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Student’s Solutions Manual and Study Guide: Chapter 5
Page 1
Chapter 5
Discrete Distributions
LEARNING OBJECTIVES
The overall learning objective of Chapter 5 is to help you understand a category of probability
distributions that produces only discrete outcomes, thereby enabling you to:
1.
2.
3.
4.
5.
Define a random variable in order to differentiate between a discrete distribution and a
continuous distribution.
Determine the mean, variance, and standard deviation of a discrete distribution.
Solve problems involving the binomial distribution using the binomial formula and the
binomial table.
Solve problems involving the Poisson distribution using the Poisson formula and the
Poisson table.
Solve problems involving the hypergeometric distribution using the hypergeometric
formula.
CHAPTER OUTLINE
5.1
Discrete Versus Continuous Distributions
5.2
Describing a Discrete Distribution
Mean, Variance, and Standard Deviation of Discrete Distributions
Mean or Expected Value
Variance and Standard Deviation of a Discrete Distribution
5.3
Binomial Distribution
Solving a Binomial Problem
Using the Binomial Table
Using the Computer to Produce a Binomial Distribution
Mean and Standard Deviation of the Binomial Distribution
Graphing Binomial Distributions
5.4
Poisson Distribution
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 2
Working Poisson Problems by Formula
Using the Poisson Tables
Mean and Standard Deviation of a Poisson Distribution
Graphing Poisson Distributions
Using the Computer to Generate Poisson Distributions
Approximating Binomial Problems by the Poisson Distribution
5.5
Hypergeometric Distribution
Using the Computer to Solve for Hypergeometric Distribution
Probabilities
KEY TERMS
Binomial Distribution
Continuous Distribution
Continuous Random Variable
Discrete Distribution
Discrete Random Variables
Hypergeometric Distribution
Lambda ()
Mean, or Expected Value
Poisson Distribution
Random Variable
STUDY QUESTIONS
1. Variables that take on values at every point over a given interval are called
_______________ _______________ variables.
2. If the set of all possible values of a variable is at most finite or a countably infinite number of
possible values, then the variable is called a _______________ _______________ variable.
3. An experiment in which a die is rolled six times will likely produce values of a
_______________ random variable.
4. An experiment in which a researcher counts the number of customers arriving at a
grocery store checkout counter every two minutes produces values of a _______________
random variable.
5. An experiment in which the time it takes to assemble a product is measured is likely to
produce values of a _______________ random variable.
6. A binomial distribution is an example of a _______________ distribution.
7. The normal distribution is an example of a _______________ distribution.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 3
8. The long-run average of a discrete distribution is called the __________________ or
______________________________________.
Use the following discrete distribution to answer 9 and 10
x
P(x)
1
2
3
4
.435
.241
.216
.108
9. The mean of the discrete distribution above is ____________.
10.
The variance of the discrete distribution above is _______________.
11.
On any one trial of a binomial experiment, there can be only _______________ possible outcomes.
12.
Suppose the probability that a given part is defective is .10. If four such parts are randomly drawn
from a large population, the probability that exactly two parts are defective is ________.
13.
Suppose the probability that a given part is defective is .04. If thirteen such parts are randomly
drawn from a large population, the expected value or mean of the binomial distribution that
describes this experiment is ________.
