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					                                                                             Chapter 4
                                                              Discrete Random Variables

True/False


1. The binomial experiment consists of n independent, identical trials, each of which results in
either success or failure and is such that the probability of success on any trial is the same.
Answer: True Difficulty: Medium


2. A Poisson random variable is a continuous variable that can be used to describe the number
of occurrences of an event over a specified interval of time or space.
Answer: False Difficulty: Medium


3. The hypergeometric distribution is appropriate when sampling with replacement from small
populations
Answer: False Difficulty: Medium


4. A discrete random variable may assume a countable number of outcome values.
Answer: True Difficulty: Easy


5. The variable “home ownership” can take on one of two values, one if the person living in a
home owns the home and zero if the person living in a home does not own the home is an
example of a discrete random variable.
Answer: True Difficulty: Easy


6. If the number of surface nonconformities on a specific size of a metal piece is the discrete
random variable in question, then the appropriate probability distribution that can describe the
probability of a specific size metal sheet containing 3 nonconformities is most likely given by the
binomial distribution.
Answer: False Difficulty: Medium


7. The mean of the binomial distribution is np(1-p)
Answer: False Difficulty: Easy


8. In a binomial experiment, the results of one trial are dependent on the results of other trials.
Answer: False Difficulty: Easy



114                                                         Bowerman, Essentials of Business Statistics, 2/e
9. In a binomial distribution the random variable X is continuous.
Answer: False Difficulty: Easy


Multiple Choice


10. When using hypergeometric distribution, it is assumed that the trials are __________ and the
probability of success ________ from trial to trial.
A) Independent, does not change
B) Dependent, does not change
C) Independent, changes
D) Dependent, changes
Answer: D Difficulty: Medium


11. If p = .1 and n = 5, then the corresponding binomial distribution is
A) Right skewed
B) Left skewed
C) Symmetric
D) Bimodal
Answer: A Difficulty: Medium


12. If p = .5 and n = 4, then the corresponding binomial distribution is
A) Right skewed
B) Left skewed
C) Symmetric
D) Bimodal
Answer: C Difficulty: Medium


13. The requirement that the probability of success remains constant from trial to trial is a
property of the _________________ distribution.
A) Binomial
B) Uniform
C) Normal
D) Poisson
E) Hypergeometric
Answer: A Difficulty: Medium (REF)




Bowerman, Essentials of Business Statistics, 2/e                                                115
14. If the number of surface nonconformities on a specific size of metal piece is the discrete
random variable in question, then the appropriate probability distribution that can describe the
probability of a specific size metal sheet containing 3 defectives is given most likely by
___________________ distribution
A) Binomial
B) Poisson
C) Hypergeometric
D) Both A and B
Answer: B Difficulty: Medium (REF)


15. Which of the following distributions can be used to solve the following problem?
The average number of cars arriving at a drive-thru fast food restaurant is 3 in ten minutes, what
is the probability that exactly 4 cars will arrive in a five minute interval?
A) Binomial
B) Poisson
C) Both of the above
D) None of the above
Answer: B Difficulty: Medium (REF)


16. The mean of the binomial distribution is equal to:
A) p
B) np
C) px(1-p)n-x
D) (n)(p)(1-p)
E) n. p(1  p)
Answer: B Difficulty: Medium


17. A company's annual profit, y, is described by y = -1,500,000 + 60x where x is the number of
units sold. If the expected number of units sold is 100,000, and X = 12,000, what are and Y?
A) 6,000,000 and 12,000
B) 6,000,000 and 720,000
C) 4,500,000 and 518,400,000,000
D) 4,500,000 and 720,000
Answer: D Difficulty: Medium




116                                                       Bowerman, Essentials of Business Statistics, 2/e
18. The number of ways to arrange x successes among n trials is equal to
        n!
A) x !(n  x)!
         n!
B)    (n  x)!
      n
C)        x
   n!
D)    x!
Answer: A Difficulty: Medium


19. Which of the following is a valid probability value for a discrete random variable?
A) .2
B) 1.01
C) -.7
D) All of the above
Answer: A Difficulty: Medium


