Chapter 5: Discrete Distributions
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.
Distinguish between discrete random variables and continuous random variables.
2.
Know how to determine the mean and variance of a discrete distribution.
3.
Identify the type of statistical experiments that can be described by the binomial
distribution and know how to work such problems.
4.
Decide when to use the Poisson distribution in analyzing statistical experiments and
know how to work such problems.
5.
Decide when binomial distribution problems can be approximated by the Poisson
distribution and know how to work such problems.
6.
Decide when to use the hypergeometric distribution and know how to work such
problems
.
CHAPTER TEACHING STRATEGY
Chapters 5 and 6 introduce the student to several statistical distributions. It is important
to differentiate between the discrete distributions of chapter 5 and the continuous distributions of
chapter 6.
The approach taken in presenting the binomial distribution is to build on techniques
presented in chapter 4. It can be helpful to take the time to apply the law of multiplication for
independent events to a problem and demonstrate to students that sequence is important. From
there, the student will more easily understand that by using combinations, one can more quickly
determine the number of sequences and weigh the probability of obtaining a single sequence by
that number. In a sense, we are developing the binomial formula through an inductive process.
Chapter 5: Discrete Distributions
2
Thus, the binomial formula becomes more of a summary device than a statistical "trick". The
binomial tables presented in this text are noncumulative. This makes it easier for the student to
recognize that the table is but a listing of a series of binomial formula computations. In addition,
it lends itself more readily to the graphing of a binomial distribution.
It is important to differentiate applications of the Poisson distribution from binomial
distribution problems. It is often difficult for students to determine which type of distribution to
apply to a problem. The Poisson distribution applies to rare occurrences over some interval. The
parameters involved in the binomial distribution (n and p) are different from the parameter
(Lambda) of a Poisson distribution.
It is sometimes difficult for students to know how to handle Poisson problems where the
interval for the problem is different than the interval for which Lambda was developed. If they
can view Lambda as a long run average which can be appropriately adjusted for various
intervals, then students can be more successful with these types of problems.
Solving for the mean and standard deviation of binomial distributions prepares the
students for chapter 6 where the normal distribution is sometimes used to solve binomial
distribution problems. In addition, graphing binomial and Poisson distributions affords the
student the opportunity to visualize the meaning and impact of a particular set of parameters for a
distribution.
It can be emphasized that the hypergeometric distribution is an exact distribution.
However, it is cumbersome to determine probabilities using the hypergeometric formula
particularly when computing cumulative probabilities. The hypergeometric distribution can be
presented as a fall-back position to be used when the binomial distribution should not be applied
because of the non independence of trials and size of sample.
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
Chapter 5: Discrete Distributions
3
5.4
Poisson Distribution
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 Distributions
Continuous Random Variables
Discrete Distributions
Discrete Random Variables
Hypergeometric Distribution
Lambda ()
Mean, or Expected Value
Poisson Distribution
Random Variable
SOLUTIONS TO PROBLEMS IN CHAPTER 5
5.1
x
P(x)
x·P(x)
1
2
3
4
5
.238
.290
.177
.158
.137
.238
.580
.531
.632
.685
µ = [x·P(x)] = 2.666
 =
(x-µ)2
2.775556
0.443556
0.111556
1.779556
5.447556
(x-µ)2·P(x)
0.6605823
0.1286312
0.0197454
