solution Chapter 4 Mandenhall book

Chapter 4 50

CHAPTER 4 ………………………………………………………….. Discrete Random Variables

4.2 a. p(0) + p(1) + p(2) + p(3) + p(4) = .09 + .30 + .37 + .20 + .04 = 1.00 b. P(y = 3 or 4) = p(3) + p(4) = .20 + .04 = .24 c. P(y < 2) = p(0) + p(1) = .09 + .30 = .39 4.4 a. The list of all possible pairs of beach hotspots is shown below:

Beach Hotspot Pair MBF , CINY MBF , SCA MBF , MBNJ MBF , OCNJ MBF , SLNJ CINY , SCA CINY , MBNJ CINY , OCNJ CINY , SLNJ SCA , MBNJ SCA , OCNJ SCA , SLNJ MBNJ , OCNJ MBNJ , SLNJ OCNJ , SLNJ

b. The probabilities of each of these outcomes should all be equal if a random sampling

technique is employed. Therefore, they all have probability 1/15. c. The value of Y is found by determining the total number of beach hotspots in the

sample with a planar nearshore bar condition.

Discrete Random Variables 51

Beach Hotspot Pair P(sample) YMBF , CINY 1/15 0 MBF , SCA 1/15 0 MBF , MBNJ 1/15 1 MBF , OCNJ 1/15 0 MBF , SLNJ 1/15 1 CINY , SCA 1/15 0 CINY , MBNJ 1/15 1 CINY , OCNJ 1/15 0 CINY , SLNJ 1/15 1 SCA , MBNJ 1/15 1 SCA , OCNJ 1/15 0 SCA , SLNJ 1/15 1 MBNJ , OCNJ 1/15 1 MBNJ , SLNJ 1/15 2 OCNJ , SLNJ 1/15 1

d. The probability distribution for Y is found by grouping similar values of Y together in

the table.

Y = y P(Y) 0 6/15 1 8/15 2 1/15

e. P(Y ≥ 1) = P(1) + P(2) = 8/15 + 1/15 = 9/15 4.6 a. p(1) = p(2) = p(3) = .5 Where 1: relay 1 works 2: relay 2 works 3: relay 3 works The sample events for this experiment as well as the value of y are: y y 1 2 3 3 c1 2 3 2 1 2 c3 2 c1 2 c3 1 1 c2 3 2 c1 c2 3 1 1 c2 c3 1 c1 c2 c3 0

Since p(1) = p(2) = p(3) = .5, each simple event has probability 18

of occurring.

Chapter 4 52

b. P(current flows from A to B) = 7( 1)8

P y ≥ =

4.8 Let y = the number of the three machining conditions with steel material and .25 in drill size

that will detect the flaw. Since two of the eight machining conditions will detect the flaw, we list all possible

combinations of machining conditions which detect the flaw. (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8), (5, 6), (5, 7), (5, 8), (6, 7), (6, 8), (7, 8) There are 28 such combinations. We must assume that each of the 28 combinations are

equally likely. ( 0) (0) 10 / 28P y p= = = ( 1) (1) 15/ 28P y p= = = ( 2) (2) 3/ 28P y p= = = ( 3) (3) 0P y p= = = 4.10 The probability distribution for Y = number of delphacid eggs on a blade of water hyacinth is

shown here:

Y = y 1 2 3 4 P(Y) 0.4 0.54 0.02 0.04

( ) ( ) 1(.4) 2(.54) 3(.02) 4(.04) 1.70E Y yp yμ = = = + + + =∑ If repeated samples of water hyacinth blades were sampled, the average number of delphacid

eggs on the blades would be 1.7 eggs. 4.12 a. To find the probabilities associated with each value of y, we divide the frequency

associated with each value by the total sample size, 743. The probabilities appear in the table:

y 0 1 2 3

p(y) 18

38

38

18

y 0 1 2 3 p(y) 10/28 15/28 3/28 0


First DigitFrequency

of Occurrence Probability 1 109 109 / 743 = .14672 75 75 / 743 = .10093 77 77 / 743 = .10364 99 99 / 743 = .13325 72 72 / 743 = .09696 117 117 / 743 = .15757 89 89 / 743 = .11988 62 62 / 743 = .08349 43 43 / 743 = .0579

Total 743 1.0000

b. ( ) ( ) 1(.1467) 2(.1009) 3(.1036) 4(.1332) 5(.0969) 6(.1575) 7(.1198) 8(.0834) 9(.0579) 4.6485

E y yp yμ = = = + + + +∑+ + + + =

4.14 a. For ARC a1: μ = E(y) = ( )yp y∑ = 0(.05) + 1(.10) + 2(.25) + 3(.60) = 2.4 The mean capacity for ARC a1 is 2.4

For ARC a2: μ = E(y) = ( )yp y∑ = 0(.10) + 1(.30) + 2(.60) = 1.5 The mean capacity for ARC a2 is 1.5

For ARC a3: μ = E(y) = ( )yp y∑ = 0(.10) + 1(.90) = .90 The mean capacity for ARC a3 is .90



For ARC a6: μ = E(y) = ( )yp y∑ = 0(.05) + 1(.25) + 2(.70) = 1.65 The mean capacity for ARC a6 is 1.65 b. For ARC a1: σ2 = E(y − μ)2 = 2( ) ( )y p yμ−∑

= (3 − 2.4)2(.60) + (2 − 2.4)2(.25) + (1 − 2.4)2(.10) + (0 − 2.4)2(.05) = .216 + .040 + .196 + .288 = .74 σ = 2 .74σ = = .8602

For ARC a2: σ2 = E(y − μ)2 = 2( ) ( )y p yμ−∑

= (2 − 1.5)2(.60) + (1 − 1.5)2(.30) + (0 − 1.5)2(.10) = .15 + .075 + .225 = .45 σ = 2 .45σ = = .6708