14.
Suppose a binomial experiment is conducted by randomly selecting 20 items where p = .30. The
standard deviation of the binomial distribution is _______________.
15.
Suppose 47 percent of the workers in a large corporation are under 35 years of age. If 15 workers
are randomly selected from this corporation, the probability of selecting exactly 10 who are under
35 years of age is _______________.
16.
Suppose that 23 percent of all adult Americans fly at least once a year. If 12 adult Americans are
randomly selected, the probability that exactly four have flown at least once last year is
_______________.
17.
Suppose that 60 percent of all voters support the Prime Minister of Canada at this time. If 20 voters
are randomly selected, the probability that at least 11 support the Prime Minister is
_______________.
18.
The Poisson distribution was named after the French mathematician _______________.
19.
The Poisson distribution focuses on the number of discrete occurrences per _______________.
20.
The Poisson distribution tends to describe _______________ occurrences.
21.
The long-run average or mean of a Poisson distribution is _______________.
22.
The variance of a Poisson distribution is equal to _______________.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 4
23.
If Lambda is 2.6 occurrences over an interval of 5 minutes, the probability of getting 6 occurrences
over one 5 minute interval is _______________.
24.
Suppose that in the long-run a company determines that there are 1.2 flaws per every 20 pages of
typing paper produced. If 10 pages of typing paper are randomly selected, the probability that more
than 2 flaws are found is _______________.
25.
If Lambda is 1.8 for a four minute interval, an adjusted new Lambda of _______ would be used to
analyze the number of occurrences for a twelve minute interval.
26.
Suppose a binomial distribution problem has an n = 200 and a p = .03. If this problem is worked
using the Poisson distribution, the value of Lambda is ________.
27.
The hypergeometric distribution should be used when a binomial type experiment is being
conducted without replacement and the sample size is greater than or equal to ________% of the
population.
28.
Suppose a population contains sixteen items of which seven are X and nine are Y. If a random
sample of five of these population items is selected, the probability that exactly three of the five are
X is ________.
29.
Suppose a population contains twenty people of which eight are members of the Catholic church. If
a sample of four of the population is taken, the probability that at least three of the four are
members of the Catholic church is ________.
30.
Suppose a lot of fifteen personal computer printers contains two defective printers. If three of the
fifteen printers are randomly selected for testing, the probability that no defective printers are
selected is _______________.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 5
ANSWERS TO STUDY QUESTIONS
1. Continuous Random
16. .1712
2. Discrete Random
17. .755
3. Discrete
18. Poisson
4. Discrete
19. Interval
5. Continuous
20. Rare
6. Discrete
21. Lambda
7. Continuous
22. Lambda
8. Mean, Expected Value
23. .0319
9. 1.997
24. .0232
10. 1.083
25. 5.4
11. Two
26. 6.0
12. .0486
27. 5
13. 0.52
28. .2885
14. 2.049
29. .1531
15. .0661
30. .6286
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 6
SOLUTIONS TO PROBLEMS IN CHAPTER 5
P(x)
x·P(x)
(x – µ)2
(x – µ)2·P(x)
.238
.238
2.775556
0.6605823
.290
.580
0.443556
0.1286312
.177
.531
0.111556
0.0197454
.158
.632
1.779556
0.2811698
.137
.685
5.447556
0.7463152
2
2
µ = [x·P(x)] = 2.666
 = [(x – µ) ·P(x)] = 1.8364439
 = 1.8364439 = 1.355155
5.1
x
1
2
3
4
5
5.3
x
P(x)
x·P(x)
0
.461
.000
1
.285
.285
2
.129
.258
3
.087
.261
4
.038
.152
Expected number = µ = [x·P(x)]= 0.956
(x – µ)2
0.913936
