean of 21. What percentage of the student body is at least 21 years old?

a.	It could be any value, depending on the magnitude of the standard deviation
b.	50%
c.	21%
d.	1.96%

 

 

ANS:  B                    PTS:   1

 

Exhibit 6-1

Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.

 

56. Refer to Exhibit 6-1. The probability density function has what value in the interval between 20 and 28?

a.	0
b.	0.050
c.	0.125
d.	1.000

 

 

ANS:  C                    PTS:   1

 

57. Refer to Exhibit 6-1. The probability that x will take on a value between 21 and 25 is

a.	0.125
b.	0.250
c.	0.500
d.	1.000

 

 

ANS:  C                    PTS:   1

 

58. Refer to Exhibit 6-1. The probability that x will take on a value of at least 26 is

a.	0.000
b.	0.125
c.	0.250
d.	1.000

 

 

ANS:  C                    PTS:   1

 

59. Refer to Exhibit 6-1. The mean of x is

a.	0.000
b.	0.125
c.	23
d.	24

 

 

ANS:  D                    PTS:   1

 

60. Refer to Exhibit 6-1. The variance of x is approximately

a.	2.309
b.	5.333
c.	32
d.	0.667

 

 

ANS:  B                    PTS:   1

 

Exhibit 6-2

The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.

 

61. Refer to Exhibit 6-2. What is the random variable in this experiment?

a.	the uniform distribution
b.	40 minutes
c.	90 minutes
d.	the travel time

 

 

ANS:  D                    PTS:   1

 

62. Refer to Exhibit 6-2. The probability that she will finish her trip in 80 minutes or less is

a.	0.02
b.	0.8
c.	0.2
d.	1.00

 

 

ANS:  B                    PTS:   1

 

63. Refer to Exhibit 6-2. The probability that her trip will take longer than 60 minutes is

a.	1.00
b.	0.40
c.	0.02
d.	0.600

 

 

ANS:  D                    PTS:   1

 

64. Refer to Exhibit 6-2. The probability that her trip will take exactly 50 minutes is

a.	zero
b.	0.02
c.	0.06
d.	0.20

 

 

ANS:  A                    PTS:   1

 

Exhibit 6-3

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

 

65. Refer to Exhibit 6-3. What is the random variable in this experiment?

a.	the weight of football players
b.	200 pounds
c.	25 pounds
d.	the normal distribution

 

 

ANS:  A                    PTS:   1

 

66. Refer to Exhibit 6-3. The probability of a player weighing more than 241.25 pounds is

a.	0.4505
b.	0.0495
c.	0.9505
d.	0.9010

 

 

ANS:  B                    PTS:   1

 

67. Refer to Exhibit 6-3. The probability of a player weighing less than 250 pounds is

a.	0.4772
b.	0.9772
c.	0.0528
d.	0.5000

 

 

ANS:  B                    PTS:   1

 

68. Refer to Exhibit 6-3. What percent of players weigh between 180 and 220 pounds?

a.	34.13%
b.	68.26%
c.	0.3413%
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

69. Refer to Exhibit 6-3. What is the minimum weight of the middle 95% of the players?

a.	196
b.	151
c.	249
d.	None of the alternative answers is correct.

 

 

ANS:  B                    PTS:   1

 

Exhibit 6-4

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.

 

70. Refer to Exhibit 6-4. What is the random variable in this experiment?

a.	the starting salaries
b.	the normal distribution
c.	$40,000
d.	$5,000

 

 

ANS:  A                    PTS:   1

 

71. Refer to Exhibit 6-4. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

a.	0.4772
b.	0.9772
c.	0.0228
d.	0.5000

 

 

ANS:  B                    PTS:   1

 

72. Refer to Exhibit 6-4. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500?

a.	0.4332
b.	0.9332
c.	0.0668
d.	0.5000

 

 

ANS:  C                    PTS:   1

 

73. Refer to Exhibit 6-4. What percentage of MBA’s will have starting salaries of $34,000 to $46,000?

a.	38.49%
b.	38.59%
c.	50%
d.	76.98%

 

 

ANS:  D                    PTS:   1

 

Exhibit 6-5

The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

 

74. Refer to Exhibit 6-5. What is the random variable in this experiment?

a.	the weight of items produced by a machine
b.	8 ounces
c.	2 ounces
d.	the normal distribution

