Applied Statistics and Probability for Engineers, 5th edition Chapter3 PDF

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Applied Statistics and Probability for Engineers, 5th edition Chapter3
December 21, 2009...

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Applied Statistics and Probability for Engineers, 5 th edition

December 21, 2009

CHAPTER 3 Section 3-1

{0,1,2,...,1000}

3-1.

The range of X is

3-2.

, ,...,50} The range of X is { 0,12

3-3.

The range of X is { 0,12 , ,...,99999}

3-4.

, ,3,4,5} The range of X is { 0,12

3-5.

The range of X is

{1,2,...,491} . Because 490 parts are conforming, a nonconforming part must be selected in 491

selections. 3-6.

, ,...,100} . Although the range actually obtained from lots typically might not exceed 10%. The range of X is { 0,12

3-7.

The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is{0,12 , ,...}

3-8.

, ,...} The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is{0,12

3-9.

The range of X is

3-10.

The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8.  1 3 1 5 6 Therefore the range of X is  , , , ,  4 8 2 8 8 

3-11.

The range of X is

{0,1,2,...,15}

{0,1,2, K,10000}

3-12.The range of X is {100, 101, …, 150} 3-13.The range of X is {0,1,2,…, 40000) Section 3-2 3-14.

f X (0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 f X (1.5) = P( X = 1.5) = 1 / 3 f X (2) = 1 / 6 f X (3) = 1 / 6 a) P(X = 1.5) = 1/3 b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P (0  X < 2) = P ( X = 0) + P ( X = 1.5) = 1/ 3+ e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2 3-15.

1/ 3= 2 / 3

All probabilities are greater than or equal to zero and sum to one. a) P(X  2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8

3-1

c) P(-1  X  1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X  -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-16.

All probabilities are greater than or equal to zero and sum to one. a) P(X 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286 c) P(2 10) = 1

3-18.

Probabilities are nonnegative and sum to one. a) P(X = 2) = 3/4(1/4)2 = 3/64 b) P(X  2) = 3/4[1+1/4+(1/4)2] = 63/64 c) P(X > 2) = 1  P(X  2) = 1/64 d) P(X  1) = 1  P(X  0) = 1  (3/4) = 1/4

3-19.

X = number of successful surgeries. P(X=0)=0.1(0.33)=0.033 P(X=1)=0.9(0.33)+0.1(0.67)=0.364 P(X=2)=0.9(0.67)=0.603

3-20.

P(X = 0) = 0.023 = 8 x 10 -6 P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012 P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576 P(X = 3) = 0.983 = 0.9412

3-21.

X = number of wafers that pass P(X=0) = (0.2)3 = 0.008 P(X=1) = 3(0.2)2(0.8) = 0.096 P(X=2) = 3(0.2)(0.8)2 = 0.384 P(X=3) = (0.8)3 = 0.512

3-22.

X: the number of computers that vote for a left roll when a right roll is appropriate. p=0.0001. P(X=0)=(1-p)4=0.9999 4=0.9996 P(X=1)=4*(1-p)3p=4*0.999930.0001=0.0003999 P(X=2)=C42(1-p)2p2=5.999*10-8 P(X=3)=C43(1-p)1p3=3.9996*10 -12 P(X=4)=C40(1-p)0p4=1*10 -16

3-23.

P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2

3-24.

P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1

3-25.

P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1

3-26.

X = number of components that meet specifications P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 P(X=2) = (0.95)(0.98) = 0.931

3-27.

X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663

3-2

P(X=3) = (0.95)(0.98)(0.99) = 0.92169 3.28.