0.2811700
0.7463152
2 = (x-µ)2·P(x) = 1.836444
1.836444 = 1.355155
Chapter 5: Discrete Distributions
5.2
x
P(x)
x·P(x)
0
1
2
3
4
5
6
7
.103
.118
.246
.229
.138
.094
.071
.001
.000
.118
.492
.687
.552
.470
.426
.007
µ = [x·P(x)] = 2.752
 =
5.3
x
(x-µ)2
(x-µ)2·P(x)
7.573504
3.069504
0.565504
0.061504
1.557504
5.053504
10.549500
18.045500
0.780071
0.362201
0.139114
0.014084
0.214936
0.475029
0.749015
0.018046
2 = (x-µ)2·P(x) = 2.752496
2.752496 = 1.6591
P(x)
0
1
2
3
4
x·P(x)
.461
.285
.129
.087
.038
.000
.285
.258
.261
.152
(x-µ)2
(x-µ)2·P(x)
0.913936
0.001936
1.089936
4.177936
9.265936
 =
0.421324
0.000552
0.140602
0.363480
0.352106
2 = (x-µ)2·P(x) = 1.278064
E(x)=µ= [x·P(x)]= 0.956
5.4
4
1.278064 = 1.1305
x
P(x)
x·P(x)
0
1
2
3
4
5
6
.262
.393
.246
.082
.015
.002
.000
.000
.393
.492
.246
.060
.010
.000
µ = [x·P(x)] = 1.201
 =
.96260 = .98112
(x-µ)2
1.4424
0.0404
0.6384
3.2364
7.8344
14.4324
23.0304
(x-µ)2·P(x)
0.37791
0.01588
0.15705
0.26538
0.11752
0.02886
0.00000
2 = (x-µ)2·P(x) = 0.96260
Chapter 5: Discrete Distributions
5
5.5
a)
n=4
P(x=3) =
b)
n=7
p = .10
3
1
4C3(.10) (.90)
p = .80
q = .90
= 4(.001)(.90) = .0036
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
d)
n = 12
p = .45
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
5.6
By Table A.2:
a)
n = 20
p = .50
P(x=12) = .120
b)
n = 20
p = .30
P(x > 8) = P(x=9) + P(x=10) + P(x=11) + ...+ P(x=20) =
=
Chapter 5: Discrete Distributions
6
.065 + .031 + .012 + .004 + .001 + .000 = .113
c)
n = 20
p = .70
P(x < 12) =
P(x=11) + P(x=10) + P(x=9) + ... + (Px=0) =
.065 + .031 + .012 + .004 + .001 + .000 = .113
d)
n = 20
p = .90
P(x < 16) = P(x=16) + P(x=15) + P(x=14) + ...+ P(x=0) =
.090 + .032 + .009 + .002 + .000 = .133
e)
n = 15
p = .40
P(4 < x < 9) =
P(x=4) + P(x=5) + P(x=6) + P(x=7) + P(x=8) + P(x=9) =
.127 + .186 + .207 + .177 + .118 + .061 = .876
f)
n = 10
p = .60
P(x > 7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) =
.215 + .122 + .040 + .006 = .382
5.7
a)
n = 20
p = .70
q = .30
µ = np = 20(.70) = 14
 =
b)
n = 70
n  p  q  20(.70)(.30)  4.2 = 2.05
p = .35
µ = np = 70(.35) = 24.5
q = .65
Chapter 5: Discrete Distributions
 =
c)
7
n  p  q  70(.35)(.65)  15.925 = 3.99
n = 100
p = .50
q = .50
µ = np = 100(.50) = 50
 =
n  p  q  100(.50)(.50)  25 = 5
5.8
a)
b)
n=6
n = 20
p = .70
p = .50
x
0
1
2
3
4
5
6
Prob
.001
.010
.060
.185
.324
.303
.118
x
0
1
2
3
4
5
6
7
Prob
.000
.000
.000
.001
.005
.015
.037
.074
Chapter 5: Discrete Distributions
c)
n=8
p = .80
8
8
9
10
11
12
13
14
15
16
17
18
19
20
.120
.160
.176
.160
.120
.074
.037
.015
.005
.001
.000
.000
.000
x
0
1
2
3
4
5
6
7
8
Prob
.000
.000
.001
.009
.046
.147
.294
.336
.168
Chapter 5: Discrete Distributions
9
5.9
a)
n = 20
20C14
b)
x = 14
(.78)14(.22)6 = 38,760(.030855)(.00011338) = .1356
n = 20
20C20
c)
p = .78
p = .75
x = 20
(.75)20(.25)0 = (1)(.0031712)(1) = .0032
n = 20
p = .70
x < 12
Use table A.2:
P(x=0) + P(x=1) + . . . + P(x=11)=
.000 + .000 + .000 + .000 + .000 + .000 + .000 +
.001 + .004 + .012 + .031 + .065 = .113
5.10
n = 16
p = .40
P(x > 9): from Table A.2:
x Prob
9
10
11
12
13
.084
.039
.014
.004
.001
.142
Chapter 5: Discrete Distributions
10
P(3 < x < 6):
x Prob
3
4
5
6
.047
.101
.162
.198
.508
n = 13
p = .88
P(x = 10) = 13C10(.88)10(.12)3 = 286(.278500976)(.001728) = .1376
P(x = 13) = 13C13(.88)13(.12)0 = (1)(.1897906171)(1) = .1898
Expected Value = µ = n p = 13(.88) = 11.44
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
Prob
15
16
17
18
19
20
21
22
.161
.151
.120
.080
.044
.020
.007
.002
n = 25, p = .80
.585
b) x > 20
from a): P(x > 20) = P(x = 21) + P(x = 22) + P(x = 23) +
P(x = 24) + P(x = 25) =
.007 + .002 + .000 + .000 + .000 = .009
Chapter 5: Discrete Distributions
11
c) P(x < 10)
from Table A.2, x
Prob.
9
8
7
<6
.009
.003
.001
.000
.013
5.12
The highest probability values are for x = 15, 16, 14, 17, 13, 18, and 12.
The expected value is 25(.60) = 15. The standard deviation is 2.45.
15 + 2(2.45) = 15 + 4.90 gives a range that goes from 10.10 to 19.90. From
table A.2, the sum of the probabilities of the values in this range (11 through