Chapter 4 54


= (1 − .9)2(.90) + (0 − .9)2(.10) = .009 + .081 = .09 σ = 2 .09σ = = .3


= (1 − .9)2(.90) + (0 − .9)2(.10) = .009 + .081 = .09 σ = 2 .09σ = = .3


= (1 − .9)2(.90) + (0 − .9)2(.10) = .009 + .081 = .09 σ = 2 .09σ = = .3 For ARC a6: σ2 = E(y − μ)2 = 2( ) ( )y p yμ−∑

= (2 − 1.65)2(.70) + (1 − 1.65)2(.25) + (0 − 1.65)2(.05) = .08575 + .105625 + .136125 = .3275 σ = 2 .09σ = = .5723 4.16 Since the cost of firing each pin is $200, we get the following probability distribution for the

cost:

cost $200 $400 $600

p(cost) 610

310

110

a. [ ] cost (cost)E cost pμ = =∑

= 6 3 1$200 $400 $60010 10 10⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= $300

b. 2 2 2( ) (cost mean) (cost)E y pσ μ⎡ ⎤= − = −⎣ ⎦ ∑

= 2 2 26 3 1(200 300) (400 300) (600 300)10 10 10⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 18,000 c. We would expect the inspection cost to fall within a range of 2 .μ σ±

2 18000 $134.164σ σ= = = 2 300 2(134.164) 300 268.328μ σ± ⇒ ± ⇒ ± ⇒ $31.672 to $568.328


4.18 From Exercise 4.4, the probability distribution for Y is: Using Theorem 4.4, 2 2 2( )E Yσ μ= − ( ) ( ) 0(6 /15) 1(8/15) 2(1/15) 10 /15 .6667E Y yp yμ = = = + + = =∑ 2 2 2 2 2( ) ( ) 0 (6 /15) 1 (8 /15) 2 (1/15) 12 /15 0.80E Y Y p Y= = + + = =∑ 2 2 2 2( ) 0.80 .6667 0.3556E Yσ μ= − = − =

2 0.3556 0.5963σ σ= = = We expect the number of beach hotspots in the sample with a planar nearshore bar condition

to fall between 2 .6667 2(.5963) .6667 1.1926μ σ± ⇒ ± ⇒ ± ( 0.5259,1.8593).⇒ − 4.20 From Exercise 4.9, the probability distribution of y is: c = 200 + 100y Using Theorems 4.1, 4.2, and 4.3, ( ) (200 100 ) (200) (100 ) 200 100 ( )E c E y E E y E y= + = + = + Where

all

( ) ( ) 1(.6) 2(.3) 3(.1) .6 .6 .3 .15y

E y yp yμ= = = + + = + + =∑

Thus, ( )E c = 200 + 100(.15) = 350 ( ) ( )2 2( ) 200 100 350V c E c E yμ⎡ ⎤ ⎡ ⎤= − = + −⎣ ⎦ ⎣ ⎦

= ( ) ( )2 2

all

100 150 100 150 ( )y

E y y p y⎡ ⎤− = −⎣ ⎦ ∑

= ( ) ( ) ( )2 2 2100(1) 150 (.6) 100(2) 150 (.3) 100(3) 150 (.1)− + − + − = 1500 + 750 + 2250 = 4500

Y = y P(Y) 0 6/15 1 8/15 2 1/15

y p(y) 1 .6 2 .3 3 .1

Chapter 4 56

4.22 Theorem 4.2 says: ( ) ( )E cy cE y= if c is a constant. From Definition 4.5,

all all

( ) ( ) ( ) ( )y y

E cy cyp y c yp y cE y= = =∑ ∑

4.24 a. The experiment consists of n = 20 trials. Each trial results in an S (guppy survived after

5 days) or an F (guppy did not survive after 5 days). The probability of success, p, is .60 and q = 1 − .60 = .40. We assume the trials are independent. Therefore, y has a binomial distribution with n = 20 and p = .60. The probability distribution for y is:

2020( ) .60 (.40)y n y y yn

p y p qy y

− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

b. 7 20 7 7 1320 20!( 7) .60 (.40) .60 (.40) 0.01467 7!13!

p Y −⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

c. ( 10) 1 ( 9) 1 0.1275 0.8725p Y P Y≥ = − ≤ = − = from Table 2 in Appendix B.

4.26 a. The experiment consists of n = 10 trials. Each trial results in an S (bridge will have an

inspection rating of 4 or below in 2020) or an F (bridge does not have an inspection rating of 4 or below in 2020). The probability of success, p, is .09 and q = 1 − .09 = .91. We assume the trials are independent. Therefore, y has a binomial distribution with

n = 10 and p = .09. The probability distribution for y is:

1010( ) .09 (.91)y n y y yn

p y p qy y

− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( 3) 1 ( 2) 1 [ ( 0) ( 1) ( 2)]p Y P Y P y P y P y≥ = − ≤ = − = + = + =

0 10 0 1 10 1 2 10 210 10 101 .09 (.91) .09 (.91) .09 (.91)

0 1 2− − −⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= − + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

1 0.9460 .0540= − =

b. Since the probability of observing this outcome is small, we would question the validity of the engineer’s forecast of 9%.

4.28 a. Let y = number of beach trees damaged by fungi in 20 trials. Then y is a binomial

random variable with n = 20 and p = .25.