 

 

ANS:  A                    PTS:   1

 

75. Refer to Exhibit 6-5. What is the probability that a randomly selected item will weigh more than 10 ounces?

a.	0.3413
b.	0.8413
c.	0.1587
d.	0.5000

 

 

ANS:  C                    PTS:   1

 

76. Refer to Exhibit 6-5. What is the probability that a randomly selected item will weigh between 11 and 12 ounces?

a.	0.4772
b.	0.4332
c.	0.9104
d.	0.0440

 

 

ANS:  D                    PTS:   1

 

77. Refer to Exhibit 6-5. What percentage of items will weigh at least 11.7 ounces?

a.	46.78%
b.	96.78%
c.	3.22%
d.	53.22%

 

 

ANS:  C                    PTS:   1

 

78. Refer to Exhibit 6-5. What percentage of items will weigh between 6.4 and 8.9 ounces?

a.	0.1145
b.	0.2881
c.	0.1736
d.	0.4617

 

 

ANS:  D                    PTS:   1

 

79. Refer to Exhibit 6-5. What is the probability that a randomly selected item weighs exactly 8 ounces?

a.	0.5
b.	1.0
c.	0.3413
d.	None of the alternative answers is correct.

 

 

ANS:  D                    PTS:   1

 

Exhibit 6-6

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.

 

80. Refer to Exhibit 6-6. What is the random variable in this experiment?

a.	the life expectancy of this brand of tire
b.	the normal distribution
c.	40,000 miles
d.	None of the alternative answers is correct.

 

 

ANS:  A                    PTS:   1

 

81. Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 30,000 miles?

a.	0.4772
b.	0.9772
c.	0.0228
d.	None of the alternative answers is correct.

 

 

ANS:  B                    PTS:   1

 

82. Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 47,500 miles?

a.	0.4332
b.	0.9332
c.	0.0668
d.	None of the alternative answers is correct.

 

 

ANS:  C                    PTS:   1

 

83. Refer to Exhibit 6-6. What percentage of tires will have a life of 34,000 to 46,000 miles?

a.	38.49%
b.	76.98%
c.	50%
d.	None of the alternative answers is correct.

 

 

ANS:  B                    PTS:   1

 

84. Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of exactly 47,500 miles?

a.	0.4332
b.	0.9332
c.	0.0668
d.	zero

 

 

ANS:  D                    PTS:   1

 

Exhibit 6-7

 

f(x) = (1/10) e-x/10

x ³ 0

 

 

85. Refer to Exhibit 6-7. The mean of x is

a.	0.10
b.	10
c.	100
d.	1,000

 

 

ANS:  B                    PTS:   1

 

86. Refer to Exhibit 6-7. The probability that x is less than 5 is

a.	0.6065
b.	0.0606
c.	0.3935
d.	0.9393

 

 

ANS:  C                    PTS:   1

 

87. Refer to Exhibit 6-7. The probability that x is between 3 and 6 is

a.	0.4512
b.	0.1920
c.	0.2592
d.	0.6065

 

 

ANS:  B                    PTS:   1

 

88. Excel’s NORM.S.DIST function can be used to compute

a.	cumulative probabilities for a standard normal z value
b.	the standard normal z value given a cumulative probability
c.	cumulative probabilities for a normally distributed x value
d.	the normally distributed x value given a cumulative probability

 

 

ANS:  A                    PTS:   1

 

89. Excel’s NORM.S.INV function can be used to compute

a.	cumulative probabilities for a standard normal z value
b.	the standard normal z value given a cumulative probability
c.	cumulative probabilities for a normally distributed x value
d.	the normally distributed x value given a cumulative probability

 

 

ANS:  B                    PTS:   1

 

90. Excel’s NORM.DIST function can be used to compute

a.	cumulative probabilities for a standard normal z value
b.	the standard normal z value given a cumulative probability
c.	cumulative probabilities for a normally distributed x value
d.	the normally distributed x value given a cumulative probability

 

 

ANS:  C                    PTS:   1

 

91. Excel’s NORM.INV function can be used to compute

a.	cumulative probabilities for a standard normal z value
b.	the standard normal z value given a cumulative probability
c.	cumulative probabilities for a normally distributed x value
d.	the normally distributed x value given a cumulative probability