0.001936
1.089936
4.177936
9.265936
(x – µ)2·P(x)
0.421324
0.000552
0.140602
0.363480
0.352106
2
2
 = [(x – µ) ·P(x)] =
1.278064
 =
5.5
a)
n=4
P(x=3) =
b)
n=7
1.278064 = 1.1305
p = .10
3
1
4C3(.10) (.90)
q = .90
= 4(.001)(.90) = .0036
p = .80
q = .20
P(x=4) = 7C4(.80)4(.20)3 = 35(.4096)(.008) = .1147
c)
n = 10
p = .60
q = .40
P(x > 7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) =
7
3
8
2
9
1
10
0
10C7(.60) (.40) + 10C8(.60) (.40) + 10C9(.60) (.40) +10C10(.60) (.40)
120(.0280)(.064) + 45(.0168)(.16) + 10(.0101)(.40) + 1(.0060)(1) =
.2150 + .1209 + .0403 + .0060 = .3822
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
=
Student’s Solutions Manual and Study Guide: Chapter 5
d)
n = 12
p = .45
Page 7
q = .55
P(5 < x < 7) = P(x=5) + P(x=6) + P(x=7) =
5
7
6
6
7
5
12C5(.45) (.55) + 12C6(.45) (.55) + 12C7(.45) (.55)
=
792(.0185)(.0152) + 924(.0083)(.0277) + 792(.0037)(.0503) =
.2225 + .2124 + .1489 = .5838
5.7
a)
n = 20
p = .70
q = .30
µ = np = 20(.70) = 14
 =
b)
n  p  q  20(.70)(.30)  4.2 = 2.05
n = 70
p = .35
q = .65
µ = np = 70(.35) = 24.5
 =
c)
n  p  q  70(.35)(.65)  15.925 = 3.99
n = 100
p = .50
q = .50
µ = np = 100(.50) = 50
 =
5.9
a)
n = 20
n  p  q  100(.50)(.50)  25 = 5
p = .78
x = 14
P(x=14) = 20C14 (.78)14(.22)6 = 38,760(.030855)(.00011338) = .1356
b)
n = 20
p = .75
x = 20
P(x=20) = 20C20 (.75)20(.25)0 = (1)(.0031712)(1) = .0032
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
c)
n = 20
Page 8
p = .70
x < 12
Use table A.2:
P(x<12) = P(x=0) + P(x=1) + . . . + P(x=11)=
.000 + .000 + .000 + .000 + .000 + .000 + .000 +
.001 + .004 + .012 + .031 + .065 = .113
5.11
n = 25
p = .60
a) x > 15
P(x > 15) = P(x = 15) + P(x = 16) + · · · + P(x = 25)
Using Table A.2
x
15
16
17
18
19
20
21
22
n = 25, p = .60
Prob
.161
.151
.120
.080
.044
.020
.007
.002
.585
b) x > 20
P(x > 20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25) =
Using Table A.2
n = 25, p = .60
.007 + .002 + .000 + .000 + .000 = .009
c) P(x < 10)
Using Table A.2
n = 25, p = .60 and x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
x
9
Prob.
.009
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 9
8
7
<6
5.13
n = 15
.003
.001
.000
.013
p = .20
a) P(x = 5) =
5
10
15C5(.20) (.80)
= 3003(.00032)(.1073742) = .1032
b) P(x > 9): Using Table A.2
P(x = 10) + P(x = 11) + . . . + P(x = 15) = .000 + .000 + . . . + .000 = .000
c) P(x = 0) =
0
15
15C0(.20) (.80)
= (1)(1)(.035184) = .0352
d) P(4 < x < 7): Using Table A.2
P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) = .188 + .103 + .043 + .014 = .348
e)
The distribution is skewed to the right. Values of x near   3 have the highest probabilities.
For the large values of x the probabilities are very small.
5.14 n = 18
a)
b)
p =.22
µ = 18(.22) = 3.96 (earn between $100,000 and $149,999)
p = .32
µ = 18(.32) = 5.76 (earn $150,000 or more)
n = 18
p = .22
P(x > 8) = P(x = 8) + P(x = 9) + ...+P(x = 18)
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 10
P(x = 8) =
8
10
18C8(.22) (.78)
P(x = 9) =
9
9
18C9(.22) (.78)
= .0200
= .0063
P(x = 10) =
10
8
18C10(.22) (.78)
= .0016
P(x = 11) =
11
7
18C11(.22) (.78)
= .0003
P(x = 12) =
12
6
18C12(.22) (.78)
= .0001
P(x) for x > 13 are negligible. Therefore
P(x > 8) = .0200 + .0063 + .0016 + .0003 + .0001 = .0283
c) n = 18
p = .32
P(2 < x < 4) = P(x = 2) + P(x = 3) + P(x = 4) =
2
16
18C2(.32) (.68)
+
3
15
18C3(.32) (.68)
+
4
14
18C4(.32) (.68)
=
.0327 + .0822 + .1450 = .2599
d) n = 18
p = .22
x=0
P(none earn between $100,000 and $149,999) = 18C0(.22)0(.78)18 = .01142
n = 18
p = .34
x=0
P(none earn $150,000 or more) = 18C0(.32)0(.68)18 = .00097
Since only 22% (compared to 32%) fall in the $100,000 to $149,999 category, it is
more likely that none of the 18 CPFs would fall in this category.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
5.15
Page 11
a) P(x=5 = 2.3) =
2.35  e  2.3 (64.36343)(.100259)
= .0538