20. A total of 50 raffle tickets are sold for a contest to win a car. If you purchase one ticket, what
are your odds against winning?
A) 49 to 1
B) 50 to 1
C) .05
D) .01
Answer: A Difficulty: Medium


21. Which one of the following statements is not an assumption of the binomial distribution?
A) Sampling is with replacement.
B) The experiment consists of n identical trials.
C) The probability of success remains constant from trial to trial.
D) Trials are independent of each other.
E) Each trial results in one of two mutually exclusive outcomes.
Answer: A Difficulty: Medium (REF)


22. The binomial distribution is characterized by situations that are analogous to
A) Drawing balls from and urn
B) Coin tossing
C) Counting defects on an item
D) Measuring the length of an item.
Answer: B Difficulty: Medium




Bowerman, Essentials of Business Statistics, 2/e                                                  117
23. A random variable is said to be discrete if
A) Its outcomes are countable
B) It can assume any real number within a given interval
C) The rules of probability apply
D) It can be represented graphically
Answer: A Difficulty: Easy


24. Two characteristics/assumptions of the Poisson distribution include:
A) Probability of success remains constant from trial to trial and the random variable of interest
is continuous.
B) The event occurring in one interval is independent of the event occurring in any other non-
overlapping interval, and the random variable of interest is continuous.
C) The event occurring in one interval is independent of the event occurring in any other non-
overlapping interval, and the random variable of interest is discrete.
D) The event occurring in one interval is dependent of the event occurring in any other non-
overlapping interval, and the random variable of interest is continuous.
Answer: C Difficulty: Medium (REF)


Use the following information to answer questions 25-27:
The Securities and Exchange Commission has determined that the number of companies listed in
NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per
month.


25. Find the probability that exactly 4 bankruptcies occur next month.
A) .8774
B) .1414
C) .1557
D) .2176
Answer: B Difficulty: Medium


26. Find the probability that more than 1 bankruptcy occur next month.
A) .1931
B) .9257
C) .7326
D) .4816
E) .2674
Answer: C Difficulty: Medium




118                                                        Bowerman, Essentials of Business Statistics, 2/e
27. Find the probability that no more than one bankruptcy occurs next month.
A) .1931
B) .9257
C) .7326
D) .4816
E) .2674
Answer: E Difficulty: Medium


28. A fair die is rolled 10 times. What is the probability that an odd number (1, 3, or 5) will
occur less than 3 times?
A) .0547
B) .1172
C) .1550
D) .7752
E) .8450
Answer: A Difficulty: Hard


29. A fair die is rolled 10 times. What is the probability that an even number (2, 4, or 6) will
occur between 2 and 4 times?
A) .6123
B) .1709
C) .1611
D) .3662
E) .3223
Answer: D Difficulty: Hard


30. A fair die is rolled 10 times. What is the average number of even number (2, 4, or 6)
outcomes?
A) 3
B) 4
C) 5
D) 6
E) 7
Answer: C Difficulty: Easy




Bowerman, Essentials of Business Statistics, 2/e                                                   119
31. A fair die is rolled 36 times. What is the standard deviation of the even number (2, 4, or 6)
outcomes?
A) 18
B) 9
C) 5
D) 3
E) 1.732
Answer: D Difficulty: Medium


32. Whenever p = .5, the binomial distribution will _________ be symmetric.
A) Always
B) Sometimes
C) Never
Answer: A Difficulty: Medium


33. Which of the following statements about the binomial distribution is not correct?
A) Each trial results in a success or failure.
B) Trials are independent of each other.
C) The probability of success remains constant from trial to trial.
D) The random variable of interest is continuous.
E) The experiment consists of n identical trials.
Answer: D Difficulty: Easy


34. If n = 20 and p = .4, then the mean of the binomial distribution is
A) .4
B) 4.8
C) 8
D) 12
Answer: C Difficulty: Easy


35. If n = 15 and p = .4, then the standard deviation of the binomial distribution is
A) 9
B) 6
C) 3.6
D) 1.8974
E) .4
Answer: D Difficulty: Easy




120                                                        Bowerman, Essentials of Business Statistics, 2/e
36. The equation for the variance of the binomial distribution is given by:
A) px(1 – p)n-x
B) np
C) np(1-p)
      np(1  p)
D)
Answer: C Difficulty: Easy


Use the following information to answer questions 37-38:
The manager of the local grocery store has determined that, on the average, 4 customers use the
service desk every half-hour. Assume that the number of customers using the service desk has a
Poisson distribution.