 

 

ANS:  D                    PTS:   1

 

92. A continuous probability distribution that is useful in describing the time, or space, between occurrences of an event is a(n)

a.	normal probability distribution
b.	uniform probability distribution
c.	exponential probability distribution
d.	Poisson probability distribution

 

 

ANS:  C                    PTS:   1

 

93. The exponential probability distribution is used with

a.	a discrete random variable
b.	a continuous random variable
c.	any probability distribution with an exponential term
d.	an approximation of the binomial probability distribution

 

 

ANS:  B                    PTS:   1

 

94. An exponential probability distribution

a.	is a continuous distribution
b.	is a discrete distribution
c.	can be either continuous or discrete
d.	must be normally distributed

 

 

ANS:  A                    PTS:   1

 

95. Excel’s EXPON.DIST function can be used to compute

a.	exponents
b.	exponential probabilities
c.	cumulative exponential probabilities
d.	Both exponential probabilities and cumulative exponential probabilities are correct.

 

 

ANS:  D                    PTS:   1

 

96. Excel’s EXPON.DIST function has how many inputs?

a.	2
b.	3
c.	4
d.	5

 

 

ANS:  B                    PTS:   1

 

97. When using Excel’s EXPON.DIST function, one should choose TRUE for the third input if

a.	a probability is desired
b.	a cumulative probability is desired
c.	the expected value is desired
d.	the correct answer is desired

 

 

ANS:  B                    PTS:   1

 

PROBLEM

 

1. A random variable x is uniformly distributed between 45 and 150.

a.	Determine the probability of x = 48.
b.	What is the probability of x £ 60?
c.	What is the probability of x ³ 50?
d.	Determine the expected vale of x and its standard deviation.

 

 

ANS:

a.	0.000
b.	0.1429
c.	0.9524
d.	97.5, 30.31

 

 

PTS:   1

 

2. The price of a bond is uniformly distributed between $80 and $85.

a.	What is the probability that the bond price will be at least $83?
b.	What is the probability that the bond price will be between $81 and $90?
c.	Determine the expected price of the bond.
d.	Compute the standard deviation for the bond price.

 

 

ANS:

a.	0.4
b.	0.8
c.	$82.50
d.	$1.44

 

 

PTS:   1

 

3. The price of a stock is uniformly distributed between $30 and $40.

a.	What is the probability that the stock price will be more than $37?
b.	What is the probability that the stock price will be less than or equal to $32?
c.	What is the probability that the stock price will be between $34 and $38?
d.	Determine the expected price of the stock.
e.	Determine the standard deviation for the stock price.

 

 

ANS:

a.	0.3
b.	0.2
c.	0.4
d.	$35
e.	$2.89

 

 

PTS:   1

 

4. The time it takes to hand carve a guitar neck is uniformly distributed between 110 and 190 minutes.

a.	What is the probability that a guitar neck can be carved between 95 and 165 minutes?
b.	What is the probability that the guitar neck can be carved between 120 and 200 minutes?
c.	Determine the expected completion time for carving the guitar neck.
d.	Compute the standard deviation.

 

 

ANS:

a.	.6875
b.	.875
c.	150 minutes
d.	23.09 minutes

 

 

PTS:   1

 

5. The length of time it takes students to complete a statistics examination is uniformly distributed and varies between 40 and 60 minutes.

a.	Find the mathematical expression for the probability density function.
b.	Compute the probability that a student will take between 45 and 50 minutes to complete the examination.
c.	Compute the probability that a student will take no more than 40 minutes to complete the examination.
d.	What is the expected amount of time it takes a student to complete the examination?
e.	What is the variance for the amount of time it takes a student to complete the examination?

 

 

ANS:

a.	f(x) = 0.05 for 40 £ x £ 60; zero elsewhere
b.	0.25
c.	0.00
d.	50 minutes
e.	33.33 (minutes)2

 

 

PTS:   1

 

6. The advertised weight on a can of soup is 10 ounces. The actual weight in the cans follows a uniform distribution and varies between 9.3 and 10.3 ounces.

a.	Give the mathematical expression for the probability density function.
b.	What is the probability that a can of soup will have between 9.4 and 10.3 ounces?
c.	What is the mean weight of a can of soup?
d.	What is the standard deviation of the weight?