0 20 1 19 2 18 9 11

( 10) ( 0) ( 1) ( 9)20 20 20 20

.25 .75 .25 .75 .25 .75 .25 .750 1 2 9

.0032 .0211 .0669 .1339 .1897 .2023 .1686 .1124 .0609 .0271

P y P y P y P y< = = + = + ⋅⋅⋅ + =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + ⋅⋅⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + + + + + + + + + .9861=

16 4 17 3 18 2 20 0

b. ( 15) ( 16) ( 17) ( 20)20 20 20 20

.25 .75 .25 .75 .25 .75 .25 .7516 17 18 20

.000000356 .000000027 .0

P y P y P y P y> = = + = + ⋅⋅⋅ + =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + ⋅⋅⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= + + 00000001 0 0 .000000384+ + =

c. To find the number of trees that we expect to be damaged by fungi, we find

n pμ = ⋅ = 20(.25) = 5 4.30 Let y = the number of foreign students in a random sample of 25 engineering students who

recently earned their Ph.D. Then y is a binomial random variable with n = 25 and p = .70.

a. 10 25 10 10 1525 25!( 10) (.70) (.30) (.70) (.30) .001324910 10!15!

P y −⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

b. ( 5)P y ≤ =.0000 from Table 2 in Appendix B. c. 25(.7) 17.5npμ = = =

2 25(.7)(.3) 5.25npqσ = = = 5.25 2.29σ = =

d. We expect the number of foreign students in the 25 sampled engineering students to

fall between 2μ σ− and 2μ σ+ . 2μ σ− = 17.5 − 2(2.29) = 17.5 − 4.58 = 12.92 ⇒ 13 2μ σ+ = 17.5 + 2(2.29) = 17.5 + 4.58 = 21.98 ⇒ 21 4.32 a. The experiment consists of n = 10 trials. Each trial results in an S (contain shipping

order files in their computerized data base) or an F (do not contain shipping order files in their computerized data base). The probability of success, p, is .99 and q = 1 − .99 = .01. We assume the trials are independent. Therefore, y has a binomial distribution with n = 10 and p = .99.

b. 7 10 7 7 310 10!( 7) .99 (.01) .99 (.01) 0.00011187 7!3!

p Y −⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

c. ( 5) 1 ( 5) 1 .0000 1p Y p Y> = − ≤ = − = from Table 2 in Appendix B.

Chapter 4 58

d. 10(.99) 9.9npμ = = = 2 10(.99)(.01) .099npqσ = = =

2 .099 0.3146σ σ= = = We expect most of the observations to fall within 3 9.9 3(.3146)μ σ± ⇒ ± 9.9 0.9439 (8.9561,10.8439).⇒ ± ⇒ 4.34 a. A sample of n = 4 particles are released. A “success” is when the particle is absorbed

into the inner duct wall. ( )P s = .84. The random variable y is a binomial random variable with n = 4, p = .84.

4 0 4 0 44 4!( 4) (4) (.84) (1 .84) (.84) (.16) 1(.84) (1) .49794 0!4!

P y p⎛ ⎞

= = = − = = =⎜ ⎟⎝ ⎠

3 4 3 3 14 4!( 3) (3) (.84) (.16) (.84) (.16) .37933 1!3!

P y p −⎛ ⎞= = = = =⎜ ⎟

⎝ ⎠

b. Letting a “success” be a particle reflected by the inner wall duct means y is a binomial

random variable with n = 20, p = .16. P(at least 10 are released) = ( 10)P y ≥ = p(10) + p(11) + p(12) + ⎭ + p(19) + p(20)

= 10 20 10 11 20 11 .20 20 2020 20 20(.16) (.84) (.16) (.84) (.16) (.84)

10 11 20− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= .0004267

P(exactly 10 are released) = 10 20 1020( 10) (.16) (.84) .0003553

10P y −⎛ ⎞

= = =⎜ ⎟⎝ ⎠

4.36 1 2 2( )0 1 2

n n n n nn n n nq p q q p q p p

n− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ = + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= (0) (1) (2) ( )p p p p n+ + + +

= 0

( ) 1n

y

p y=

=∑


4.38 ( ) 2 2 21 ( ) ( ) ( ) ( )E y y E y y E y E y E y μ⎡ − ⎤ = − = − = −⎣ ⎦ From Exercise 4.37, ( ) 21E y y npq μ μ⎡ − ⎤ = + −⎣ ⎦

2 2( )npq E yμ μ μ⇒ + − = − 2 2( )E y npq μ⇒ = + 4.40 This experiment consists of 100 identical trials. There are four possible outcomes on each

trial with the probabilities indicated in the table below. Assuming the trials are independent, this is a multinomial experiment with n = 100, k = 4, p1 = .40, p2 = .54, p3 = .02, and p4 = .04.

Result Proportion One Egg 0.40 Two Eggs 0.54 Three Eggs 0.02 Four Eggs 0.04

P(50, 50, 0, 0) = 50 50 0 0100! (.40) (.54) (.02) (.04) 0.000053350!50!0!0!

=

4.42 a. This experiment consists of 100 identical trials. There are four possible outcomes on

each trial with the probabilities indicated in the table below. Assuming the trials are independent, this is a multinomial experiment with n = 100, k = 4, p1 = .29, p2 = .32,

p3 = .09, and p4 = .30.

Job Match ProportionYes, My job is a close match 0.29 No. It’s engineering, but not what I studied 0.32 Job is not engineering related 0.09 Currently unemployed 0.30

P(40, 30, 10, 20) = 40 30 10 20100! (.29) (.32) (.09) (.30) 0.000026640!30!10!20!