37. What is the probability that during a randomly selected half-hour period exactly 2 customers
use the service desk?
A) .0183
B) .0733
C) .1465
D) .9084
E) .7619
Answer: C Difficulty: Medium


38. What is the probability that during a randomly selected half-hour period no more than 2
customers use the service desk?
A) .2381
B) .1465
C) .7619
D) .8535
E) .0916
Answer: A Difficulty: Medium


Fill-in-the-Blank


39. The variable "employment status" which can take on one of two values: 1 for "employed"
and 0 for "unemployed" is and example of a(n) _____________ random variable.
Answer: Discrete Difficulty: Medium


40. If x is a binomial random variable, then the standard deviation of x is given by
Answer: square root (npq) Difficulty: Medium




Bowerman, Essentials of Business Statistics, 2/e                                              121
41. A random variable that is defined to be the total number of successes in n trials is a(n) _____
random variable.
Answer: Binomial Difficulty: Medium


42. A discrete variable that can often be used to describe the number of occurrences of an event
over a specified interval of time or space is a(n) _____ random variable.
Answer: Poisson Difficulty: Medium


43. The requirement that the probability of success remains constant from trial to trial is a
property of the _______________ distribution.
Answer: Binomial Difficulty: Medium


44. The hypergeometric distribution is used when sampling _____ replacement from small
populations
Answer: Without Difficulty: Medium


45. The distribution whose mean is equal to its variance is the _________ distribution.
Answer: Poisson Difficulty: Easy


46. For a random variable X, the mean value of the squared deviations of its values from their
expected value is called its
Answer: Variance Difficulty: Hard


Essay


Use the following information to answer questions 47-50:
The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a
defective calculator is 10%.


47. If 3 calculators are selected at random, what is the probability that one of the calculators will
be defective?
Answer: .243

    3!
           (.1)1 (.9)2  .243
1!(3  1)!
Difficulty: Medium




122                                                        Bowerman, Essentials of Business Statistics, 2/e
48. If 10 calculators are selected at random, what is the probability that 3 or more of the
calculators will be defective?
Answer: .0702
P(X  3) = 1 – P(X  2) = 1 - .9298 = .0702 Difficulty: Medium


49. If 100 calculators are selected at random, what is the expected number of defectives?
Answer: 10
E[X] =  x = (.10)(100) = 10 Difficulty: Medium


50. If 100 calculators are selected at random, what is the standard deviation of the number of
defectives?
Answer: 3
 X  (.1)(.9)(100)  3
Difficulty: Medium


Use the following information to answer questions 51-52:
Historical data shows that the average number of patient arrivals at the intensive care unit of
General Hospital is 3 patients every 2 hours. Assume that the patient arrivals are distributed
according to a Poisson distribution.


51. Determine the probability of 6 patients arriving in a five-hour period.
Answer: .136
             (e7.5 )(7.5)6
P( X  6)                   .1359
                    6!
Difficulty: Medium


52. Determine the probability of at least 4 but no more than 8 patients arriving in a three-hour
period.
Answer: .6174
P(4  X  8) = P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4)
P(4  X  8) = (.0463) + (.0824) + (.1281) + (.1708) + (.1898) = .6174
Difficulty: Medium




Bowerman, Essentials of Business Statistics, 2/e                                                   123
Use the following information to answer questions 53-55:
If the probability distribution of X is:

      X        P(X)
      3         1/8
      4         1/8
      5         3/8
      6         3/8


53. What is the expected value of X?
Answer: 5.0
                           
E  X   (3) 1  (4) 1  5 3  6 3  40  5.0
               8        8       8    8  8
Difficulty: Medium


54. What is the variance of X?
Answer: 1.0
                           
E  X   (3) 1  (4) 1  5 3  6 3  40  5.0
               8        8      8   8    8

              1             1              3            3
 X  (3  5)2    (4  5) 2    (5  5) 2    (6  5) 2    1
  2

               
                8              
                                8              
                                                8             8