19) is .936 or 93.6% of the values which compares quite favorably with the
95% suggested by the empirical rule.
5.13
n = 15 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) =
Chapter 5: Discrete Distributions
.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)
5.14
n = 18
a)
b)
p =.30
µ = 18(.30) = 5.4
p = .34
µ = 18(.34) = 6.12
P(x > 8)
n = 18
from Table A.2
x
8
9
10
11
12
Prob
.081
.039
.015
.005
.001
.141
p = .30
12
Chapter 5: Discrete Distributions
c) n = 18
13
p = .34
P(2 < x < 4) = P(x = 2) + P(x = 3) + P(x = 4) =
2
16
18C2(.34) (.66)
+
3
15
18C3(.34) (.66)
+
4
14
18C4(.34) (.66)
=
.0229 + .0630 + .1217 = .2076
d) n = 18
p = .30
0
18
18C0(.30) (.70)
n = 18
= .00163
p = .34
0
18
18C0(.34) (.66)
x=0
x=0
= .00056
The probability that none are in the $500,000 to $1,000,000 is higher because
there is a smaller percentage in that category which is closer to zero.
5.15
a) Prob(x=5 = 2.3)=
(2.35)(e-2.3) = (64.36343)(.1002588) = .0538
5!
(120)
b) Prob(x=2 = 3.9) =
(3.92)(e-3.9) = (15.21)(.02024) = .1539
2!
(2)
c) Prob(x < 3 = 4.1) =
Prob(x=3) + Prob(x=2) + Prob(x=1) + Prob(x=0) =
(4.13)(e-4.1) = (68.921)(.016574) = .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
Chapter 5: Discrete Distributions
14
+ (4.10)(e-4.1) = (1)(.016573) = .0166
0!
1
.1904 + .1393 + .0679 + .0166 = .4142
d) Prob(x=0 = 2.7) =
(2.70)(e-2.7) = (1)(.0672) = .0672
0!
1
e) Prob(x=1  = 5.4)=
(5.41)(e-5.4) = (5.4)(.0045) = .0244
1!
1
f) Prob(4 < x < 8  = 4.4) =
Prob(x=5  = 4.4) + Prob(x=6  = 4.4) + Prob(x=7  = 4.4)=
(4.45)(e-4.4) + (4.46)(e-4.4) + (4.47)(e-4.4) =
5!
6!
7!
(1649.162)(.012277) + (7256.314)(.012277) + (31927.781)(.012277)
120
720
5040
= .1687 + .1237 + .0778 = .3702
5.16
a) Prob(x=6  = 3.8) = .0936
b) Prob(x>7  = 2.9):
x
8
9
10
11
12
Prob
.0068
.0022
.0006
.0002
.0000
x > 7 .0098
Chapter 5: Discrete Distributions
15
c) Prob(3 < x < 9  = 4.2)=
x
3
4
5
6
7
8
9
3<x<9
Prob
.1852
.1944
.1633
.1143
.0686
.0360
.0168
.7786
d) Prob(x=0  = 1.9) = .1496
e) Prob(x < 6  = 2.9)=
x
0
1
2
3
4
5
6
x< 6
Prob
.0050
.1596
.2314
.2237
.1622
.0940
.0455
.9214
f) Prob(5 < x < 8  = 5.7) =
x
6
7
8
5<x < 8
5.17 a)  = 6.3
Prob
.1594
.1298
.0925
.3817
mean = 6.3
x
0
1
2
3
4
5
6
Standard deviation =
Prob
.0018
.0116
.0364
.0765
.1205
.1519
.1595
6.3 = 2.51
Chapter 5: Discrete Distributions
7
8
9
10
11
12
13
14
15
16
17
18
19
b)  = 1.3
mean = 1.3
x
0
1
2
3
4
5
6
7
8
9
16
.1435
.1130
.0791
.0498
.0285
.0150
.0073
.0033
.0014
.0005
.0002
.0001
.0000
standard deviation =
Prob
.2725
.3542
.2303
.0998
.0324
.0084
.0018
.0003
.0001
.0000
1.3 = 1.14
Chapter 5: Discrete Distributions
c)  = 8.9
mean = 8.9
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
17
standard deviation =
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
Chapter 5: Discrete Distributions
d)  = 0.6
mean = 0.6
x
0
1
2
3
4
5
6
5.18
 = 2.84 minutes
a) Prob(x=6  = 2.8)
from Table A.3 .0407
18
standard deviation =
Prob
.5488
.3293
.0988
.0198
.0030
.0004
.0000
0.6 = .775
Chapter 5: Discrete Distributions
19
b) Prob(x=0  = 2.8) =
from Table A.3 .0608
c) Unable to meet demand if x > 44 minutes:
x
5
6
7
8
9
10
11
x>4
.1523
Prob.
.0872
.0407
.0163
.0057
.0018
.0005
.0001
.1523
probability of being unable to meet the demand.
Probability of meeting the demand = 1 - (.1523) = .8477
15.23% of the time a second window will need to be opened.
d)  = 2.8 arrivals4 minutes
Prob(x=3) arrivals2 minutes = ??
Lambda must be changed to the same interval (½ the size)
New lambda=1.4 arrivals2 minutes
Prob(x=3)  =1.4) = from Table A.3 = .1128
Prob(x > 5 8 minutes) = ??
Lambda must be changed to the same interval(twice the size):
New lambda= 5.6 arrivals8 minutes
Prob(x > 5   = 5.6):
Chapter 5: Discrete Distributions
From Table A.3:
x
5
6
7
8
9
10
11
12
13
14
15
16
17
x> 5
20
Prob.
.1697
.1584
.1267
.0887
.0552
.0309
.0157
.0073
.0032
.0013
.0005
.0002
.0001
.6579
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)
 = 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
Chapter 5: Discrete Distributions
21
e) P(x = 8  15 minutes)
 = 10.5  15 minutes
P(x = 8  15 minutes) =
5.20
X  e  
X!