=

b. The number of readers we expect to answer “Yes. My job is a close match” is 1 1 100(.29) 29npμ = = = c. The number of readers we expect to answer “No. It’s engineering, but not what I

studied” is 2 2 100(.32) 32npμ = = = 4.44 a. This experiment consists of 200 identical trials. There are seven possible outcomes on

each trial with the probabilities indicated in the table below. Assuming the trials are independent, this is a multinomial experiment with n = 200, k = 7, p1 = .22, p2 = .20,

p3 = .17, p4 = .17, p5 = .12, p6 = .05, and p7 = .07

Chapter 4 60

Day of the week ProportionMonday 0.22 Tuesday 0.20 Wednesday 0.17 Thursday 0.17 Friday 0.12 Saturday 0.05 Sunday 0.07

P(50, 50, 30, 30, 20, 10, 10)

= 50 50 30 30 20 10 10200! (.22) (.20) (.17) (.17) (.12) (.05) (.07)50!50!30!30!20!10!10

0.00000004= b. The experiment consists of n = 200 trials. Each trial results in an S (detected on a

Monday) or an F (not detected on a Monday). The probability of success, p, is .22 and q = 1 − .22 = .78. We assume the trials are independent. Therefore, y has a binomial

distribution with n = 200 and p = .22. ( 50) 1 ( 49) 1 0.82664 0.17336p Y p Y≥ = − ≤ = − = using a computer program to find the

cumulative binomial probability. 4.46 This experiment consists of 10 identical trials. There are 3 paths available on each trial with

path probabilities of .25, .30, and .45. Assuming the trials are independent, this is a multinomial experiment with n = 10, k = 3, 1 2 3.25, .30, .45.p p p= = =

a. 2 4 410!(2, 4, 4) (.25) (.30) (.45) .065392!4!4!

P = =

b. 2 2( ) 10(.30) 3E y n p= ⋅ = = 2

2 2 2 2( ) (1 ) (1 ) 10(.30)(.70) 2.1i iV y n p p n p pσ = = ⋅ − = ⋅ − = = 2

2 2 2.1 1.45σ σ= = = We expect 2 ,y the number of times path two is used, to fall within two standard

deviations of its mean. 2 3 2(1.45) 3 2.90μ σ± ⇒ ± ⇒ ± ⇒ .10 to 5.90


4.48 [ ]2 2 1 22 2 2( ) ( ) ( )

0 1 2a b c a a b c b c

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + = + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 2 2 22 2 2 2 2 2 20 1 1 2 0 1 2

a ab ac b bc c⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= 2 1 1 1 1 2 22! 2! 2! 2! 2 2!2!0! 1!1! 1!1! 0!2! 1!1! 2!0!

a a b a c b bc c+ + + + +

= 2 0 0 1 1 0 1 0 1 0 2 02! 2! 2! 2!2!0!0! 1!1!0! 1!0!1! 0!2!0!

a b c a b c a b c a b c+ + +

0 1 1 0 0 02! 2!0!1!1! 0!0!2!

a b c a b c+ +

Substituting 1 2 3, ,a p b p c p= = = yields: = P(2, 0, 0) + P(1, 1, 0) + P(1, 0, 1) + P(0, 2, 0) + P(0, 1, 1) + P(0, 0, 2) = 1 4.50 Let S = a charged shower particle is observed and F = a charged shower particle is not

observed. We are given that the number of charged particles that must be observed in order to detect r charged shower particles follows a negative binomial distribution with p = .75. The probability that five charged particles must be observed in order to detect three charged shower particles is:

3 5 3 3 2 3 25 1 4 4!( 5) (.75) (.25) (.75) (.25) (.75) (.25) 0.15823 1 2 2!2!

P Y −−⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

4.52 Let y = the number of shuttle flights until a “critical item” fails. Then y is a geometric

random variable with p = 1 .63

a. 1 1 631/ 63p

μ = = =

b. 22 2

62 / 63(1/ 63)

qp

σ = = = 3906

2 3906σ σ= = = 62.5 c. The interval 2μ σ± will capture the number of missions before a critical item failure

occurs with probability of approximately .95. 2 63 2(62.5) 63 125 ( 62, 188) (0, 188)μ σ± ⇒ ± ⇒ ± ⇒ − ⇒

Chapter 4 62

4.54 a. Y has a geometric distribution with p = .48.

2

3

4

5

6

( 1) (.48) 0.48( 2) (.48)(.52) 0.2496

( 3) (.48)(.52) 0.1298

( 4) (.48)(.52) 0.0675

( 5) (.48)(.52) 0.0351

( 6) (.48)(.52) 0.0182

( 7) (.48)(.52) 0.0095

P YP Y

P Y

P Y

P Y

P Y

P Y

= = == = =

= = =

= = =

= = =

= = =

= = =

b. The formula for P(Y = y) = pqy-1 y = 1, 2, 3, … .

c. 1 1.48p

μ = = = 2.083

22 2

.52.48

qp

σ = = = 2.2569

2.2569σ = = 1.502 d. We know from Chebysheff’s Theorem that at least 3/4 of the observations are within 2

standard deviations of the mean. 2 2.083 2(1.502) 2.083 3.004 ( .921, 5.087)μ σ± ⇒ ± ⇒ ± ⇒ − Since we know y cannot be negative, the interval should be (0, 5.087). 4.56 Let S = particle is reflected and F = particle is absorbed. From Exercise 4.34, p = .16 and q = .84. Let y = number of trials until the second particle is reflected. Then y has a negative

binomial distribution with r = 2. ( 5) 1 ( 5) 1 (2) (3) (4) (5)P y P y p p p p> = − ≤ = − − − −

= 2 2 2 2 3 22 1 3 11 .16 (.84) .16 (.84)

2 2 3 2− −− −⎛ ⎞ ⎛ ⎞

− −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

2 4 2 2 5 24 1 5 1.16 (.84) .16 (.84)

4 2 5 2− −− −⎛ ⎞ ⎛ ⎞

− −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

= 2 0 2 2 21 2 31 .16 (.84) .16 (.84) .16 (.84)

0 1 2⎛ ⎞ ⎛ ⎞ ⎛ ⎞

− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 34.16 (.84)

3⎛ ⎞

−⎜ ⎟⎝ ⎠


= 2 2 2 21! 2! 3!1 .16 .16 (.84) .16 (.84)0!(1 0)! 1!(2 1)! 2!(3 2)!