Difficulty: Medium


55. Assume the number of trucks passing an intersection has a Poisson distribution with mean
of 5 trucks per minute. What is the probability of 0 or 1 trucks in one minute?
Answer: .0404

e5 50 e5 51
              .0404
  0!       1!
Difficulty: Medium


Use the following information to answer questions 56-57:
A vaccine is 95 percent effective. What is the probability that it is not effective for:


56. One and only one individual out of 20 individuals?
Answer: .3774
(20) .95  .05  = .3774
           19     1


Difficulty: Medium



124                                                             Bowerman, Essentials of Business Statistics, 2/e
57. More than one out of 20 individuals?
Answer: .2641
P(X  2) = 1 – [P(X = 0) + p(X =1)]
P(X  2) = 1 – [(.3585) + (.3774)] = .2641
Difficulty: Medium


58. If the probability of a success on a single trial is .2, what is the probability of obtaining 3
successes in 10 trials if the number of successes is binomial?
Answer: .2013

    10!
            (.2)3 (.8)7  .2013
3!(10  3)!
Difficulty: Medium


59. The number of calls coming into a PBX follows a Poisson process with a mean of 120 calls
per hour. What is the probability of no calls in a one-minute interval?
Answer: .1353
            e2 20
P( X  0)          e2  .1353
              0!
Difficulty: Medium (AS)


60. If x is a Poisson random variable with a mean of 10, what is the probability that x is greater
than 6?
Answer: .8698
P(X  7) = 1 – P(X  6) = 1 – (.1302) = .8698
Difficulty: Medium


61. An archer hits a target 95% of the time. What is the probability he/she will miss the target
for the first time on the 15th shot?
Answer: .0244
.95 14 .05 1 =.0244
Difficulty: Easy




Bowerman, Essentials of Business Statistics, 2/e                                                      125
62. Three candidates run for different offices in different counties. Each has a one in three
chance of being elected in his/her county. What is the probability that at least one of them will
be elected?
Answer: .7037

P( X  1)  1  P( X  0)
                      0     3
              3!  1   2     8
P( X  0)       
              3!  3   3  27
                        19
P( X  1)  1  8           .7037
                  27 27
Difficulty: Hard


63. A test has 6 multiple choice questions, each with 4 alternatives. What is the probability of
guessing 5 or more questions correctly?
Answer: .0046
P(X  5) = P(X = 5) + P(X = 6)
P(X  5) = (.0044) + (.0002) = .0046
Difficulty: Hard


64. If x is a Poisson random variable with a mean of 10, what is the probability that x is greater
than or equal to 2?
Answer: .9995
P(X  2) = 1 – [P(X = 0) + P(X = 1)]
P(X  2) = 1 – (0 + .0005) = .9995
Difficulty: Medium (AS)


65. If x is a Poisson random variable with a mean of 10, what is the probability that x is equal to
8?
Answer: .1126
             e10 108
P( X  8)             .1126
                8!
Difficulty: Medium


66. Twenty coins are tossed. What is the probability of getting exactly 10 heads?
Answer: .1762
                  20!
P( X  10)                (.5)10 (.5)10  .1762
             10!(20  10)!
Difficulty: Medium




126                                                       Bowerman, Essentials of Business Statistics, 2/e
67. Determine the probability that a 3 will appear twice, if a fair die is rolled 10 times.
Answer: .2907
              10!   1   5 
                           2      8

P ( X  2)              .2907
              2!8!   6   6 
Difficulty: Hard


Use the following information to answer questions 68-69:
During off hours, cars arrive at a tollbooth on the East-West toll road at an average rate of 0.5
cars per minute. The arrivals are distributed according to the Poisson distribution.


68. What is the probability that during the next minute three cars will arrive?
Answer: .0126
            e.5 .53 (.6065)(.125)
P( X  3)                          .0126
               3!          6
Difficulty: Medium


69. What is the probability that during the next five minutes three cars will arrive?
Answer: .2138
            e2.5 2.53 (.0821)(15.625)
P( X  3)                               .2138
                3!             6
Difficulty: Medium


Use the following information to answer questions 70-75:
For a binomial process, the probability of success is 40% and the number of trials is 5.