(10.58 )(e10.5 )
8!
= .1009
 = 5.6 days3 weeks
a) Prob(x=0   = 5.6):
from Table A.3 = .0037
b) Prob(x=6   = 5.6):
from Table A.3 = .1584
c) Prob(x > 15   = 5.6):
x
15
16
17
x > 15
Prob.
.0005
.0002
.0001
.0008
Because this probability is so low, if it actually occurred, the researcher would
actually have to question the Lambda value as too low for this period.
5.21
 = 0.6 trips 1 year
a) Prob(x=0   = 0.6):
from Table A.3 = .5488
b) Prob(x=1   = 0.6):
from Table A.3 = .3293
c) Prob(x > 2   = 0.6):
Chapter 5: Discrete Distributions
from Table A.3
22
x
2
3
4
5
6
Prob.
.0988
.0198
.0030
.0004
.0000
x > 2
.1220
d) Prob(x < 3  3 year period):
The interval length has been increased (3 times)
New Lambda =  = 1.8 trips3 years
Prob(x < 3   = 1.8):
from Table A.3
x
0
1
2
3
x < 3
e) Prob(x=4  6 years):
The interval has been increased (6 times)
New Lambda =  = 3.6 trips6 years
Prob(x=4   = 3.6):
from Table A.3 = .1912
5.22
 = 1.2 collisions4 months
a) Prob(x=0   = 1.2):
from Table A.3 = .3012
b) Prob(x=2 2 months):
Prob.
.1653
.2975
.2678
.1607
.8913
Chapter 5: Discrete Distributions
23
The interval has been decreased (by ½)
New Lambda =  = 0.6 collisions2 months
Prob(x=2   = 0.6):
from Table A.3 = .0988
c) Prob (x < 1 collision6 months):
The interval length has been increased (by 1.5)
New Lambda =  = 1.8 collisions6 months
Prob(x < 1  = 1.8):
from Table A.3
x
0
1
Prob.
.1653
.2975
x< 1
.4628
The result is likely to happen almost half the time (46.26%). Ship channel and
weather conditions are about normal for this period. Safety awareness is
about normal for this period. There is no compelling reason to reject the
lambda value of 0.6 collisions per 4 months based on an outcome of 0 or 1
collisions per 6 months.
5.23
 = 1.2 penscarton
a) Prob(x=0   = 1.2):
from Table A.3 = .3012
b) Prob(x > 8   = 1.2):
from Table A.3 = .0000
c) Prob(x > 3   = 1.2):
from Table A.3
x
4
5
Prob.
.0260
.0062
Chapter 5: Discrete Distributions
5.24
n = 100,000
24
6
7
8
.0012
.0002
.0000
x>3
.0336
p = .00004
Prob(x > 7 n = 100,000 p = .00004):
 = µ = np = 100,000(.00004) = 4.0
Since n > 20 and np < 7, the Poisson approximation to this binomial problem is
close enough.
Prob(x > 7   = 4):
Using Table A.3
x
7
8
9
10
11
12
13
14
Prob.
.0595
.0298
.0132
.0053
.0019
.0006
.0002
.0001
x > 7
.1106
x
11
12
13
14
Prob.
.0019
.0006
.0002
.0001
x > 10
.0028
Prob(x>10   = 4):
Using Table A.3
Since getting more than 10 is a rare occurrence, this particular geographic region
appears to have a higher average rate than other regions. An investigation of
particular characteristics of this region might be warranted.
Chapter 5: Discrete Distributions
5.25
p = .009
25
n = 200
Use the Poisson Distribution:
 = np = 200(.009) = 1.8
P(x > 6) from Table A.3 =
P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + . . . =
.0078 + .0020 + .0005 + .0001 = .0104
P(x > 10) = .0000
P(x = 0) = .1653
P(x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P( x = 3) + P(x = 4) =
.1653 + .2975 + .2678 + .1607 + .0723 = .9636
5.26
n = 300,
p = .01,
 = n(p) = 300(.01) = 3
a) Prob(x = 5):
Using  = 3 and Table A.3 = .1008
b) Prob (x < 4) = Prob.(x = 0) + Prob.(x = 1) + Prob.(x = 2) + Prob.(x = 3) =
.0498 + .1494 + .2240 + .2240 = .6472
c) The expected number = µ =  = 3
5.27
a) Prob(x = 3 N = 11, A = 8, n = 4)
8
C3 3 C1 (56)(3)
= .5091