5!
120
b) P(x=2 = 3.9) =
3.9 2  e  3.9 (15.21)(.020242)
= .1539

2!
2
c) P(x < 3 = 4.1) = P(x=3) + P(x=2) + P(x=1) + P(x=0) =
4.13  e  4.1 (68.921)(.016573)
= .1904

3!
6
4.12  e  4.1 (16.81)(.016573)
= .1393

2!
2
4.11  e  4.1 (4.1)(.016573)
= .0679

1!
1
4.10  e  4.1 (1)(.016573)
= .0166

0!
1
.1904 + .1393 + .0679 + .0166 = .4142
d) P(x=0 = 2.7) =
2.7 0  e  2.7 (1)(.06721)
= .0672

0!
1
e) P(x=1  = 5.4)=
5.41  e  5.4 (5.4)(.0045166)
= .0244

1!
1
f) P(4 < x < 8  = 4.4): P(x=5 = 4.4) + P(x=6 = 4.4) + P(x=7 = 4.4)=
4.4 5  e  4.4
4.4 6  e  4.4
4.4 7  e  4.4
+
+
=
6!
7!
5!
(1649.1622)(.01227734) (7256.3139)(.01227734) (31,927.781)(.01227734)
+
+
120
720
5040
= .1687 + .1237 + .0778 = .3702
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
5.17 a)  = 6.3
mean = 6.3
Page 12
Standard deviation =
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Prob
.0018
.0116
.0364
.0765
.1205
.1519
.1595
.1435
.1130
.0791
.0498
.0285
.0150
.0073
.0033
.0014
.0005
.0002
.0001
.0000
6.3 = 2.51
Student’s Solutions Manual and Study Guide: Chapter 5
b)  = 1.3
mean = 1.3
Page 13
standard deviation =
x
0
1
2
3
4
5
6
7
8
9
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Prob
.2725
.3543
.2303
.0998
.0324
.0084
.0018
.0003
.0001
.0000
1.3 = 1.14
Student’s Solutions Manual and Study Guide: Chapter 5
c)  = 8.9
mean = 8.9
Page 14
standard deviation =
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Prob
.0001
.0012
.0054
.0160
.0357
.0635
.0941
.1197
.1332
.1317
.1172
.0948
.0703
.0481
.0306
.0182
.0101
.0053
.0026
.0012
.0005
.0002
.0001
8.9 = 2.98
Student’s Solutions Manual and Study Guide: Chapter 5
d)  = 0.6
mean = 0.6
x
0
1
2
3
4
5
6
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Page 15
standard deviation =
Prob
.5488
.3293
.0988
.0198
.0030
.0004
.0000
0.6 = .775
Student’s Solutions Manual and Study Guide: Chapter 5
Page 16
5.19  = x/n = 126/36 = 3.5
Using Table A.3
a) P(x = 0) = .0302
b) P(x > 6) = P(x = 6) + P(x = 7) + . . . =
.0771 + .0385 + .0169 + .0066 + .0023 +
.0007 + .0002 + .0001 = .1424
c) P(x < 4 10 minutes)
Double Lambda to  = 7.010 minutes
P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) =
.0009 + .0064 + .0223 + .0521 = .0817
d) P(3 < x < 6 10 minutes)
 = 7.0  10 minutes
P(3 < x < 6) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)
= .0521 + .0912 + .1277 + .1490 = .42
e) P(x = 8  15 minutes)
Change Lambda for a 15 minute interval by multiplying the original Lambda by 3.
 = 10.5  15 minutes
P(x = 815 minutes) =
5.21
x  e 
x!
 = 0.6 trips1 year
a) P(x=0  = 0.6):
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
(10.58 )(e 10.5 )