70. Find the expected value.
Answer: 2
E[X] = (5)(.40) = 2
Difficulty: Medium


71. Find the variance.
Answer: 1.2
 2 x = (5)(.4)(.6) = 1.2
Difficulty: Medium




Bowerman, Essentials of Business Statistics, 2/e                                                    127
72. Find the standard deviation.
Answer: 1.0954

 X  (5)(.4)(.6)  1.2  1.0954
Difficulty: Medium


73. Find P(X  1).
Answer: .3370
P(X  1) = [P(X = 0) + P(X = 1)
P(X  1) = (.0778) + (.2592) = .337
Difficulty: Medium


74. Find P(X > 4).
Answer: .0102
P(X = 5) = (.4)5 = .0102
Difficulty: Medium


75. Find the P(X = 2).
Answer: .3456
              5! 
P ( X  2)         (.4) (.6)  .3456
                          2    3

              2!3! 
Difficulty: Medium


76. Consider a Poisson distribution with an average of 3 customers per minute at the local
grocery store. Determine the expected number of customer arrivals for a five-minute period
Answer: 15
 = (3)(5) = 15 Difficulty: Easy (AS)


77. Consider a Poisson distribution with an average of 3 customers per minute at the local
grocery store. Write a formula for the probability of X, where X = the number of arrivals per
minute.
Answer:
         e3 3X
P( X ) 
           X!   Difficulty: Medium


Use the following information to answer questions 78-83:
Consider a Poisson distribution with an average of 3 customers per minute at the local grocery
store. If X = the number of arrivals per minute:



128                                                      Bowerman, Essentials of Business Statistics, 2/e
78. Find the expected value of X.
Answer: 3 Difficulty: Easy


79. Consider a Poisson distribution with an average of 3 customers per minute at the local
grocery store. If X = the number of arrivals per minute, find the variance of X.
Answer: 3
 2    3 Difficulty: Easy


80. Consider a Poisson distribution with an average of 3 customers per minute at the local
grocery store. If X = the number of arrivals per minute, find the standard deviation of X.
Answer: 1.732
 X  3  1.7321
Difficulty: Easy


81. Find the probability of 10 customers or less arriving within a minute.
Answer: .9997
P(X  10) = 1 – P(X  11) = 1 – (.0002 + .0001) = .9997
Difficulty: Medium


82. Find the probability of more than 7 customers arriving within a minute.
Answer: .0119
P(X  8) = .0081 + .0027 + .0008 + .0002 + .0001 = .0119
Difficulty: Medium


83. Find the probability of 3 customers arriving within a minute.
Answer: .224
            e3 33 (.0498)(27)
P( X  3)                       .224
              3!          6
Difficulty: Medium


84. One die is thrown. What is the expected value of the number of dots on the top face of the
die?
Answer: 3.5
           1 1 1 1 1 1
E  X   1   2    3    4    5    6  
           6 6 6 6 6 6
Difficulty: Medium




Bowerman, Essentials of Business Statistics, 2/e                                             129
85. If X has the probability distribution

       X     P(X)
      -1      .2
      0       .3
      1       .5

compute the expected value of X.
Answer: .3
E[X] = -1(.2) + 0(.3) + 1(.5) = .3
Difficulty: Medium


86. If X has the probability distribution

X          -2       -1        9         1         2
P(X)       .2       .2        .2        .2        .2

compute the expected value of X.
Answer: 1.8
E[X] = (-2)(.2) + (-1)(.2) + (1)(.2) + (2)(.2) + (9)(.2) = 1.8
Difficulty: Medium (AS)



Use the following information to answer questions 87-88:
X has the following probability distribution P(X):

  X          1           2         3          4
  P(X)       .1          .5        .2        .2


87. Compute the expected value of X.
Answer: 2.5
E[X] = (1)(.1) + (2)(.5) + (3)(.2) + (4)(.2) = 2.5
Difficulty: Medium


88. Compute the variance value of X.
Answer: .9125
E[X] = (1)(.1) + (2)(.5) + (3)(.2) + (4)(.2) = 2.5
 2 x  = (1 – 2.5) 2 (.1) + (2 – 2.5) 2 (.5) + (3 – 2.5) 2 (.2) + (4 – 2.5) 2 (.2) = .9125
Difficulty: Medium




130                                                            Bowerman, Essentials of Business Statistics, 2/e
Use the following information to answer questions 89-91:
Historical data for a local steel manufacturing company shows that the average number of
defects per standard sheet of steel is 2. In addition, the number of defects per unit is distributed
according to Poisson distribution.