− − −− − −

2 34! .16 (.84)3!(4 3)!

−−

= 1 − .0256 − .0430 − .0542 − .0607 = .8165 4.58 a. Let x= number of facilities chosen that treat hazardous waster on-site in 10 trials. For

this problem, N = 209, r = 8, and n = 10.

E(x) = 10(8)209

nrN

μ = = = .383

b. P(x = 4) =

8 201 8! 201!4 6 4!4! 6!95! .0002209!209

10!99!10

r N rx n x

Nn

−⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

4.60 a. Let y = number of defective items in a sample of size 4. For this problem, y is a

hypergeometric random variable with N = 10, n = 4, and r = 1. You will accept the lot if you observe no defectives.

P(y = 0) =

1 10 1 1! 9!0 4 0 1(84)0!1! 4!5!

10!10 2104!6!4

r N ry n y

Nn

− −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= .4

b. If r = 2,

P(y = 0) =

2 10 2 2! 8!0 4 0 1(70)0!2! 4!4!

10!10 2104!6!4

r N ry n y

Nn

− −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= = =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= .333

4.62 a. ( )

r N ry n y

p yNn

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠=

⎛ ⎞⎜ ⎟⎝ ⎠

7 3 7! 3!2 2 632!5! 2!1!(2) .3010!10 210

4!6!4

p

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠= = = =⎛ ⎞⎜ ⎟⎝ ⎠

Chapter 4 64

b.

7 30 4

( 1) 1 (0) 1 1 0 1104

P y p

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠≥ = − = − = − =⎛ ⎞⎜ ⎟⎝ ⎠

4.64 Let y = number of firms operating in violation of regulations in 20 firms. Then y has a

hypergeometric distribution with N = 20, r = 5, and n = 3.

a.

5 20 50 3 0 1(455)( 0) .399

20 11403

P y

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠= = = =

⎛ ⎞⎜ ⎟⎝ ⎠

b.

5 20 53 3 3 10(1)( 3) .009

20 11403

P y

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠= = = =

⎛ ⎞⎜ ⎟⎝ ⎠

c. ( 1) 1 ( 0) 1 .399 .601P y P y≥ = − = = − =

4.66 Show that the mean of a hypergeometric distribution is .nrN

First, 1! ( 1)!1!( )! ( 1)!( )!

r rr r ry y ry yy r y y r y

−⎛ ⎞ ⎛ ⎞−= = =⎜ ⎟ ⎜ ⎟−− − −⎝ ⎠ ⎝ ⎠

We can write 1 ( 1)

as 1 ( 1)

N r N rn y n y− − − −⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠

Also, 1! ( 1)!1!( )! ( 1)!( )!

N NN N N Nn nn N n n n N n n

−⎛ ⎞ ⎛ ⎞−= = =⎜ ⎟ ⎜ ⎟−− − −⎝ ⎠ ⎝ ⎠

Thus,

1 1 ( 1)1 1 ( 1)

11

r N r r N ry r

y n y y n yN NNn n n

− − − − −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− − − − −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠=

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

=

1 1 ( 1)1 1 ( 1)

11

r N rrnN y n y

Nn

− − − −⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟− − − −⎝ ⎠⎝ ⎠

−⎛ ⎞⎜ ⎟−⎝ ⎠


Let z = y − 1. For N − 1 elements, r − 1 successes in N − 1 elements, and n − 1 trials, z has a hypergeometric distribution.

Thus, all all

1 1 ( 1)1 1 ( 1)

( )11

y y

r N r r N ry

y n y y n yrn rnE yN NN Nn n

− − − − −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− − − − −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠= = =

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

∑ ∑

4.68 a. 0 1.15 1 1.15 2 1.151.15 1.15 1.15( 2) (0) (1) (2) 0.890150! 1! 2!e e eP Y P P P− − −

≤ = + + = + + =

b. 2 1.15σ λ= = c. We would not expect the driver to exceed trips in the interval 3 1.15 3(1.15)μ σ± ⇒ ±

⇒ 1.15 3.45 ( 2.30, 4.6)± ⇒ − . The driver is not likely to exceed four trips.

4.70 a. 0111.0!0

5.4)0(5.40

===−eYP

b. 04999.0!1

5.4)1(5.41

===−eYP

c. 5.4)( === λμyE 5.42 == λσ 121.25.4 ==σ 4.72 a. 4 2σ λ= = = b. P(y > 10) = 1 − P(y ≤ 10) = 1 − .997 = .003 from Table 3, Appendix B, with λ = 4. Since the probability is so small (.003), it would be very unlikely that the plant

would yield a value that would exceed the EPA limit. 4.74 a. 1.57μ λ= = 2 1.57σ λ= = 1.57 1.253σ λ= = = b. [ ]( 3) 1 ( 2) 1 (0) (1) (2)P y P y p p p≥ = − ≤ = − + +

= 0 1.57 1 1.57 2 1.571.57 1.57 1.5710! 1! 2!