8!
= .1009
Student’s Solutions Manual and Study Guide: Chapter 5
Page 17
from Table A.3 = .5488
b) P(x=1  = 0.6):
from Table A.3 = .3293
c) P(x > 2  = 0.6):
from Table A.3
x
2
3
4
5
6
Prob.
.0988
.0198
.0030
.0004
.0000
.1220
d) P(x < 3 3 year period):
The interval length has been increased (3 times)
New Lambda =  = 1.8 trips3 years
P(x < 3  = 1.8):
from Table A.3
x
0
1
2
3
e) P(x=46 years):
The interval has been increased (6 times)
New Lambda =  = 3.6 trips6 years
P(x=4 = 3.6):
from Table A.3 = .1912
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Prob.
.1653
.2975
.2678
.1607
.8913
Student’s Solutions Manual and Study Guide: Chapter 5
5.23
Page 18
 = 1.2 penscarton
a) P(x=0   = 1.2):
from Table A.3 = .3012
b) P(x > 8   = 1.2):
from Table A.3 = .0000
c) P(x > 3   = 1.2):
from Table A.3
5.25
p = .009
x
4
5
6
7
8
Prob.
.0260
.0062
.0012
.0002
.0000
.0336
n = 200
Use the Poisson Distribution:
 = np = 200(.009) = 1.8
a) P(x > 6) from Table A.3 =
P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + . . . =
.0078 + .0020 + .0005 + .0001 = .0104
b) P(x > 10) = .0000
c) P(x = 0) = .1653
d) P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P( x = 3) + P(x = 4) =
.1653 + .2975 + .2678 + .1607 + .0723 = .9636
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
5.27
a) P(x = 3 N = 11, A = 8, n = 4)
8
C3 3 C1 (56)(3)
= .5091

C
330
11 4
b) P(x < 2)N = 15, A = 5, n = 6)
P(x = 1) + P (x = 0) =
5
C1 10 C5
+
15 C 6
5
C0 10 C6
(5)( 252) (1)( 210)

=
5005
5005
15 C 6
.2517 + .0420 = .2937
c) P(x=0 N = 9, A = 2, n = 3)
2
C0 7 C3 (1)(35)
= .4167

84
9 C3
d) P(x > 4 N = 20, A = 5, n = 7) =
P(x = 5) =
5
C5 15 C 2
(1)(105)
=
= .0014
77520
20 C 7
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Page 19
Student’s Solutions Manual and Study Guide: Chapter 5
5.29
N = 20
A=4
4
a) P(x = 0) =
b) P(x = 4) =
4
N = 10
n=4
C0 16 C4
(1)(1820)
=
= .3756
4845
20 C 4
C4 16 C0
(1)(1)
=
= .0002
4845
20 C 4
4
c) P(x = 2) =
5.31
Page 20
C2 16 C 2
(6)(120)
=
= .1486
4845
C
20 4
n=4
a) A = 4 (British Columbia, Alberta, Saskatchewan, Manitoba)
P(x = 2) =
4
x=2
C2 6 C2 (6)(15)
= .4286

210
10 C 4
b) A = 4 (Nova Scotia, New Brunswick, Newfoundland and Labrador, Prince Edward
Island) x = 0
P(x = 0) =
4
C0 6 C 4 (1)(15)
= .0714

210
10 C 4
c) A = 4 (Ontario, British Columbia, Alberta, Quebec)
P(x = 3) =
4
C3 6 C1 (4)(6)
= .1143

210
10 C 4
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
x=3
Student’s Solutions Manual and Study Guide: Chapter 5
5.33
N = 18
A = 11 men
Page 21
n=5
P(x < 1) = P(1) + P(0) =
11
C1 7 C 4
+
18 C 5
11
C 0 7 C5
(11)(35) (1)( 21)