330
11 C 4
b) Prob(x < 2)N = 15, A = 5, n = 6)
Prob(x = 1) + Prob (x = 0) =
5
C1 10 C5
+
15 C 6
5
C0 10 C6
(5)( 252) (1)( 210)

=
5005
5005
15 C 6
Chapter 5: Discrete Distributions
26
.2517 + .0420 = .2937
c) Prob(x=0 N = 9, A = 2, n = 3)
2
C0 7 C3 (1)(35)
= .4167

C
84
9 3
d) Prob(x > 4 N = 20, A = 5, n = 7) =
Prob(x = 5) + Prob(x = 6) + Prob(x = 7) =
5
C5 15 C 2
+
20 C 7
5
C 6 15 C1
+
20 C 7
5
C7 15 C0
=
20 C 7
(1)(105)
+ 5C6 (impossible) + 5C7(impossible) = .0014
77520
5.28
N = 19 n = 6
a) P(x = 1 private)
11
A = 11
C1 8 C5 (11)(56)
= .0227

27,132
19 C 6
b) P(x = 4 private)
11
C 4 8 C 2 (330)( 28)
= .3406

C
27
,
132
19 6
c) P(x = 6 private)
11
C6 8 C0 (462)(1)
= .0170

27,132
19 C 6
d) P(x = 0 private)
11
C0 8 C6 (1)( 28)
= .0010

27,132
19 C 6
Chapter 5: Discrete Distributions
5.29
N = 17
A=8
8
a) P(x = 0) =
b) P(x = 4) =
8
n=4
C 0 9 C 4
(1)(126)
=
= .0529
2380
17 C 4
C 4 9 C 0
(70)(1)
=
= .0294
2380
17 C 4
c) P(x = 2 non computer) =
5.30
N = 20
27
9
C 2 8 C 2
(36)( 28)
=
= .4235
2380
17 C 4
A = 16 white
a) Prob(x = 4 white) =
b) Prob(x = 4 red) =
4
16
N - A = 4 red
C 4  4 C1
(1820)( 4)
=
15504
20 C 5
n=5
= .4696
C 4 16 C1
(1)(16)
=
= .0010
15504
20 C 5
C5 16 C0
= .0000 because 4C5 is impossible to determine
20 C5
The participant cannot draw 5 red beads if there are only 4 to draw from.
c) Prob(x = 5 red) =
5.31
N = 10
4
n=4
a) A = 3 x = 2
3
C 2 7 C 2 (3)( 21)
= .30

210
10 C 4
b) A = 5 x = 0
5
C0 5 C 4 (1)(5)
= .0238

210
10 C 4
c) A = 5 x = 3
5
C3 5 C1 (10)(5)
= .2381

210
10 C 4
Chapter 5: Discrete Distributions
5.32
N = 16
A = 4 defective
a) Prob(x = 0) =
4
b) Prob(x = 3) =
4
28
n=3
C0 12 C3 (1)( 220)
= .3929

560
16 C 3
C3 12 C0 (4)(1)
= .0071

560
16 C 3
c) Prob(x > 2) = Prob(x=2) + Prob(x=3) =
4
C 2 12 C1
+ .0071 (from part b.) =
16 C 3
(6)(12)
+ .0071 = .1286 + .0071 = .1357
560
d) Prob(x < 1) = Prob(x=1) + Prob(x=0) =
4
5.33
C1 12 C 2
(4)(66)
+ .3929 (from part a.) =
+ .3929 = .4714 + .3929 = .8643
560
16 C 3
N = 18
A = 11 Hispanic
n=5
Prob(x < 1) = Prob(1) + Prob(0) =
11
C1 7 C 4
+
18 C 5
11
C 0 7 C5
(11)(35) (1)( 21)