e e e− − −⎡ ⎤⋅ ⋅ ⋅− + +⎢ ⎥⎣ ⎦

= 1 − [.2080 + .3266 + .2564] = 1 − .7910 = .2090

Chapter 4 66

4.76 a. ( 20) (0) (1) (20)P Y P P P≤ = + + +

0 18 1 18 2 18 20 1818 18 18 18 0.73070! 1! 2! 20!e e e e− − − −

= + + + + =

b. (5 10) (5) (6) (10)P y p p p≤ ≤ = + + +

= 5 18 6 18 10 1818 18 185! 6! 10!e e e− − −⋅ ⋅ ⋅

+ + + = .03028

c. 2 18σ λ= =

2 18 4.24σ σ= = = We would expect y to fall within 2μ σ± 18 2(4.24) 18 8.48 9.52 to 26.48⇒ ± ⇒ ± ⇒ d. The trend would indicate that the number of occurrences were dependent with one

another. This casts doubts on the independence characteristic of the Poisson. 4.78 a. Show for Poisson random variable y that 0 ( ) 1.p y≤ ≤ The probability function for y

is:

( )!

yep yy

λλ −

=

For a Poisson random variable, 0 and 0. Thus, ( ) 0.y p yλ > ≥ ≥

The Taylor expansion for 1 2 3 4

is 11! 2! 3! 4!

eλ λ λ λ λ+ + + + +

Thus, any 1 term of the Taylor expansion is less than eλ , so a term of the expansion

times 1. Thus, 0 ( ) 1.e p yλ− < ≤ ≤

b. Show for Poisson random variable, y, 0

( ) 1.y

p y∞

=

=∑

0 1 2 3

0

( )0! 1! 2! 3!y

e e e ep yλ λ λ λλ λ λ λ∞ − − − −

=

= + + + +∑

= 0 1 2 3

( ) 10! 1! 2! 3!

e e eλ λ λλ λ λ λ− −⎛ ⎞+ + + + = =⎜ ⎟

⎝ ⎠

As shown above, the terms inside the parentheses are the Taylor expansion for eλ .


c. [ ]0 0

( 1) ( 1) ( ) ( 1)!

y

y y

eE y y y y p y y yy

λλ∞ ∞ −

= =

− = − = −∑ ∑

= 2

2

2 2( 2)! ( 2)!

y y

y y

e ey y

λ λλ λλ∞ ∞− − −

= =

=− −∑ ∑

Let 2 2

0

2, ,!

z

z

ez yz

λλλ λ∞ −

=

= − ⇒ =∑

since 0

1!

z

z

ez

λλ∞ −

=

=∑

[ ] 2 2( 1) ( ) ( )E y y E y E yλ− = = −

2 2 2( ) ( )E y E yλ λ λ⇒ = + = +

4.80 0 0 0

( ) ( )( ) ( )! ! !

tt

y t y t y ety ty e

y y y

e e e em t E e e e e ey y y

λ λλ λ λλ λ λ∞ ∞ ∞− −

− −

= = =

= = = =∑ ∑ ∑

Let ( ) ( )1 1

0

( )!

t tt ty

e et e e

y

ee e e e e e m t ey

βλ λλ λ λ λββ λ

∞ −− −− −

=

= = = = ⇒ =∑

4.82 ( )1 (1 ) 1

t t

t t tpe pem t

p e e pe= =

− − − +

( ) ( )

1 20

0

1( )

1

t t t t t t

t ttt

pe e pe pe e pedm tdt e pe

μ=

=

⎤− + − − +⎤ ⎥′ = =⎥ ⎥⎦ ⎡ ⎤− + ⎥⎣ ⎦ ⎦

= 2 2 2 2 2 2

2

0

( ) ( ) ( ) ( )

1

t t t t t

t t

t

pe p e p e p e p e

e pe=

⎤− + + − ⎥

⎥⎡ ⎤− + ⎥⎣ ⎦ ⎦

= 2 2

0

1

1

t

t t

t

pe pppe pe

=

⎤⎥ = =⎥⎡ ⎤− + ⎥⎣ ⎦ ⎦

Chapter 4 68

( )2

2

2 40

0

1 (2) 1( )

1

t t t t t t t t

t tt

t

pe e pe pe e pe e ped m tdt e pe

μ=

=

⎤⎡ ⎤ ⎡ ⎤− + − − + − +⎤ ⎣ ⎦ ⎣ ⎦ ⎥′ = =⎥ ⎥⎡ ⎤⎦ − + ⎥⎣ ⎦ ⎦

= [ ] [ ][ ]

2

4

1 1 (2) 1 1 ( 1 )

1 1

p p p p p

p

− + − − + − +

− +

= [ ]23 2

4 4

2(1 )2 (1 )[ ]

p p pp p pp p

+ −+ −=

= 2 22 2 2p p pp p

+ − −=

( )222 1 2 2 2

2 1 1p pp p p

σ μ μ − −′ ′= − = − =

4.84 Let y = the number of female fence lizards that will be resting from the sample of 20. Then y is a binomial random variable with n = 20 and p = .95. a. ( 15) 1 ( 14) 1 .0003 .9997P y P y≥ = − ≤ = − = using Table 2 in Appendix B. b. ( 10) ( 9) .0000P y P y< = ≤ = using Table 2 in Appendix B. c. 200(.95) 190npμ = = = 2 200(.95)(.05) 9.5npqσ = = = 2 9.5 3.08σ σ= = = We would expect to observe a number in the interval 2 to 2μ σ μ σ− + . 2μ σ− = 190 – 2(3.08) = 190 – 6.16 = 183.84 ⇒ 184 2μ σ+ = 190 + 2(3.08) = 190 + 6.16 = 196.16 ⇒ 196 Observing fewer than 190 would be expected based on the interval above. 4.86 a. If the number of respondents with symptoms does not depend on the daily amount of

water consumed, each of the 4 categories would be equally likely. Each would have a probability of 1/4 or .25.

b. 6 11 13 101 2 3 4

40!( 6, 11, 13, 10) (.25) (.25) (.25) (.25)6!11!13!10!