89. What is the probability that there will be a total of 7 defects on four standard sheets of steel?
Answer: 0.1396
             e8 87 (.0003)(2, 097,152)
P( X  7)                                .1396
               7!           5040
Difficulty: Medium


90. A batch has just been completed. What is the probability that the first three units
manufactured in this batch will contain at least a total of 4 defects?
Answer: 0.8488
P(X  4) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
P(X  4) = 1 – (.0025 + .0149 + .0446 + .0892) = .8488
Difficulty: Medium


91. Determine the standard deviation of the number of defective units for 32 sheets of metal.
Answer: 8
  (2)(32)  64  8
Difficulty: Easy


92. Consider the experiment of tossing a fair coin three times and observing the number of
heads that result (X = number of heads). Construct the probability distribution for the random
variable X.
Answer:
     X          P(X)
     0           1/8
     1           3/8
     2           3/8
     3           1/8
Difficulty: Hard (AS)


Use the following information to answer questions 93-97:
Consider the experiment of tossing a fair coin three times and observing the number of heads that
result (X = number of heads).




Bowerman, Essentials of Business Statistics, 2/e                                                   131
93. Determine the expected number of heads.
Answer: 1.5
                                          
E  X    X  (0) 1  (1) 3  (2) 3  3 1  1.5
                     8       8        8     8
Difficulty: Hard


94. What is the variance for this distribution?
Answer: .75
                                          
E  X    X  (0) 1  (1) 3  (2) 3  3 1  1.5
                     8            8          8        8
                 1               3              3             1
 X  (0  1.5)2    (1  1.5)2    (2  1.5)2    (3  1.5)2    .75
  2

                 8              8              8              8
Difficulty: Hard


95. What is the standard deviation for this distribution?
Answer: .866
                                          
E  X    X  (0) 1  (1) 3  (2) 3  3 1  1.5
                      8             8          8         8
                  1               3                3              1
 X  (0  1.5) 2    (1  1.5) 2    (2  1.5) 2    (3  1.5) 2    .75
  2

                  8               8               8               8
 X  .75  .866
Difficulty: Hard


96. If you were asked to play a game in which you tossed a fair coin three times and were given
$2 for every head you threw, how much would you expect to win on average?
Answer: $3
                                                       
Expected return  (0) 1  ($2)(1) 3  ($2)(2) 3  ($2)(3) 1  3.0
                         8            8             8               8       
Difficulty: Hard




132                                                               Bowerman, Essentials of Business Statistics, 2/e
97. A pharmaceutical company has determined that if a new cholesterol-reducing drug is
manufactured (introduced to the market), the following probability distribution will describe this
drug's contribution to the company's profits during the next six months.

  Profit Contribution         Probability of profit contribution
      -$ 30,000                              .20
       $ 50,000                              .50
       $200,000                              .30

The company management has decided to market this product if the expected contribution to
profit for the next six months is more than $90,000. Based on the information given above,
should the company begin manufacturing the new drug?
Answer: No (79,000 < 90,000)
 x = .2(-30,000) + .5(50,000) + .3(200,000) = 79,000
Difficulty: Medium


Use the following information to answer questions 98-102:
According to data from the state blood program, 40% of all individuals have group A blood. If
six (6) individuals give blood, find the probability


98. None of the individuals has group A blood?
Answer: .0467
Difficulty: Easy


99. Exactly three of the individuals has group A blood?
Answer: .2765
Difficulty: Easy


100. At least 3 of the individuals have group A blood
Answer: .4557
P(x≥3) = P(x=3) + p(x=4) + p(x=5) + p(x=6)=.4557
Difficulty: Medium