 

 

ANS:

a.	f(x) = 1.000 for 9.3 £ x £ 10.3; zero elsewhere
b.	0.90
c.	9.8 ounces
d.	0.289 ounce

 

 

PTS:   1

 

7. The length of time patients must wait to see a doctor in a local clinic is uniformly distributed between 15 minutes and 2 1/2 hours.

a.	Define the random variable in words.
b.	What is the probability of a patient waiting exactly 50 minutes?
c.	What is the probability that a patient would have to wait between 45 minutes and 2 hours?
d.	Compute the probability that a patient would have to wait over 2 hours.
e.	Determine the expected waiting time and its standard deviation.

 

 

ANS:

a.	the length of time patients must wait
b.	0.000
c.	0.556
d.	0.222
e.	82.5 minutes, 38.97 minutes

 

 

PTS:   1

 

8. For the standard normal distribution, determine the probability of obtaining a z value

a.	greater than zero.
b.	between -2.34 to -2.55
c.	less than 1.86.
d.	between -1.95 to 2.7.
e.	between 1.5 to 2.75.

 

 

ANS:

a.	0.5000
b.	0.0042
c.	0.9686
d.	0.9709
e.	0.0638

 

 

PTS:   1

 

9. Z is a standard normal random variable. Compute the following probabilities.

a.	P(-1.33 £ z £ 1.67)
b.	P(1.23 £ z £ 1.55)
c.	P(z ³ 2.32)
d.	P(z ³ -2.08)
e.	P(z ³ -1.08)

 

 

ANS:

a.	0.8607
b.	0.0487
c.	0.0102
d.	0.9812
e.	0.1401

 

 

PTS:   1

 

10. Z is a standard normal random variable. Compute the following probabilities.

a.	P(-1.23 £ z £ 2.58)
b.	P(1.83 £ z £ 1.96)
c.	P(z ³ 1.32)
d.	P(z £ 2.52)
e.	P(z ³ -1.63)
f.	P(z £ -1.38)
g.	P(-2.37 £ z £ -1.54)
h.	P(z = 2.56)

 

 

ANS:

a.	0.8858
b.	0.0086
c.	0.0934
d.	0.9941
e.	0.9484
f.	0.0838
g.	0.0529
h.	0.0000

 

 

PTS:   1

 

11. Z is a standard normal variable. Find the value of z in the following.

a.	The area between 0 and z is 0.4678.
b.	The area to the right of z is 0.1112.
c.	The area to the left of z is 0.8554
d.	The area between –z and z is 0.754.
e.	The area to the left of –z is 0.0681.
f.	The area to the right of –z is 0.9803.

 

 

ANS:

a.	1.85
b.	1.22
c.	1.06
d.	1.16
e.	1.49
f.	2.06

 

 

PTS:   1

 

12. The miles-per-gallon obtained by the 1995 model Q cars is normally distributed with a mean of 22 miles-per-gallon and a standard deviation of 5 miles-per-gallon.

a.	What is the probability that a car will get between 13.35 and 35.1 miles-per-gallon?
b.	What is the probability that a car will get more than 29.6 miles-per-gallon?
c.	What is the probability that a car will get less than 21 miles-per-gallon?
d.	What is the probability that a car will get exactly 22 miles-per-gallon?

 

 

ANS:

a.	0.9538
b.	0.0643
c.	0.4207
d.	0.0000

 

 

PTS:   1

 

13. The salaries at a corporation are normally distributed with an average salary of $19,000 and a standard deviation of $4,000.

a.	What is the probability that an employee will have a salary between $12,520 and $13,480?
b.	What is the probability that an employee will have a salary more than $11,880?
c.	What is the probability that an employee will have a salary less than $28,440?

 

 

ANS:

a.	0.0312
b.	0.9625
c.	0.9909

 

 

PTS:   1

 

14. A major department store has determined that its customers charge an average of $500 per month, with a standard deviation of $80. Assume the amounts of charges are normally distributed.

a.	What percentage of customers charges more than $380 per month?
b.	What percentage of customers charges less than $340 per month?
c.	What percentage of customers charges between $644 and $700 per month?