P y y y y= = = = =

= .00104


4.88 Let y be the number of engineers you choose in the sample with experience. y follows a

hypergeometric distribution with N = 5, r = 2, and n = 2.

a.

2 5 2 2! 3!2 2 2 12!0! 0!3!( 2) (2) .105!5 10

2!3!2

P y p

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⋅−⎝ ⎠⎝ ⎠= = = = = =

⎛ ⎞⎜ ⎟⎝ ⎠

b. ( 1) (1) (2)P y p p≥ = +

2 5 2 2! 3!1 2 1 61!1! 1!2!(1) .605!5 10

2!3!2

p

−⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⋅−⎝ ⎠⎝ ⎠= = = =

⎛ ⎞⎜ ⎟⎝ ⎠

( 1)P y ≥ = .60 + .10 = .70 4.90 a. This is a multinomial experiment with n = 10 and 1 2 3.20, .15, .20,p p p= = = 4 5 6.30, .10, and .05.p p p= = =

p(1, 2, 2, 4, 1, 0) = 1 2 2 4 1 010! (.20) (.15) (.20) (.30) (.10) (.05)1!2!2!4!1!0!

= .0055112

b. 100(.15) 15i inpμ = = =

2 (1 ) 100(.15)(.85) 12.75i inp pσ = − = =

2 12.75 3.571σ σ= = = We expect the number of specimens from the Eocene era to fall within 2 .μ σ± 2 15 2(3.571) 15 7.142μ σ± ⇒ ± ⇒ ± ⇒ 7.858 to 22.142 4.92 Let y = number of arrivals in a 1 minute interval. Then y has a Poisson distribution with λ =1. a. ( 3) 1 ( 0) ( 1) ( 2)P y P y P y P y≥ = − = − = − =

= 0 1 1 1 2 11 1 110! 1! 2!e e e− − −

− − − = 1 − .3679 − .3679 − .1839 = .0803

b. ( 3) 1 ( 0) ( 1) ( 2) ( 3)P y P y P y P y P y> = − = − = − = − =

= .0803 −3 113!e− = .0803 − .0613 = .019

Yes. Since the probability of observing more than 3 arrivals is so small (.019), one can assure the engineer that the number of arrivals will rarely exceed 3 per minute.

Chapter 4 70

4.94 a. The sample space, values of x (in thousands) and associated probabilities are:

Sample Space x p(x)

−50, −50 −100 .6(.6) = .36 −50, −20 −70 .6(.1) = .06 −50, 30 −20 .6(.15) = .09 −50, 430 380 .6(.1) = .06 −50, 950 900 .6(.05) = .03 −20, −50 −70 .1(.6) = .06 −20, −20 −40 .1(.1) = .01 −20, 30 10 .1(.15) = .015 −20, 430 410 .1(.1) = .01 −20, 950 930 .1(.05) = .005

30, −50 −20 .15(.6) = .09 30, −20 10 .15(.1) = .015 30, 30 60 .15(.15) = .002530, 430 460 .15(.1) = .015 30, 950 980 .15(.05) = .0075

430, −50 380 .1(.6) = .06 430, −20 410 .1(.1) = .01 430, 30 460 .1(.15) = .015 430, 430 860 .1(.1) = .01 430, 950 1380 .1(.05) = .005 950, −50 900 .05(.6) = .03 950, −20 930 .05(.1) = .005 950, 30 980 .05(.15) = .0075950, 430 1380 .05(.1) = .005 950, 950 1900 .05(.05) = .0025

The probability distribution of x is:

x p(x) −100,000 .36 −70,000 .12 −40,000 .01 −20,000 .18

10,000 .03 60,000 .0225

380,000 .12 410,000 .02 460,000 .03 860,000 .01 900,000 .06 930,000 .01 980,000 .015

1,380,000 .01 1,900,000 .0025


b. all

( ) ( )x

E x xp x= =∑ −100,000(.36) − 70,000(.12) − 40,000(.01) + ⎭ + 1,900,000(.0025)

= 126,000 2 2 2 2

all

( ) ( ) ( )x

V x E x x p xμ μ= − = −∑

= (−100,000)2(.36) + (−70,000)2(.12) + (−40,000)2(.01) + ⎭ + 1,900,0002(.0025) − 126,0002 = 122,642,000,000 c. The probability of doubling the $100,000 investment is: ( 200,000) ( 380,000) ( 410,000 ( 1,900,000)P x P x P x P x≥ = = + = + + = = .12 + .02 + .03 + .01 + .06 + .01 + .015 + .01 + .0025 = .2775 d. The probability of 2 dry holes is ( 100,000) .36P x = − = From Exercise 4.93, ( 50,000) .6P y = − = Thus, the probability of one dry hole is much greater than the probability of two dry

holes. 4.96 Let y be the number of the five blips that resulted in enemy aircraft. Then y has a binomial

distribution with n = 5 and p = .60.

a. 5 05( 5) (.60) (.40) .0078

5P y

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

b. ( 3) 1 ( 2) 1 .317 .683P y P y≥ = − ≤ = − = using Table 2 in Appendix B.

c. 0 55( 0) (.60) (.40) .01024

0P y

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

4.98 Let y be the number of vehicles using the acceleration lane per minute. y has a Poisson

distribution with λ = 1.1. a. [ ]( 2) 1 ( 2) 1 (0) (1) (2)P y P y p p p> = − ≤ = − + +

= 0 1.1 1 1.1 2 1.11.1 1.1 1.110! 1! 2!

e e e− − −⎡ ⎤⋅ ⋅ ⋅− + +⎢ ⎥⎣ ⎦

= 1 − [.3329 + .3662 + .2014] = 1 − .9005 = .0995

Chapter 4 72

b. 3 1.11.1( 3) (3)3!