101. Find the mean number of individuals having group A blood.
Answer: 2.4
 x = np = (.6)(.4)=2.4
Difficult: Easy




Bowerman, Essentials of Business Statistics, 2/e                                               133
102. Suppose that of the six randomly selected individuals, 3 have group A blood. Based on the
probability in question 100, would you believe the data from the state blood program? Why or
why not?
Answer: Yes because probability > .05
Difficulty: Medium


103. A lawyer believes that the probability is .3 that she can win a discrimination suit. If she
wins the case she will make $40,000, but if she loses she gets nothing. Assume that she has to
spend $5000 preparing the case. What is her expected gain?
Answer: $7000
 x = (.7)(-5000) + (.3)(35000)=7000
Difficulty: Hard


Use the following information to answer questions 104-36:
Your company’s internal auditor believes that 10% of the company’s invoices contain errors. To
check this theory, 20 invoices are randomly selected and 5 are found to have errors.


104. What is the probability that of the 20 invoices written, five or more would contain errors if
the theory is valid?
Answer: .0433
P(x≥5)= (.0319)+(.0089)+(.0020)+(.0004)+(0001)=.0433
Difficulty: Easy


105. Would you accept or reject the claim? Explain.
Answer: Reject claim because P < .05


Use the following information to answer questions 106-109:
An important part of the customer service responsibilities of a cable company relates to the speed
with which trouble in service can be repaired. Historically, the data show that the likelihood is
0.75 that troubles in a residential service can be repaired on the same day. For the first five
troubles reported on a given day, what is the probability that:


106. All five will be repaired on the same day
Answer: .2373
P(x = 5) = .2373
Difficulty: Easy




134                                                       Bowerman, Essentials of Business Statistics, 2/e
107. Fewer than two troubles will be repaired on the same day?
Answer: .0046
P(x<2) = P(x=0) + P(x=1) = .0156
Difficult: Medium


108. At least 3 troubles will be repaired on the same day.
Answer: .9624
P(x≥3) = 1-(P≤2) = .8965Difficulty: Medium


109. Find the mean number of troubles repaired on the same day
Answer: 3.75
 x = np = (5)(.75) = 3.75Difficulty: Easy


Use the following information to answer questions 110-113:
The Post Office has established a record in a major Midwestern city for delivering 90% of its
local mail the next working day.


110. If you mail eight local letters, what is the probability that all of then will be delivered the
next day.
Answer: .4305
P(x=8) = .4305
Difficulty Easy (AS)


111. Of the eight, what is the average number you expect to be delivered the next day?
Answer: 7.2
 x = np =(8)(.9) = 7.2
Difficulty: Easy (AS)


112. Calculate the standard deviation of the number delivered.
Answer: .85
Σ= npq = (7.2)(.1) = .72 = .85
Difficulty: Medium (AS)




Bowerman, Essentials of Business Statistics, 2/e                                                   135
113. What is the probability that the number of delivered will be within 2 standard deviations of
the mean
Answer: .9619
P(7.2 ±2(.85)) = p(7.2 ±1.7) = P(5.5≤x≤8)= P(6≤x≤8)=(.4305)+(.3826)+(.1488)=9616
Difficulty: Hard (AS)


114. A car wash loses $30 on rainy days and makes $120 on days when is does not rain. If the
probability of rain is 0.15, calculate the car wash’s expected profit.
Answer: $97.50
 x = (-30)(.15) + (120)(.85)= -4.50+102= 97.50
Difficulty: Medium


Use the following information to answer questions 115-116:
An insurance company will insure a $75,000 Hummer for its full value against theft at a
premium of $1500 per year. Suppose that the probability that the Hummer will be stolen is
0.0075.