X = final temperature P(X=266) = 48/200 = 0.24 P(X=271) = 60/200 = 0.30 P(X=274) = 92/200 = 0.46

 0.24,  f ( x) =  0.30,  0.46,  3.29.

x = 274

X = waiting time (hours) P(X=1) = 19/500 = 0.038 P(X=2) = 51/500 = 0.102 P(X=3) = 86/500 = 0.172 P(X=4) = 102/500 = 0.204 P(X=5) = 87/500 = 0.174 P(X=6) = 62/500 = 0.124 P(X=7) = 40/500 = 0.08 P(X=8) = 18/500 = 0.036 P(X=9) = 14/500 = 0.028 P(X=10) = 11/500 = 0.022 P(X=15) = 10/500 = 0.020

 0.038,  0.102,   0.172,   0.204,  0.174,  f ( x) =  0.124,  0.080,   0.036,   0.028,  0.022,   0.020, 3.30.

x = 266 x = 271

x=1 x=2 x=3 x=4 x=5 x=6 x=7 x=8 x=9 x = 10 x = 15

X = days until change P(X=1.5) = 0.05 P(X=3) = 0.25 P(X=4.5) = 0.35 P(X=5) = 0.20 P(X=7) = 0.15

 0.05,  0.25,  f (x ) =  0.35,  0.20,   0.15,

x = 1.5 x=3 x = 4.5 x=5 x=7

3-3

3.31.

X = Non-failed well depth P(X=255) = (1515+1343)/7726 = 0.370 P(X=218) = 26/7726 = 0.003 P(X=317) = 3290/7726 = 0.426 P(X=231) = 349/7726 = 0.045 P(X=267) = (280+887)/7726 = 0.151 P(X=217) = 36/7726 = 0.005

 0.005,  0.003,   0.045, f ( x) =   0.370,  0.151,   0.426,

x = 217 x = 218 x = 231 x = 255 x = 267 x = 317

Section 3-3

3-32.

x< 0   0,  1 / 3 0  x < 1.5    F ( x) =  2 / 3 1.5  x < 2  5 / 6 2  x < 3     1 3  x 

where

f X (0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 f X (1.5) = P ( X = 1.5) = 1 / 3 f X (2) = 1 / 6 f X (3) = 1 / 6

3-33.

x < 2   0,   1/ 8  2  x < 1   3/8  1  x < 0  F ( x) =   0  x 0) = 1  P(X  0) = 1  5/8 = 3/8 3-34.

 0 x< 1    4 7 1  x < 2  F( x) =   6 7 2  x < 3  1 3  x  a) b) c) d)

P(X < 1.5) = 4/7 P(X  3) = 1 P(X > 2) = 1 – P(X  2) = 1 – 6/7 = 1/7 P(1 < X  2) = P(X  2) – P(X  1) = 6/7 – 4/7 = 2/7

3-4

3-35.

x 4) = 0.1 e) P(X2) = 0.7

3-41.

The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(-10) = 0.25, f(30) = 0.5, f(50) = 0.25 a) P(X50) = 1 b) P(X40) = 0.75 c) P(40  X  60) = P(X=50)=0.25 d) P(X 5) =   0 .02 00.985 = 0.9039 0

c) E(X) = 1/0.02 = 50 3-106.

X = number of opponents until the player is defeated. p=0.8, the probability of the opponent defeating the player.

(a) f(x) = (1 – p)x – 1p = 0.8(x – 1)*0.2 (b) P(X>2) = 1 – P(X=1) – P(X=2) = 0.64 (c) µ = E(X) = 1/p = 5

3-18

(d) P(X≥4) = 1-P(X=1)-P(X=2)-P(X=3) = 0.512 (e) The probability that a player contests four or more opponents is obtained in part (d), which is po = 0.512. Let Y represent the number of game plays until a player contests four or more opponents. Then, f(y) = (1-po)y-1po. µY = E(Y) = 1/po = 1.95 3-107.

3-108.

p=0.13

(a) P(X=1) = (1-0.13)1-1*0.13=0.13. (b) P(X=3)=(1-0.13) 3-1*0.13 =0.098 (c) µ=E(X)= 1/p=7.69≈8 X = number of attempts before the hacker selects a user password.