=
= .0449 + .0025 = .0474
8568
8568
18 C 5
It is fairly unlikely that these results occur by chance. A researcher might
investigate further the causes of this result. Were officers selected based on
leadership, years of service, dedication, prejudice, or what?
5.34
a) Prob(x=4 n = 11 and p = .23)
4
7
11C4(.23) (.77)
= 330(.0028)(.1605) = .1482
b) Prob(x > 1n = 6 and p = .50) =
1 - Prob(x < 1) = 1 - Prob(x = 0) =
1-6C0(.50)0(.50)6 = 1-(1)(1)(.0156) = .9844
Chapter 5: Discrete Distributions
29
c) Prob(x > 7 n = 9 and p = .85) = Prob(x = 8) + Prob(x = 9) =
8
1
9C8(.85) (.15)
+ 9C9(.85)9(.15)0 =
(9)(.2725)(.15) + (1)(.2316)(1) = .3679 + .2316 = .5995
d) Prob(x < 3 n = 14 and p = .70) =
Prob(x = 3) + Prob(x = 2) + Prob(x = 1) + Prob(x = 0) =
3
11
14C3(.70) (.30)
+ 14C2(.70)2(.30)12 +
1
13
14C1(.70) (.30)
+ 14C0(.70)0(.30)14 =
(364)(.3430)(.00000177) + (91)(.49)(.000000047)=
(14)(.70)(.00000016) + (1)(1)(.000000047) =
.0002 + .0000 + .0000 + .0000 = .0002
5.35
a) Prob(x = 14 n = 20 and p = .60) = .124
b) Prob(x < 5 n = 10 and p =.30) =
Prob(x = 4) + Prob(x = 3) + Prob(x = 2) + Prob(x = 1) + Prob(x=0) =
x
0
1
2
3
4
x<5
Prob.
.028
.121
.233
.267
.200
.849
c) Prob(x > 12 n = 15 and p = .60) =
Prob(x = 12) + Prob(x = 13) + Prob(x = 14) + Prob(x = 15)
x
12
13
Prob.
.063
.022
Chapter 5: Discrete Distributions
14
15
x > 12
30
.005
.000
.090
d) Prob(x > 20 n = 25 and p = .40) = Prob(x = 21) + Prob(x = 22) +
Prob(x = 23) + Prob(x = 24) + Prob(x=25) =
x
21
22
23
24
25
x > 20
Prob.
.000
.000
.000
.000
.000
.000
5.36
a) Prob(x=4   = 1.25)
(1.254)(e-1.25) = (2.4414)(.2865) = .0291
4!
24
b) Prob(x < 1  = 6.37) = Prob(x = 1) + Prob(x = 0) =
(6.37)1(e-6.37) + (6.37)0(e-6.37) = (6.37)(.0017) + (1)(.0017) =
1!
0!
1
1
.0109 + .0017 = .0126
c) Prob(x > 5  = 2.4) = Prob(x = 6) + Prob(x = 7) + ... =
(2.4)6(e-2.4) + (2.4)7(e-2.4) + (2.4)8(e-2.4) + (2.4)9(e-2.4) + (2.4)10(e-2.4) + ...
6!
7!
8!
9!
10!
.0241 + .0083 + .0025 + .0007 + .0002 = .0358
for values x > 11 the probabilities are each .0000 when rounded off to 4
decimal places.
Chapter 5: Discrete Distributions
5.37
31
a) Prob(x = 3   = 1.8) = .1607
b) Prob(x < 5  = 3.3) =
Prob(x = 4) + Prob(x = 3) + Prob(x = 2) + Prob(x = 1) + Prob(x = 0) =
x
0
1
2
3
4
x<5
Prob.
.0369
.1217
.2008
.2209
.1823
.7626
c) Prob(x > 3  = 2.1) =
x
3
4
5
6
7
8
9
10
11
x> 5
Prob.
.1890
.0992
.0417
.0146
.0044
.0011
.0003
.0001
.0000
.3504
d) Prob(2 < x < 5  = 4.2) = Prob(x=3) + Prob(x=4) + Prob(x=5) =
5.38
x
3
4
5
Prob.
.1852
.1944
.1633
2<x <5
.5429
a) Prob(x = 3 N = 6, n = 4, A = 5) =
5
C3 1 C1 (10)(1)
= .6667

C
15
6 4
Chapter 5: Discrete Distributions
32
b) Prob(x < 1 N = 10, n = 3, A = 5) = Prob(x = 1) + Prob(x = 0) =
5
C1 5 C 2
+
10 C 3
5
C 0 5 C 3
(5)(10) (1)(10)

=
120
120
10 C 3
= .4167 + .0833 = .5000
c) Prob(x > 2  N = 13, n = 5, A = 3) =
Prob(x=2) + Prob(x=3) Note: only 3 x's in population
3
C 2 10 C3
+
13 C 5
3
C3 10 C 2
(3)(120) (1)( 45)