115. Calculate the insurance company’s expected net profit.
Answer:$937.50
 x = (-73500)(.0075) + (1500)(.9925)=-551.25+1488.75=937.50
Difficulty: Hard


116. Find the premium that the insurance company should charge if its wants its expected net
profit to be $2000?
Answer: $2,562.50
2000 = (x-75000)(.0075) + x(.9925)=.0075x – 562.5 + .9925x=2562.5
Difficulty: Hard


Use the following information to answer questions 117-119:
A large disaster cleaning company estimates that 30% of the jobs it bids on are finished within
the bid time. Looking at a random sample of 8 jobs that is has contracted:


117. Calculate the probability that exactly 4 of the jobs were not completed within the bid time.
Answer: .1361
P(x=4) = .1361
Difficulty: Medium




136                                                      Bowerman, Essentials of Business Statistics, 2/e
118. Calculate the mean number of jobs completed within the bid time.
Answer: 2.4
 x = np = 8(.3) = 2.4
Difficulty: Medium


119. Find the probability that x (number of jobs finished on time) is within one standard
deviation of the mean.
Answer: .5506
  npq = (2.4)(.7) = 1.68 =1.3, 2.4±1.3= (1.1,3.7) – 2,3 (.2965)+(.2541)=.5506
Difficulty: Hard


Multiple Choice


120. The probability that a given computer chip will fail is 0.02. Find the probability that of 5
delivered chips, exactly 2 will fail.
A) .9039
B) .0922
C) .0038
D) .0000
Answer: C Difficulty: Medium


121. According to a survey of adults, 64% have money in a regular savings account. If we plan
on surveying 50 randomly selected adults, find the mean number of adults who would have
regular savings accounts.
A) 12
B) 22
C) 32
D) 42
Answer: C Difficult: Medium


Use the following information to answer questions 122-123:
In the most recent election, 19% of all eligible college students voted. If a random sample of 20
students were surveyed:


122. Find the probability that exactly half voted in the election.
A) .0000
B) .0014
C) .0148
D) .4997
Answer: B Difficulty: Medium


Bowerman, Essentials of Business Statistics, 2/e                                                137
123. Find the probability that none of the students voted
A) .0000
B) .0014
C) .0148
D) .4997
Answer: C Difficulty: Medium


124. Of all individual tax returns, 37% include errors made by the taxpayer. If IRS examiners
are assigned randomly selected returns in batches of 12, find the mean and standard deviation for
the number of erroneous returns per batch.
A)  = 2.80,  = 1.67
B)  = 4.44,  = 1.67
C)  = 4.44,  = 2.80
D)  = 7.56,  = 2.80
Answer: B Difficulty: Medium


125. In a study conducted for the State Department of Education, 30% of the teachers who left
teaching did so because they were laid off. Assume that we randomly select 10 teachers who
have recently left their profession. Find the probability that exactly 4 of them were laid off.
A) .3000
B) .2668
C) .2001
D) .0090
Answer: C Difficulty: Medium


126. A company manufactures an appliance, gives a warranty and 95% of its appliances do not
require repair before the warranty expires. If an organization buys 10 of these appliances,
calculate an interval that contains 95.44% of all the appliances that won’t require repair.
A) [8.12 10.88]
B) [7.43 11.57]
C) [8.81 10.19]
D) [8.55 10.45]
Answer: A Difficulty: Medium




138                                                         Bowerman, Essentials of Business Statistics, 2/e
127. A manufacturer tested a sample of semiconductor chips and found that 35 were defective
and 190 were good. If additional tests are to be conducted with random samples of 160
semiconductor chips, find the mean for the number of defects in these groups of 160 (rounded to
nearest whole number).
A) 56
B) 35
C) 29
D) 25
Answer: D Difficulty: Medium


128. In a study conducted by UCLA, it was found that 25% of college freshmen support
increased military spending. If 6 college freshmen are randomly selected, find the probability
that:
Fewer than 4 support increased military spending
A) .0330
B) .7844
C) .9624
D) .9954
Answer: C Difficulty: Medium (AS)


129. Exactly 3 support increased military spending
A) .0330
B) .1318
C) .7844
D) .9624
Answer: B Difficulty: Medium (AS)


130. Only 1 supports increased military spending
A) .0330
B) .1318
C) .3560
D) .7844
Answer: C Difficulty: Medium (AS)


131. A multiple-choice test has 30 questions and each one has five possible answers, of which
one is correct. If all answers were guesses, find the probability of getting exactly four correct
answers.
A) .0604
B) .1325
C) .2552
D) .8000
Answer: B Difficulty: Medium



Bowerman, Essentials of Business Statistics, 2/e                                                 139

				
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