 

 

ANS:

a.	93.32%
b.	2.28%
c.	2.97%

 

 

PTS:   1

 

15. The contents of soft drink bottles are normally distributed with a mean of twelve ounces and a standard deviation of one ounce.

a.	What is the probability that a randomly selected bottle will contain more than ten ounces of soft drink?
b.	What is the probability that a randomly selected bottle will contain between 9.5 and 11 ounces?
c.	What percentage of the bottles will contain less than 10.5 ounces of soft drink?

 

 

ANS:

a.	0.9772
b.	0.1525
c.	6.68%

 

 

PTS:   1

 

16. The life expectancy of computer terminals is normally distributed with a mean of 4 years and a standard deviation of 10 months.

a.	What is the probability that a randomly selected terminal will last more than 5 years?
b.	What percentage of terminals will last between 5 and 6 years?
c.	What percentage of terminals will last less than 4 years?
d.	What percentage of terminals will last between 2.5 and 4.5 years?
e.	If the manufacturer guarantees the terminals for 3 years (and will replace them if they malfunction), what percentage of terminals will be replaced?

 

 

ANS:

a.	0.1151
b.	10.69%
c.	50%
d.	68.98%
e.	11.51%

 

 

PTS:   1

 

17. Scores on a recent national statistics exam were normally distributed with a mean of 80 and a standard deviation of 6.

a.	What is the probability that a randomly selected exam will have a score of at least 71?
b.	What percentage of exams will have scores between 89 and 92?
c.	If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award?
d.	If there were 334 exams with scores of at least 89, how many students took the exam?

 

 

ANS:

a.	.9332
b.	.04
c.	91.76
d.	5000

 

 

PTS:   1

 

18. The average starting salary for this year’s graduates at a large university (LU) is $30,000 with a standard deviation of $8,000. Furthermore, it is known that the starting salaries are normally distributed.

a.	What is the probability that a randomly selected LU graduate will have a starting salary of at least $30,400?
b.	Individuals with starting salaries of less than $15,600 receive a low income tax break. What percentage of the graduates will receive the tax break?
c.	What are the minimum and the maximum starting salaries of the middle 95% of the LU graduates?
d.	If 303 of the recent graduates have salaries of at least $43,120, how many students graduated this year from this university?

 

 

ANS:

a.	0.4801
b.	3.59%
c.	minimum = $14320, maximum = $45,680
d.	6000

 

 

PTS:   1

 

19. The weights of items produced by a company are normally distributed with a mean of 4.5 ounces and a standard deviation of 0.3 ounces.

a.	What is the probability that a randomly selected item from the production will weigh at least 4.14 ounces?
b.	What percentage of the items weighs between 4.8 and 5.04 ounces?
c.	Determine the minimum weight of the heaviest 5% of all items produced.
d.	If 27,875 of the items of the entire production weigh at least 5.01 ounces, how many items have been produced?

 

 

ANS:

a.	0.8849
b.	12.28%
c.	4.99
d.	625,000

 

 

PTS:   1

 

20. The life expectancy of Timely brand watches is normally distributed with a mean of four years and a standard deviation of eight months.

a.	What is the probability that a randomly selected watch will be in working condition for more than five years?
b.	The company has a three-year warranty period on their watches. What percentage of their watches will be in operating condition after the warranty period?
c.	What is the minimum and the maximum life expectancy of the middle 95% of the watches?
d.	Ninety-five percent of the watches will have a life expectancy of at least how many months?

 

 

ANS:

a.	0.0668
b.	93.32%
c.	Min = 32.32 months     Max = 63.68 months
d.	34.84 months

 

 

PTS:   1

 

21. The weights of the contents of cans of tomato paste produced by a company are normally distributed with a mean of 6 ounces and a standard deviation of 0.3 ounces.

a.	What percentage of all cans produced contains more than 6.51 ounces of tomato paste?
b.	What percentage of all cans produced contains less than 5.415 ounces?
c.	What percentage of cans contains between 5.46 and 6.495 ounces?
d.	Ninety-five percent of cans will contain at least how many ounces?
e.	What percentage of cans contains between 6.3 and 6.6 ounces?