(a) p=9900/36 6=0.0000045 µ=E(X) = 1/p= 219877 V(X)= (1-p )/p 2 = 4.938*10 10 σ= V ( X ) =222222 (b) p=100/363=0.00214 µ=E(X) = 1/p= 467 V(X)= (1-p )/p 2 = 217892.39 σ= V ( X ) =466.78

Based on the answers to (a) and (b) above, it is clearly more secure to use a 6 character password. 3-109.

p = 0.005 , r = 8

P (X = 8) = 0.0058 = 3.91x10 19 1 b). µ = E ( X ) = = 200 days 0 .005

a.)

c) Mean number of days until all 8 computers fail. Now we use p=3.91x10-19

µ = E (Y ) =

1 = 2.56 x1018 days  91 3. 91x10

or 7.01 x1015 years

3-110.

Let Y denote the number of samples needed to exceed 1 in Exercise 3-66. Then Y has a geometric distribution with p = 0.0169. a) P(Y = 10) = (1  0.0169)9(0.0169) = 0.0145 b) Y is a geometric random variable with p = 0.1897 from Exercise 3-66. P(Y = 10) = (1  0.1897)9(0.1897) = 0.0286 c) E(Y) = 1/0.1897 = 5.27

3-111.

Let X denote the number of transactions until all computers have failed. Then, X is negative binomial random variable with p = 10-8 and r = 3.

a) E(X) = 3 x 10 8 b) V(X) = [3(110-80]/(10-16) = 3.0 x 1016

3-112.

3-113.

(a) p 6=0.6, p=0.918 (b) 0.6*p2=0.4, p=0.816

 x  1  (1 p) x r p r.  r 1  

Negative binomial random variable: f(x; p, r) = 

When r = 1, this reduces to f(x) = (1p)x-1p, which is the pdf of a geometric random variable. Also, E(X) = r/p and V(X) = [r(1p)]/p2 reduce to E(X) = 1/p and V(X) = (1p)/p2, respectively. 3-114. a)

3-19

b) c)

d) 3-115.

a) Probability that color printer will be discounted = 1/10 = 0.01 days b) c) Lack of memory property implies the answer equals d)

3-116. a) b) c) d) Section 3-8 3-117.

X has a hypergeometric distribution N=100, n=4, K=20 a) P( X

= 1) =

( )( ) = ( ) 20 1

80 3 100 4

20(82160) = 0.4191 3921225

b)

P ( X = 6) = 0 , the sample size is only 4

c)

P (X = 4) =

( )( ) = 4845(1) = 0.001236 ( ) 3921225 20 4

80 0 100 4

K  20  = 4  = 0.8 N  100   96 N n V ( X ) = np(1  p)   = 4(0.2)(0.8)   = 0.6206  99  N 1 

d) E ( X )

3-118.

= np = n

( )( ) = (4 ×16 ×15 ×14) / 6 = 0.4623 ( ) ( 20 ×19 ×18 ×17) / 24 ( )( ) = 1 = 0.00021 b) P (X = 4) = ( ) (20 ×19 ×18 ×17) / 24 a)

P ( X = 1) =

4 1

16 3 20 4 4 16 4 0 20 4

c)

3-20

P(X  2) = P ( X = 0) + P ( X = 1) + P(X = 2) =

( )( ) + ( )( ) + ( )( ) ( ) ( ) ( )

=

 16× 15× 14×13 4×16×15×14 6×16×15  + +   24 6 2    20× 19×18×17    24  

4 0

16 4 20 4

4 1

16 3 20 4

4 2

16 2 20 4

= 0.9866

d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-119.

N=10, n=3 and K=4

0.5

0.4

P(x)

0.3

0.2

0.1 0.0 0

1

2

3

x

 24 12   x  3  x  /   

3-120. (a) f(x) =

36    3 

(b) µ=E(X) = np= 3*24/36=2 V(X)= np(1 -p)(N-n)/(N-1) =2*(1-24/36)(36-3)/(36-1)=0.629 (c) P(X≤2) =1-P(X=3) =0.717 3-121.

Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high blood pressure. N=800, K=240 n=10 a) n=10

P (X = 1) =

( )( ) = ( )( ( ) 240 1

560 9 800 10

240! 560! 1!239! 9!551! 800! 10!790!