=
1287
1287
13 C 5
= .2797 + .0350 = .3147
5.39
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 Value = µ = np = 25(.20) = 5
n = 20 p = .40 mutual funds
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
x=8
Expected Number = µ = n p = 20(.40) = 8
Chapter 5: Discrete Distributions
5.40
33
 = 3.2 cars2 hours
a)
Prob(x=3) cars per 1 hour) = ??
The interval has been decreased by ½.
The new  = 1.6 cars1 hour.
Prob(x = 3  = 1.6) = (from Table A.3) .1378
b) Prob(x = 0 cars per ½ hour) = ??
The interval has been decreased by ¼ the original amount.
The new  = 0.8 cars½ hour.
Prob(x = 0  = 0.8) = (from Table A.3) .4493
c) Prob(x > 5  = 1.6) = (from Table A.3)
x
5
6
7
8
Prob.
.0176
.0047
.0011
.0002
.0236
Either a rare event occurred or perhaps the long-run average, , has changed
(increased).
5.41
N = 32
A = 10
a) P(x = 3) =
10
b) P(x = 6) =
10
c) P(x = 0) =
10
n = 12
C3  22 C9
(120)( 497,420)
=
= .2644
225,792,840
32 C12
C6  22 C6
(210)(74,613)
=
= .0694
225,792,840
32 C12
C0  22 C12
(1)(646,646)
=
= .0029
225,792,840
32 C12
Chapter 5: Discrete Distributions
34
d) A = 22
P(7 < x < 9) =
=
22
C7 10 C5
+
32 C12
22
C8 10 C 4
+
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
5.42
 = 1.4 defects 1 lot
If x > 3, buyer rejects
If x < 3, buyer accepts
Prob(x < 3  = 1.4) = (from Table A.3)
5.43
x
0
1
2
3
Prob.
.2466
.3452
.2417
.1128
x<3
.9463
a) n = 20 and p = .25
The expected number = µ = np = (20)(.25) = 5.00
b) Prob(x < 1 n = 20 and p = .25) =
Prob(x = 1) + Prob(x = 0) = 20C1(.25)1(.75)19 + 20C0(.25)0(.75)20
= (20)(.25)(.00423) + (1)(1)(.0032) = .0212 +. 0032 = .0244
Since the probability is so low, the population of your state may have a lower
percentage of chronic heart conditions than those of other states.
5.44
a) Prob(x > 7 n = 10 and p = .70) = (from Table A.2):
x
8
9
10
Prob.
.233
.121
.028
x>7
.382
Chapter 5: Discrete Distributions
35
Expected number = µ = np = 10(.70) = 7
b) Expected number = µ = np = 15 (1/3) = 5
Prob(x=0 n = 15 and p = 1/3) =
0
15
15C0() ()
= .0023
c) Prob(x = 7 n = 7 and p = .53) = 7C7(.53)7(.47)0 = .0117
Probably the 53% figure is too low for this population since the probability of
this occurrence is so low (.0117).
5.45
n = 12
a.) Prob(x = 0 long hours):
p = .20
0
12
12C0(.20) (.80)
= .0687
b.) Prob(x > 6) long hours):
p = .20
Using Table A.2:
.016 + .003 + .001 = .020
c) Prob(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
5.46
n = 100,000
p = .000014
Worked as a Poisson:
 = np = 100,000(.000014) = 1.4
a) P(x = 5):
from Table A.3 = .0111
b) P(x = 0):
Chapter 5: Discrete Distributions
36
from Table A.3 = .2466
c) P(x > 6):
x
7
8
5.47
(from Table A.3)
Prob
.0005
.0001
.0006
Prob(x < 3) n = 8 and p = .60) = (from Table A.2)
x
0
1
2
3
x< 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.
5.48
n = 25
p = .20
a) Prob(x = 8 n = 25 and p = .20) = (from Table A.2)
.062
b) Prob(x > 10) n=25 and p = .20) = (from Table A.2)
x
11
12
13
x > 10
Prob.
.004
.001
.000
.005
c) Since such a result would only occur 0.5% of the time by chance, it is likely
that the analyst's list was not representative of the entire state of Idaho or the
20% figure for the Idaho census is not correct.
5.49
 = 0.6 flats2000 miles
Prob(x = 0  = 0.6) = (from Table A.3) .5488
Chapter 5: Discrete Distributions
37
Prob(x > 3  = 0.6) = (from Table A.3)
x
3
4
5
Prob.
.0198
.0030
.0004
x>3
One trip is independent of the other.
Let F = flat tire and NF = no flat tire
P(NF1  NF2) = P(NF1)  P(NF2)
P(NF) = .5488
P(NF1  NF2) = (.5488)(.5488) = .3012
5.50
N = 25
n=8
a) P(x = 1 in NY)
4
A=4
C1  21 C7
(4)(116,280)
=
= .4300
1,081,575
25 C8
b) P(x = 4 in top 10)
10
A = 10
C 4 15 C 4 (210(1365)
= .2650

1,081,575
25 C8
c) P(x = 0 in California)
4
C0  21 C8 (1)( 203,490)
= .1881

1,081,575
25 C8
d) P(x = 3 with M)
3
5.51
A=4
A=3
C3  22 C5 (1)( 26,334)
= .0243

1,081,575
25 C8
N = 24
n=6
A=8
.0232
Chapter 5: Discrete Distributions
a) P(x = 6) =
b) P(x = 0) =
8
38
C6 16 C0
(28)(1)
= .0002