 

 

ANS:

a.	4.46%
b.	2.56%
c.	91.46%
d.	5.5067 oz
e.	13.59%

 

 

PTS:   1

 

22. A professor at a local university noted that the grades of her students were normally distributed with a mean of 78 and a standard deviation of 10.

a.	The professor has informed us that 16.6% of her students received grades of A. What is the minimum score needed to receive a grade of A?
b.	If 12.1% of her students failed the course and received Fs, what was the maximum score among those who received an F?
c.	If 33% of the students received grades of B or better (i.e., As and Bs), what is the minimum score of those who received a B?

 

 

ANS:

a.	87.7
b.	66.3
c.	82.4

 

 

PTS:   1

 

23. “DRUGS R US” is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a standard deviation of 0.3 ounces. Assume the contents of the bottles are normally distributed.

a.	What percentage of all bottles produced contains more than 6.51 ounces of vitamins?
b.	What percentage of all bottles produced contains less than 5.415 ounces?
c.	What percentage of bottles produced contains between 5.46 and 6.495 ounces?
d.	Ninety-five percent of the bottles will contain at least how many ounces?
e.	What percentage of the bottles contains between 6.3 and 6.6 ounces?

 

 

ANS:

a.	4.46%
b.	2.56%
c.	91.46%
d.	5.508 ounces
e.	13.59%

 

 

PTS:   1

 

24. The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6.

a.	Define the random variable in words.
b.	What is the probability that a randomly selected bill will be at least $39.10?
c.	What percentage of the bills will be less than $16.90?
d.	What are the minimum and maximum of the middle 95% of the bills?
e.	If twelve of one day’s bills had a value of at least $43.06, how many bills did the restaurant collect on that day?

 

 

ANS:

a.	the daily dinner bills
b.	0.0322
c.	3.22%
d.	minimum = $16.24     maximum = $39.76
e.	2,000

 

 

PTS:   1

 

25. The monthly income of residents of Daisy City is normally distributed with a mean of $3000 and a standard deviation of $500.

a.	Define the random variable in words.
b.	The mayor of Daisy City makes $2,250 a month. What percentage of Daisy City’s residents has incomes that are more than the mayor’s?
c.	Individuals with incomes of less than $1,985 per month are exempt from city taxes. What percentage of residents is exempt from city taxes?
d.	What are the minimum and the maximum incomes of the middle 95% of the residents?
e.	Two hundred residents have incomes of at least $4,440 per month. What is the population of Daisy City?

 

 

ANS:

a.	the monthly income of residents of Daisy City
b.	93.32%
c.	2.12%
d.	Min = 2020     Max = 3980
e.	100,000

 

 

PTS:   1

 

26. The average starting salary of this year’s MBA students is $35,000 with a standard deviation of $5,000. Furthermore, it is known that the starting salaries are normally distributed. What are the minimum and the maximum starting salaries of the middle 95% of MBA graduates?

 

ANS:

Min. = $25,200; Max. = $44,800

 

PTS:   1

 

27. A local bank has determined that the daily balances of the checking accounts of its customers are normally distributed with an average of $280 and a standard deviation of $20.

a.	What percentage of its customers has daily balances of more than $275?
b.	What percentage of its customers has daily balances less than $243?
c.	What percentage of its customers’ balances is between $241 and $301.60?

 

 

ANS:

a.	59.87%
b.	3.22%
c.	83.43%

 

 

PTS:   1

 

28. The weekly earnings of bus drivers are normally distributed with a mean of $395. If only 1.1% of the bus drivers have a weekly income of more than $429.35, what is the value of the standard deviation of the weekly earnings of the bus drivers?

 

ANS:

Standard Deviation = 15

 

PTS:   1

 

29. The monthly earnings of computer programmers are normally distributed with a mean of $4,000. If only 1.7 percent of programmers have monthly incomes of less than $2,834, what is the value of the standard deviation of the monthly earnings of the computer programmers?

 

ANS:

$550

 

PTS:   1

 

30. The Globe Fishery packs shrimp that weigh more than 1.91 ounces each in packages marked” large” and shrimp that weigh less than 0.47 ounces each into packages marked “small”; the remainder are packed in “medium” size packages. If a day’s catch showed that 19.77% of the shrimp were large and 6.06% were small, determine the mean and the standard deviation for the shrimp weights. Assume that the shrimps’ weights are normally distributed.