eP y p−⋅

= = = =.0738

4.100 Let y be the number of years before a nuclear war occurs. The random variable y is a

geometric random variable with p = .01. 1( ) (.01)(.99) yp y −= y = 1, 2, 3, … a. The probability of a nuclear war occurring in the next 5 years is: ( 5) (1) (2) (3) (4) (5)P y p p p p p≤ = + + + + = 0 1 2 3 4(.01)(.99) (.01)(.99) (.01)(.99) (.01)(.99) (.01)(.99)+ + + + = .01 + .0099 + .009801 + .009703 + .009606 = .049 b. In the next 10 years: ( 10) ( 5) (6) (7) (8) (9) (10)P y P y p p p p p≤ = ≤ + + + + + = 5 6 7 8 9.049 (.01)(.99) (.01)(.99) (.01)(.99) (.01)(.99) (.01)(.99)+ + + + + = .049 + .0095099 + .0094148 + .0093207 + .0092274 + .0091352 = .0956 c. In the next 15 years: ( 15) ( 10) (11) (12) (13) (14) (15)P y P y p p p p p≤ = ≤ + + + + + = .0956 + .0090438 + .0089534 + .0088638 + .0087752 + .0086875 = .1399 d. In the next 20 years: ( 20) ( 15) (16) (17) (18) (19) (20)P y P y p p p p p≤ = ≤ + + + + + = .1399 + .0086006 + .0085146 + .0084294 + .0083451 + .0082617 = .1821 e. The assumption is that the possibility of a nuclear war in any given year is independent

of the possibility from all other years. This assumption probably is not valid.

4.102 a. 55( )

! !

x xe ep xx x

λλ − −

= =

[ ]( 1) 1 ( 1) 1 (0) (1)P x P x p p> = − ≤ = − +

= 0 5 1 55 510! 1!e e− −⎡ ⎤

− +⎢ ⎥⎣ ⎦

= 1− [.0067 + .0337] = 1 − .0404 = .9596


b. 2.52.5( )

! !

x xe ep xx x

λλ − −

= =

[ ]( 1) 1 ( 1) 1 (0) (1)P x P x p p> = − ≤ = − +

= 0 2.5 1 2.52.5 2.510! 1!e e− −⎡ ⎤

− +⎢ ⎥⎣ ⎦

= 1 − [.0821 + .2052] = 1 − .2873 = .7127 c. Let a “success” be an industry exposing their workers to more than 1 ppm of the

solvent. Then y, the number of successes in the 55 industries, is a binomial random variable with n = 55 and p = .12.

0 55 0 0 5555 55!( 0) (.12) (1 .12) (.12) (.88) .00088420 0!55!

P y −⎛ ⎞= = − = =⎜ ⎟

⎝ ⎠

We can approximate this using a Poisson distribution since n is large, p is small, and 55(.12) 6.6 7.npμ = = = <

0

6.6( 0) .0013600!eP x e

λλ −−= = = =

d. We need to find μ such that ( 1)P x ≤ is close to .88. For μ = .5: ( 1)P x ≤ = .9098 For μ = 1.0: ( 1)P x ≤ = .7358 μ ≈ .5

4.104 2 31 2 2( )5 5 5

t t tm t e e e= + +

a.

2 3

100

1 2 2( ) 5 5 5

t t t

tt

d e e edm t

dt dtμ μ

==

⎤⎡ ⎤+ + ⎥⎢ ⎥⎤ ⎣ ⎦′ ⎥= = =⎥ ⎦⎦

= 2 3

0

1 2 2(2) (3)5 5 5

t t t

te e e

=

⎤+ + ⎥⎦

= 0 2(0) 3(0)1 4 6 11 2.25 5 5 5

e e e+ + = =

Chapter 4 74

b. ( )222 1σ μ μ′ ′= −

2 32

2 200

1 4 6( ) 5 5 5

t t t

tt

d e e ed m t

dtdtμ

==

⎤⎡ ⎤+ + ⎥⎢ ⎥⎤ ⎣ ⎦′ ⎥= =⎥ ⎦⎦

= 2 3

0

1 4 6(2) (3)5 5 5

t t t

te e e

=

⎤+ + ⎥⎦

= 0 2(0) 3(0)1 8 18 27 5.45 5 5 5

e e e+ + = =

( )22

2 1σ μ μ′ ′= − = 5.4 − 2.22 = .56

4.106 a. ( ) ( )( 1) ( 1)

0 0

( ) ( )! !

y y ty t t

y y

t e t eP t E t e e

y y

λ λλ λλ λ− −∞ ∞

− −

= =

= = =∑ ∑

b. ( 1)

( 1) (1 1)1

11

( )( )t

tt

tt

d edP tE y e edt dt

λλ λμ λ λ λ

−− −

==

=

⎤⎡ ⎤⎤ ⎣ ⎦ ⎥ ⎤= = = = = =⎥ ⎦⎥⎦ ⎦

2 2 2( )E yσ μ= − [ ] [ ]2 2( 1) ( ) ( ) ( ) ( 1) ( )E y y E y E y E y E y y E y− = − ⇒ = − +

[ ]( 1)2

( 1) 2 (1 1) 22 1

1 1

( )( 1)t

tt

t t

d ed P tE y y e edtdt

λλ λ

λλ λ λ λ

−− −

== =

⎤⎡ ⎤⎤ ⎣ ⎦ ⎥ ⎤− = = = = =⎥ ⎦⎥⎦ ⎦

Thus, [ ]2 2( ) ( 1) ( )E y E y y E y λ λ= − + = + Therefore 2 2 2 2 2( )E yσ μ λ λ λ λ= − = + − =

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