)

= 0.1201

b) n=10

P (X > 1) = 1 P ( X  1) = 1 [ P( X = 0) + P( X = 1)] P (X = 0) =

( )( ) = ( ( ) 240 0

560 10 800 10

)(

240! 560! 0!240! 10!550! 800! 10!790!

) = 0.0276

P( X > 1) = 1  P( X  1) = 1 [0.0276 + 0.1201] = 0.8523 3-122.

Let X denote the number of cards in the sample that are defective. a)

3-21

P (X  1) = 1  P( X = 0) P (X = 0 ) =

( )( ) = ( ) 20 0

120 20 140 20

120! 20!100! 140! 20!120!

= 0.0356

P (X  1) = 1 0.0356 = 0.9644 b)

P( X  1) = 1  P( X = 0) P (X = 0) =

( )( ) = ( ) 5 0

135 20 140 20

135! 20!115!

=

140! 20!120!

135!120! = 0.4571 115!140!

P (X  1) = 1  0.4571 = 0.5429 3-123. N=300 (a) K = 243, n = 3, P(X = 1)=0.087 (b) P(X≥1) = 0.9934 (c) K = 26 + 13 = 39, P(X = 1)=0.297 (d) K = 300-18 = 282 P(X ≥ 1) = 0.9998 3-124.

Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6.

( )( ) =  40!  = 2.61× 10 a) P( X = 6) = ( )  6!34!  ( )( ) = 6 × 34 = 5.31× 10 b) P( X = 5) = ( ) ( ) ( )( ) = 0.00219 c) P ( X = 4) = ( ) 1

6 6

34 0 40 6 6 34 5 1 40 6 6 34 4 2 40 6

7

5

40 6

d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p =

1 and 3,838,380

E(Y) = 1/p = 3,838,380 weeks. This is more than 738 centuries!

3-125.

Let X denote the number of blades in the sample that are dull. a)

P (X  1) = 1  P (X = 0) P(X

( )( ) = = 0) = ( ) 38 5

10 0

48 5

38! 5!33! 48! 5!43!

=

38!43! = 0.2931 48!33!

P (X  1) = 1 P (X = 0) = 0.7069 b) Let Y denote the number of days needed to replace the assembly. P(Y = 3) =

0.29312(0.7069) =0.0607

c) On the first day,

P( X = 0) =

( )( ) = ( ) 2 0

46 5 48 5

46! 5!41! 48! 5!43!

=

46!43! = 0.8005 48!41!

3-22

P( X = 0 ) =

On the second day,

( )( ) = ( ) 6 0

42 5 48 5

42! 5!37! 48! 5!43!

=

42!43! = 0.4968 48!37!

On the third day, P(X = 0) = 0.2931 from part a). Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811. 3-126.

a) For Exercise 3-97, the finite population correction is 96/99. For Exercise 3-98, the finite population correction is 16/19. Because the finite population correction for Exercise 3-97 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 3-97. b) Assuming X has a binomial distribution with n = 4 and p = 0.2,

( ) 0.2 0.8 = 0.4096 P (X = 4)= ( ) 0.2 0.8 = 0.0016 P (X = 1) =

4 1

4 4

1

4

3

0

The results from the binomial approximation are close to the probabilities obtained in Exercise 3-97. c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is not as close to the true answer as the results obtained in part (b) of this exercise. d) From Exercise 3-102, X is approximately binomial with n = 20 and p = 20/140 = 1/7.

P (X  1) = 1 P ( X = 0) =

( )( ) ( ) 20 0

1 0 6 7 7

20

= 1 0.0458 = 0.9542

finite population correction is 120/139=0.8633 From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28

P( X  1) = 1  P( X = 0) =

( )( ) ( ) 20 0

1 0 27 20 28 28

= 1  0.4832 = 0.5168

finite population correction is 120/139=0.8633

3-127.

a)

b)

c)

3-23

3-128.

a)

b) c) Section 3-9 3-129.