 

ANS:

Mean = 1.4; Standard Deviation = 0.6

 

PTS:   1

 

31. In grading eggs into small, medium, and large, the Linda Farms packs the eggs that weigh more than 3.6 ounces in packages marked “large” and the eggs that weigh less than 2.4 ounces into packages marked “small”; the remainder are packed in packages marked “medium.” If a day’s packaging contained 10.2% large and 4.18% small eggs, determine the mean and the standard deviation for the eggs’ weights. Assume that the distribution of the weights is normal.

 

ANS:

Mean = 3.092; Standard Deviation = 0.4

 

PTS:   1

 

32. A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produced weigh between 100 grams up to the mean and 49.18% weigh from the mean up to 190 grams. Determine the mean and the standard deviation.

 

ANS:

Mean = 113; Standard Deviation = 30

 

PTS:   1

 

33. Z is the standard normal random variable. Use Excel to calculate the following:

a.	P(z £ 2.5)
b.	P(0 £ z £ 2.5)
c.	P(-2 £ z £ 2)
d.	P(z £ -0.38)
e.	P(z ³ 1.62)
f.	z value with .05 in the lower tail
g.	z value with .05 in the upper tail

 

 

ANS:

a.	P(z £ 2.5)	=NORM.S.DIST(2.5,TRUE)
b.	P(0 £ z £ 2.5)	=NORM.S.DIST(2.5,TRUE)-NORM.S.DIST(0,TRUE)
c.	P(-2 £ z £ 2)	=NORM.S.DIST(2,TRUE)-NORM.S.DIST(-2,TRUE)
d.	P(z £ -0.38)	=NORM.S.DIST(-0.38,TRUE)
e.	P(z ³ 1.62)	=1-NORM.S.DIST(1.62,TRUE)
f.	z value with .05 in the lower tail		=NORM.S.INV(0.05)
g.	z value with .05 in the upper tail		=NORM.S.INV(0.95)

 

 

PTS:   1

 

34. X is a normally distributed random variable with a mean of 50 and a standard deviation of 5. Use Excel to calculate the following:

a.	P(x £ 45)
b.	P(45 £ x £ 55)
c.	P(x ³ 55)
d.	x value with .20 in the lower tail
e.	x value with .01 in the upper tail

 

 

ANS:

a.	P(x £ 45)	=NORM.DIST(45,50,5,TRUE)
b.	P(45 £ x £ 55)	=NORM.DIST(55,50,5,TRUE)-NORM.DIST(45,50,5,TRUE)
c.	P(x ³ 55)	=1-NORM.DIST(55,50,5,TRUE)
d.	x value with .20 in the lower tail		=NORM.INV(0.2,50,5)
e.	x value with .01 in the upper tail		=NORM.INV(0.99,50,5)

 

 

PTS:   1

 

35. The time it takes a mechanic to change the oil in a car is exponentially distributed with a mean of 5 minutes.

a.	What is the probability density function for the time it takes to change the oil?
b.	What is the probability that it will take a mechanic less than 6 minutes to change the oil?
c.	What is the probability that it will take a mechanic between 3 and 5 minutes to change the oil?

 

 

ANS:

a.	f(x) = (1/5) e–x/5 for x ³ 0
b.	0.6988
c.	0.1809

 

 

PTS:   1

 

36. The time it takes a worker on an assembly line to complete a task is exponentially distributed with a mean of 8 minutes.

a.	What is the probability density function for the time it takes to complete the task?
b.	What is the probability that it will take a worker less than 4 minutes to complete the task?
c.	What is the probability that it will take a worker between 6 and 10 minutes to complete the task?

 

 

ANS:

a.	f(x) = (1/8 ) e–x/8 for x ³ 0
b.	0.3935
c.	0.1859

 

 

PTS:   1

 

37. The time between arrivals of customers at the drive-up window of a bank follows an exponential probability distribution with a mean of 10 minutes.

a.	What is the probability that the arrival time between customers will be 7 minutes or less?
b.	What is the probability that the arrival time between customers will be between 3 and 7 minutes?

 

 

ANS:

a.	0.5034
b.	0.2442

 

 

PTS:   1

 

38. The time required to assemble a part of a machine follows an exponential probability distribution with a mean of 14 minutes.

a.	What is the probability that the part can be assembled in 7 minutes or less?
b.	What is the probability that the part can be assembled between 3.5 and 7 minutes?