C
134
,
596
24 6
8
C0 16 C6 (1)(8008)
= .0595

134,596
24 C 6
d) A = 16 East Side
P(x = 3) =
5.52
16
C3 8 C3 (560)(56)
= .2330

134,596
24 C 6
n = 25 p = .20
Expected Value = µ = np = 25(.20) = 5
µ = 25(.20) = 5
 = n  p  q  25(.20)(.80) = 2
P(x > 12) = (from Table A.2)
x
13
Prob
.0000
The values for x > 12 are so far away from the expected value that they are very
unlikely to occur.
P(x = 14) = 25C14(.20)14(.80)11 = .000063 which is very unlikely.
If this value (x = 14) actually occurred, one would doubt the validity of the
p = .20 figure or one would have experienced a very rare event.
Chapter 5: Discrete Distributions
5.53
39
 = 2.4 calls1 minute
a) Prob(x = 0  = 2.4) = (from Table A.3) .0907
b) Can handle x < 5 calls
Cannot handle x > 5 calls
Prob(x > 5  = 2.4) = (from Table A.3)
x
6
7
8
9
10
11
x>5
c) Prob(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) Prob(x < 1 calls15 seconds):
The interval has been decreased by ¼.
New Lambda =  = 0.6 calls15 seconds.
Prob(x < 1  = 0.6) = (from Table A.3)
Prob(x = 1) = .3293
Prob(x = 0) = .5488
Prob(x < 1) = .8781
Prob.
.0241
.0083
.0025
.0007
.0002
.0000
.0358
Chapter 5: Discrete Distributions
5.54
n = 160
40
p = .01
Working this problem as a Poisson problem:
a) Expected number = µ = n(p) = 160(.01) = 1.6
b) P(x > 8):
Using Table A.3:
x
8
9
Prob.
.0002
.0000
.0002
x
2
3
4
5
6
P(2 < x < 6)
Prob.
.2584
.1378
.0551
.0176
.0047
.4736
c) P(2 < x < 6):
Using Table A.3:
5.55
p = .005
n = 1,000
 = np = (1,000)(.005) = 5
a) P(x < 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) =
.0067 + .0337 + .0842 + .1404 = .265
b) P(x > 10) = P(x = 11) + P(x = 12) + . . . =
.0082 + .0034 + .0013 + .0005 + .0002 = .0136
c) P(x = 0) = .0067
Chapter 5: Discrete Distributions
5.56
n=8
P = .36
0
8
8C0(.36) (.64)
41
x = 0 Women
= (1)(1)(.0281475) = .0281
It is unlikely that a company would randomly hire 8 physicians from the U.S. pool
and none of them would be female. If this actually happened, the figures might
be used as evidence in a lawsuit.
5.57
N = 32
a) n = 5
10
A = 10
C3  22 C 2 (120)( 231)
= .1377

201,376
32 C5
b) n = 8
6
x=3
x<2
C0  26 C8
+
32 C8
A=6
6
C1  26 C7
+
32 C8
6
C 2  26 C 6
=
32 C8
(1)(1,562,275) (6)(657,800) (15)(38,760)
= .1485 + .3752 + .0553 = .5790


10,518,300
10,518,300
10,518,300
c) n = 5
x=2
p = 3/26 = .1154
2
3
5C2(.1154) (.8846)
5.58
N = 14
= (10)(.013317)(.692215) = .0922
n=4
a) P(x = 4 N = 14, n = 4, A = 10 Northside)
10
C 4  4 C0 (210((1)
= .2098

1001
14 C 4
b) P(x = 4 N = 14, n = 4, A = 4 West)
4
C 4 10 C0 (1)(1)
= .0010

1001
14 C 4
c) P(x = 2 N = 14, n = 4, A = 4 West)
Chapter 5: Discrete Distributions
4
5.59
42
C 2 10 C 2 (6)( 45)
= .2697

C
1001
14 4
a)  = 3.841,000
3.840  e3.84
P(x = 0) =
= .0215
0!
b)  = 7.682,000
P(x = 6) =
7.686  e7.68 (205,195.258)(.000461975)
= .1317

6!
720
c)  = 1.61,000 and  = 4.83,000
from Table A.3:
P(x < 7) = P(x = 0) + P(x = 1) + . . . + P(x = 6) =
.0082 + .0395 + .0948 + .1517 + .1820 + .1747 + .1398 = .7907
5.60
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.
5.61
This printout contains the probabilities for various values of x from zero to eleven from a
Poisson distribution with  = 2.78. Note that the highest probabilities are at x = 2 and x
= 3 which are near the mean. The probability is slightly higher at x = 2 than at x = 3 even
though x = 3 is nearer to the mean because of the “piling up” effect of x = 0.
5.62
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
Chapter 5: Discrete Distributions
43
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.
5.63
This is the graph of a Poisson Distribution with  = 1.784. Note the high
probabilities at x = 1 and x = 2 which are nearest to the mean. Note also that the
probabilities for values of x > 8 are near to zero because they are so far away
from the mean or expected value.