=
= .0449 + .0025 = .0474
8568
8568
18 C5
It is fairly unlikely that these results occur by chance. A researcher might want to
further investigate this result to determine causes. Were officers selected based on
leadership, years of service, dedication, prejudice, or some other reason?
5.35
a) P(x = 14 n = 20 and p = .60) = .124
b) P(x < 5 n = 10 and p =.30) =
P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x=0) =
x
0
1
2
3
4
Prob.
.028
.121
.233
.267
.200
.849
c) P(x > 12 n = 15 and p = .60) =
P(x = 12) + P(x = 13) + P(x = 14) + P(x = 15)
x
12
13
14
15
Prob.
.063
.022
.005
.000
.090
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
d) P(x > 20 n = 25 and p = .40) = P(x = 21) + P(x = 22) +
P(x = 23) + P(x = 24) + P(x=25) =
x
21
22
23
24
25
5.37
Prob.
.000
.000
.000
.000
.000
.000
a) P(x = 3 = 1.8) = .1607
b) P(x < 5 = 3.3) =
P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0) =
x
0
1
2
3
4
Prob.
.0369
.1217
.2008
.2209
.1823
.7626
c) P(x > 3  = 2.1) =
x
3
4
5
6
7
8
9
10
11
Prob.
.1890
.0992
.0417
.0146
.0044
.0011
.0003
.0001
.0000
.3504
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Page 22
Student’s Solutions Manual and Study Guide: Chapter 5
Page 23
d) P(2 < x < 5  = 4.2):
P(x=3) + P(x=4) + P(x=5) =
x
3
4
5
5.39
Prob.
.1852
.1944
.1633
.5429
n = 25 p = .20 retired
from Table A.2: P(x = 7) = .111
P(x > 10): P(x = 10) + P(x = 11) + . . . + P(x = 25) = .012 + .004 + .001 = .017
Expected number of retired people = µ = np = 25(.20) = 5
n = 20 p = .40 mutual funds
from Table A.2: P(x = 8) = .180
P(x < 6) = P(x = 0) + P(x = 1) + . . . + P(x = 5) =
.000 + .000 + .003 +.012 + .035 + .075 = .125
P(x = 0) = .000
P(x > 12) = P(x = 12) + P(x = 13) + . . . + P(x = 20) = .035 + .015 + .005 + .001 = .056
The highest probability is .180 for x = 8 .
Expected number of people invested in mutual funds = µ = n p = 20(.40) = 8
5.41
N = 32
A = 10
a) P(x = 3) =
10
n = 12
C3  22 C9
(120)( 497,420)
=
= .2644
225,792,840
32 C12
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
b) P(x = 6) =
10
c) P(x = 0) =
10
Page 24
C6  22 C6
(210)(74,613)
=
= .0694
225
,
792
,
840
C
32 12
C0  22 C12
(1)(646,646)
=
= .0029
225,792,840
32 C12
d) A = 22
P(7 < x < 9) =
=
22
C7 10 C5
+
32 C12
22
C8 10 C4
+
32 C12
22
C9 10 C3
32 C12
(170,544)( 252) (319,770)( 210) (497,420)(120)