 

 

ANS:

a.	0.3935
b.	0.1723

 

 

PTS:   1

 

39. The time it takes to completely tune an engine of an automobile follows an exponential distribution with a mean of 40 minutes.

a.	Define the random variable in words.
b.	What is the probability of tuning an engine in 30 minutes or less?
c.	What is the probability of tuning an engine between 30 and 35 minutes?

 

 

ANS:

a.	the time it takes to completely tune an engine
b.	0.5276
c.	0.0555

 

 

PTS:   1

 

40. X is a exponentially distributed random variable with a mean of 10. Use Excel to calculate the following:

a.	P(x £ 15)
b.	P(8 £ x £ 12)
c.	P(x ³ 8)

 

 

ANS:

a.	P(x £ 15)	=EXPON.DIST(15,1/10,TRUE)
b.	P(8 £ x £ 12)	=EXPON.DIST(12,1/10,TRUE)-EXPON.DIST(8,1/10,TRUE)
c.	P(x ³ 8)	=1-EXPON.DIST(8,1/10,TRUE)

 

 

PTS:   1

 

41. The Harbour Island Ferry leaves on the hour and at 15-minute intervals.  The time, x, it takes John to drive from his house to the ferry has a uniform distribution with x between 10 and 20 minutes.  One morning John leaves his house at precisely 8:00a.m.
42. What is the probability John will wait less than 5 minutes for the ferry?
43. What is the probability John will wait less than 10 minutes for the ferry?
44. What is the probability John will wait less than 15 minutes for the ferry?
45. What is the probability John will not have to wait for the ferry?
46. Suppose John leaves at 8:05a.m. What is the probability John will wait (1) less than 5 minutes for the ferry; (2) less than 10 minutes for the ferry?
47. Suppose John leaves at 8:10a.m. What is the probability John will wait (1) less than 5 minutes for the ferry; (2) less than 10 minutes for the ferry?
48. What appears to be the best time for John to leave home if he wishes to maximize the probability of waiting less than 10 minutes for the ferry?

 

ANS:

1. 0.5
2. 0.5
3. 1
4. 0
5. (1) 0, (2) .5
6. (1) .5, (2) 1
7. 8:10

 

PTS:   1

 

42. Delicious Candy markets a two-pound box of assorted chocolates.  Because of imperfections in the candy making equipment, the actual weight of the chocolate has a  uniform distribution ranging from 31.8 to 32.6 ounces.
43. Define a probability density function for the weight of the box of chocolate.
44. What is the probability that a box weighs (1) exactly 32 ounces; (2) more than 32.3 ounces; (3) less than 31.8 ounces?
45. The government requires that at least 60% of all products sold weigh at least as much as the stated weight. Is Delicious violating government regulations?

 

ANS:

1. f(x) = 1.25 for 31.8 < x < 32.6, and 0 otherwise
2. (1) 0, (2) .375, (3) 0
3. no; 75% are 32 oz. or more

 

PTS:   1

 

43. The time at which the mailman delivers the mail to Ace Bike Shop follows a normal distribution with mean 2:00 PM and standard deviation of 15 minutes.
44. What is the probability the mail will arrive after 2:30 PM?
45. What is the probability the mail will arrive before 1:36 PM?
46. What is the probability the mail will arrive between 1:48 PM and 2:09 PM?

 

ANS:

1. .0228
2. .0548
3. .5138

 

PTS:   1

 

44. The township of Middleton sets the speed limit on its roads by conducting a traffic study and determining the speed (to the nearest 5 miles per hour) at which 80% of the drivers travel at or below.  A study was done on Brown’s Dock Road that indicated driver’s speeds follow a normal distribution with a mean of 36.25 miles per hour and a variance of 6.25.
45. What should the speed limit be?
46. What percent of the drivers travel below that speed?

 

ANS:

1. 40 miles per hour
2. 93.32%

 

PTS:   1

 

45. A light bulb manufacturer claims its light bulbs will last 500 hours on the average.  The lifetime of a light bulb is assumed to follow an exponential distribution.
46. What is the probability that the light bulb will have to be replaced within 500 hours?
47. What is the probability that the light bulb will last more than 1000 hours?
48. What is the probability that the light bulb will last between 200 and 800 hours.

 

ANS:

1. .632
2. .135
3. .468

 

PTS:   1