e 4 40 = e 4 = 0 0183 . 0! b) P (X  2 ) = P( X = 0) + P( X = 1) + P( X = 2)

a) P( X = 0 ) =

e4 41 e4 42 + 1! 2! = 0.2381 = e4 +

3-130

c) P (X = 4 ) =

e 4 44 = 0.1954 4!

d) P (X = 8 ) =

e 4 48 = 0. 0298 8!

a) P( X = 0 ) = e0 .4 = 0.6703 e 0 .4 (0.4 ) e 0 .4 (0 .4 )2 + = 0. 9921 1! 2! 0 . 4 4 e  (0 .4 ) c) P (X = 4 )= = 0.000715 4! e0.4 (0.4 )8 d) P (X = 8 ) = = 1. 09× 10 8 8!

b) P (X  2 ) = e 0 .4 +

3-131.

3-132.

 P ( X = 0 ) = e = 0 .05 . Therefore,  = ln(0.05) = 2.996. Consequently, E(X) = V(X) = 2.996.

a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with  = 10.  e 1010 5 P( X = 5) = = 0.0378 . 5! e 10 10 e 1010 2 e 1010 3 b) P (X  3 ) = e 10 + + + = 0. 0103 1! 2! 3! c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with 20

2015 = 0.0516 15! d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with e5 55  = 5. P( W = 5) = = 01755 . 5! 3-133. λ=1, Poisson distribution. f(x) =e- λ λx/x!  = 20. P( Y = 15) =

e

(a) P(X≥2)= 0.264 (b) In order that P(X≥1) = 1-P(X=0)=1-e- λ exceed 0.95, we need λ=3. Therefore 3*16=48 cubic light years of space must be studied. 3-134. (a) λ=14.4, P(X=0)=6*10-7 (b) λ=14.4/5=2.88, P(X=0)=0.056 (c) λ=14.4*7*28.35/225=12.7

3-24

P(X≥1)=0.999997 (d) P(X≥28.8) =1-P(X ≤ 28) = 0.00046. Unusual. 3-135. (a) λ=0.61. P(X≥1)=0.4566 (b) λ=0.61*5=3.05, P(X=0)= 0.047. 3-136.

a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable with  = 0.1.

P (X = 2) =

e0.1 (0.1)2 = 0.0045 2!

b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable with  = 1.

P(Y = 1) =

e  111 1 = e  = 0.3679 1!

c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable

P (W = 0) = e 2 = 0.1353 P( Y  2) =1  P( Y  1) = 1  P( Y = 0)  P( Y = 1) = 1  e 1  e 1

with  = 2. d)

= 0.2642 3-137.

E ( X ) =  = 0.2 errors per test area e 0.2 0.2 e 0.2 (0.2) 2 0.2 b) P (X  2) = e  + + = 0.9989 1! 2! a)

99.89% of test areas 3-138.

a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with  = 10.

P( X = 0) = e10 = 4.54 ×10 5 b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with  = 1.

P (Y  1) = 1  P (Y = 0) = 1  e 1 = 0.6321 c) The assumptions of a Poisson process require that the probability of a event is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway. 3-139.

a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random variable with  = 0.5.

P (X = 0) = e 0.5 = 0.6065 b) Let Y denote the number of cars with no flaws,

 10 P (Y = 10) =   (0 .6065)10 (0.3935)0 = 0.0067  10 c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part (a), the probability a car contains surface flaws is 10.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935.

3-25

10  P (W = 0) =   (0.3935)0 (0.6065)10 = 0.0067 0 10  P (W = 1) =   (0.3935)1 (0.6065)9 = 0.0437 1  P (W  1) = 0.0067 + 0.0437 = 0.0504 3-140.

a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with  = 0.16. 0.16 = 0.8521 P (X = 0) = e

b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with= 0.48.

P (Y  1) = 1  P (Y = 0) = 1  e 48 = 0.3812 3-141.

a) b) c)

3-142.

a) b)

c) No, if a Poisson distribution is assumed, the intervals need not be consecutive. Supplemental Exercises

3-143.

3-144.

1 1  1  1  3  1  1  +  +   = , 8  3 4  3 8 3  4 2 2 2 2  1  1 ...