225,792,840
225,792,840
225,792,840
= .1903 + .2974 + .2644 = .7521
5.43
a) n = 20 and p = .176
The expected number = µ = np = (20)(.176) = 3.52
b) P(x < 1 n = 20 and p = .176) =
P(x = 1) + P(x = 0) =
1
19
20C1(.176) (.824)
+ 20C0(.176)0(.824)20
= (20)(.176)(.02527) + (1)(1)(.02082) = .0890 +. 0208 = .1098
Since the probability is comparatively low, the population of your province may have
a lower percentage of chronic heart conditions than those of other provinces.
5.45
n = 12
a.) P(x = 0 long hours):
p = .20
0
12
12C0(.20) (.80)
= .0687
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 25
b.) P(x > 6 long hours):
p = .20
Using Table A.2:
P(x = 6) + P(x = 7) + . . . + P(x = 12) = .016 + .003 + .001 = .020
c) P(x = 5 good financing):
p = .25,
5
7
12C5(.25) (.75)
= .1032
d.) p = .19 (good plan), expected number = µ = n(p) = 12(.19) = 2.28
5.47
P(x < 3 n = 8 and p = .60):
From Table A.2:
x
0
1
2
3
Prob.
.001
.008
.041
.124
.174
17.4% of the time in a sample of eight, three or fewer customers are walk-ins by
chance. Other reasons for such a low number of walk-ins might be that she is
retaining more old customers than before or perhaps a new competitor is
attracting walk-ins away from her.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
5.49
Page 26
 = 0.6 flats2,000 km
P(x = 0  = 0.6) = (from Table A.3) .5488
P(x > 3  = 0.6) = (from Table A.3)
x
3
4
5
Prob.
.0198
.0030
.0004
.0232
Assume one trip is independent of the other. Let F = flat tire and NF = no flat tire
P(NF1  NF2) = P(NF1)  P(NF2)
but P(NF) = .5488
P(NF1  NF2) = (.5488)(.5488) = .3012
5.51
N = 24
n=6
a) P(x = 6) =
b) P(x = 0) =
8
A=8
C6 16 C0
(28)(1)
= .0002

134,596
24 C 6
8
C0 16 C6 (1)(8008)
= .0595

134,596
24 C 6
c) The probability from part b) is greater as only 8 out of 24 commute from Airdrie.
d) A = 16 (do not commute from Airdrie)
P(x = 3) =
5.53
16
C3 8 C3 (560)(56)
= .2330

134,596
24 C 6
 = 2.4 calls1 minute
a) P(x = 0 = 2.4) = (from Table A.3) .0907
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
b) Can handle x < 5 calls
Page 27
Cannot handle x > 5 calls
P(x > 5 = 2.4) = (from Table A.3)
x
6
7
8
9
10
11
c) P(x = 3 calls2 minutes)
The interval has been increased 2 times.
New Lambda:  = 4.8 calls2 minutes.
from Table A.3: .1517
d) P(x < 1 calls15 seconds):
The interval has been decreased by ¼.
New Lambda =  = 0.6 calls15 seconds.
P(x < 1 = 0.6) = (from Table A.3)
P(x = 1) = .3293
P(x = 0) = .5488
.8781
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Prob.
.0241
.0083
.0025
.0007
.0002
.0000
.0358
Student’s Solutions Manual and Study Guide: Chapter 5
5.55
n=8
2
6
8C2(.55) (.45)
p = .55
Page 28
x = 2 women
= (28)(.3025)(.0083038) = .0703
It is not very likely that a company would randomly hire 8 physicians
and only two of them would be female.
5.57
N = 14 n = 4
a) P(x = 4 N = 14, n = 4, A = 10 north side)
10
C 4  4 C0 (210((1)
= .2098

C
1001
14 4
b) P(x = 4 N = 14, n = 4, A = 4 west)
4
C4 10 C0 (1)(1)
= .0010

1001
14 C 4
c) P(x = 2 N = 14, n = 4, A = 4 west)
4
5.59
C 2 10 C 2 (6)( 45)
= .2697

1001
14 C 4
This is a binomial distribution with n = 15 and p = .36.
 = np = 15(.36) = 5.4
 =
15(.36)(.64) = 1.86
The most likely values are near the mean, 5.4. Note from the printout that the
most probable values are at x = 5 and x = 6 which are near the mean.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
5.61
Page 29
This is a binomial distribution with n = 22 and p = .64.
The mean is np = 22(.64) = 14.08 and the standard deviation is:
 =
n  p  q  22(.64)(.36) = 2.25
The x value with the highest peak on the graph is at x = 14 followed by x = 15
and x = 13 which are nearest to the mean.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 5
Page 30
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Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition