Applied Statistical and Probability For Engineering [PDF]

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

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.

, ,...,99999 The range of X is 0,12

3-4.

The range of X is {0, 1, 2, 3, 4, 5}

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, 1, 2, …}

3-8.

The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0, 1, 2, …}

3-9.

The range of X is {0, 1, 2, …, 15}

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, …, 10,000}

3-12.

The range of X is {100, 101, …, 150}

3-13.

The range of X is {0, 1, 2, …, 40000)

3-14.

The range of X is {0, 1, 2, ..., 16}.

3-15.

The range of X is {1, 2, ..., 100}.

Section 3-2 3-16.

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  1/ 3  2 / 3 e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2

3-17.

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 c) P(-1  X  1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4

3-1

Applied Statistics and Probability for Engineers, 6th edition d) P(X  -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-18.

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

3-20.

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

3-21. a) P(X≥2)=P(X=2)+P(X=2.25)=0.2+0.1=0.3 b) P(X 0) = 1  P(X  0) = 1  5/8 = 3/8 3-40.

 0 x  1    4 7 1  x  2  F(x)    6 7 2  x  3  3  x   1  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

3-41.

3-5

March 27, 2014

Applied Statistics and Probability for Engineers, 6th edition

 0,  1 / 25   4 / 25 F ( x)    9 / 25 16 / 25  1

x0  0  x  1 1  x  2   2  x  3 3  x  4  4  x 

3-42.

x0   0,  3 / 4 0  x  1   15 / 16 1  x  2  F ( x)    63 / 64 2  x  3  ... ...    1 x    In general, F(x) = (4k-1) / 4k for k – 1 ≤ x < k 3-43.

x  1.25  0 0.2 1.25  x  1.5   0.6 1.5  x  1.75 F ( x)    0.7 1.75  x  2  0.9 2  x  2.25    1 2.25  x  3-44. Probability a patient from hospital 1 is admitted is 1277/5292 = 0.2413 Here X is the number of patents admitted in the sample. Range of X = {0,1,2} P(X=0) = (1 − 0.2413)(1 − 0.2413) = 0.576 P(X=1) = 0.2413(1 − 0.2413)+ (1 − 0.2413)(0.2413) = 0.366 P(X=2) = 0.2413×0.2413 = 0.058 The cumulative distribution function of X is

x0   0 0.576 0  x  1   F ( x)  0.942 1  x  2  1 x2      3-45.

x0   0, 0.008, 0  x  1   F ( x)  0.104, 1  x  2  0.488, 2  x  3    1, 3  x 

3-6

March 27, 2014

Applied Statistics and Probability for Engineers, 6th edition

f (0)  0.23  0.008 f (1)  3(0.2)(0.2)(0.8)  0.096 f (2)  3(0.2)(0.8)(0.8)  0.384 f (3)  (0.8)3  0.512 3-46.

x0   0,  0.9996, 0  x  1    F ( x )   0.9999, 1  x  3  0.99999, 3  x  4    4  x   1,

f (0)  0.99994  0.9996 f (1)  4(0.99993 )(0.0001)  0.000399 f (2)  5.999(10 8 ) f (3)  3.9996(10 12 ) f (4)  10 16 3-47.

x  10   0, 0.2, 10  x  25    F ( x)    0.5, 25  x  50  1, 50  x  where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-48.

x 1   0,  0.1, 1  x  5    F ( x)    0.7, 5  x  10  1, 10  x  where P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1

3-49.

3-50.

The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; f(1) = 0.5, f(3) = 0.5 a) P(X  3) = 1 b) P(X  2) = 0.5 c) P(1  X  2) = P(X=1) = 0.5 d) P(X>2) = 1  P(X2) = 0.5 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; f(1) = 0.7, f(4) = 0.2, f(7) = 0.1 a) P(X  4) = 0.9 b) P(X > 7) = 0 c) P(X  5) = 0.9

3-7

March 27, 2014

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

d) P(X>4) = 0.1 e) P(X2) = 0.7 3-51.

The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; 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(X7)=P(X=8)+P(X=9)+…+P(X=15)= 0 3-105.

(a) n = 20, p = 0.6122, P(X≥1) = 1 - P(X=0) = 1 (b)P(X≥3) = 1 - P(X 5) is the probability of no connections in 5 trials. That is,

 5 P( X  5)   0.020 0.985  0.9039  0

c) E(X) = 1/0.02 = 50 3-126.

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 (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 3-127.

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 3-128.

X = number of attempts before the hacker selects a user password.

(a) p = 9900/366 = 0.0000045 µ = E(X) = 1/p = 219877 V(X) = (1 - p)/p2 = 4.938E1010 σ=

V (X ) = 222,222

(b) p = 100/363 = 0.00214 µ = E(X) = 1/p = 467

3-22

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

V(X)= (1 - p)/p2 = 217892.39 =

V (X ) = 466.78

Based on the answers to (a) and (b) above, it is clearly more secure to use a 6 character password. 3-129.

p = 0.005 and r = 8 a) b)

P( X  8)  0.0058  3.91E1019 1   E( X )   200 days 0.005

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

  E (Y ) 

1  2.56 x1018 days or 7.01 x1015 years 3.91x10 91

3-130.

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

3-131.

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 108 b) V(X) = [3(110-80]/(10-16) = 3.0 x 1016 3-132.

3-133.

(a) p6 = 0.6, p = 0.918 (b) 0.6p2 = 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. 3-134.

𝑃(𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 < 272𝐾) = 0.54

= 10) = 0.469 0.541 = 0.0005 1 1 b) 𝜇 = 𝐸(𝑋) = = = 1.85 𝑝 0.54 a) 𝑃(𝑋

c) 𝑃(𝑋

≤ 3) = 𝑃(𝑋 = 3) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 1) = 0.462 0.541 + 0.461 0.541 + 0.460 0.541 = 0.903

d) 𝜇 3-135.

𝑟

2

= 𝐸(𝑋) = 𝑝 = 0.54 = 3.70

a) Probability that color printer will be discounted = 1/10 = 0.01 1

1

𝜇 = 𝐸(𝑋) = 𝑝 = 0.10 = 10 days b) 𝑃(𝑋

= 10) = 0.99 0.1 = 0.039

= 10) = 0.99 0.1 = 0.039 d) 𝑃(𝑋 ≤ 3) = 𝑃(𝑋 = 3) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 1) = 0.92 0.1 + 0.91 0.1 + 0.1 = 0.271 c) Lack of memory property implies the answer equals𝑃(𝑋

3-136.

𝑃(𝐿𝑊𝐵𝑆) = 0.037 a) 𝑃(𝑋 = 5) = 0.9634 0.0371 = 0.032 b) 𝑃(𝑋 = 5) + 𝑃(𝑋 = 6) = 0.9634 0.0371 + 0.9635 0.0371 = 0.062

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

c) 𝑃(𝑋

≤ 4) = 𝑃(𝑋 = 4) + 𝑃(𝑋 = 3) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 1) = 0.9633 0.0371 + 0.9632 0.0371 + 0.9631 0.0371 + 0.037 = 0.140 𝑟 3 d) 𝜇 = 𝐸(𝑋) = = = 81.08 𝑝 0.037 3-137. X = the number of cameras tested to detect two failures, X ϵ {2, 3, 4, ...}. Then X has a negative binomial distribution with p = 0.2 and r = 2. Y = the number of cameras tested to detect three failures, Y ϵ {3, 4, 5, ...}. Then Y has a negative binomial distribution with p = 0.2 and r = 3. Note that the events are described in terms of the number of failures, so p = 1-0.8 =0.2. a)

10  1 (1  0.2)102 0.2 2  0.0604 P(X  10)    2 1 

P(X  4)  P(X = 2) + P(X = 3)  P(X  4)  0.04  0.064  0.0768  0.1808 where  2  1 (1  0.2) 22 0.2 2  0.04 P(X  2)    2  1  3  1 (1  0.2) 32 0.2 2  0.064 P(X  3)    2  1

b)

 4  1 (1  0.2) 42 0.2 2  0.0768 P(X  4)    2  1 r 3 c) E[Y ]    15 p 0.2 3-138. X = the number of defective bulbs in an array of 30 LED bulbs. Here X is a binomial random variable with p = 0.001 and n = 30 a) P(X  2)  1  [P(X = 0)  P(X  1)]

 30    30  P(X  2) = 1   0.0010 (1  0.001) 30   0.0011 (1  0.001) 29   0.0004 1  0   b) Let Y = number of automotive lights tested to obtain one light with two or more defective bulbs among thirty LED bulbs. Here Y is distributed with negative binomial with success probability p = 0.0004 and r = 1.

E[Y ] 

r 1   2500 p 0.0004

3-139. X = the number of patients selected from hospital 4 in order to admit 2. Then X has a negative binomial distribution with p = 0.23 and r =2. Y = the number of patients selected from hospital 4 in order to admit 10. Then Y has a negative binomial distribution with p = 0.23 and r =10.

984  0.2273 4329 b) P(X  4)  P(X = 4)  P(X  3)  P(X = 2)  4  1 (1  0.23) 42 0.232  0.094 P(X  4)   2  1   a)



3-24

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

 3  1 (1  0.23) 32 0.232  0.062 P(X  3)    2  1  2  1 (1  0.23) 22 0.232  0.031 P(X  2)   2  1   P(X  4)  0.094  0.062  0.031  0.185 r 10 c) E[Y ]    43.99 p 0.227 3-140.

Let X denote the number of customers who visit the website to obtain the first order. a) Yes, since the customers behave independently and the probability of a success (i.e., obtaining an order) is the same for all customers. b) A = the event that a customer views five or fewer pages B = the event that the customer orders

P( B)  P( B  A)  P( B  A' )  P( B | A) P( A)  P( B | A' ) P( A' ) P( B)  0.01(1  0.25)  0.1(0.25)  0.0325

Then X has a geometric distribution with p = 0.0325 P(X = 10) = 0.0325 (1 - 0.0325)9 = 0.024

Section 3-8 3-141.

X has a hypergeometric distribution with N = 100, n = 4, K = 20

    20(82160)  0.4191 a) P( X  1)    3921225 20 80 1 3 100 4

b)

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

    4845(1)  0.001236 c) P( X  4)    3921225 20 80 4 0 100 4

K  20   4   0.8 N  100   N n  96  V ( X )  np(1  p)   4(0.2)(0.8)   0.6206  N 1   99 

d) E ( X )

3-142.

 np  n

    (4 16 15 14) / 6  0.4623 a) P ( X  1)    (20 19 18 17) / 24     1  0.00021 b) P ( X  4)    (20 19 18 17) / 24 4 1

16 3 20 4 4 16 4 0 20 4

c)

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

P( X  2)  P( X  0)  P( X  1)  P( X  2)

                  4 0



16 4 20 4

4 1

16 3 20 4

4 2

16 2 20 4

 16151413 4161514 61615      24 6 2    20191817    24  

 0.9866

d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-143.

Here N = 10, n = 3, 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 

3-144. (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 3-145.

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

P( X  1) 

        0.1201   240 560 1 9 800 10

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

b) n = 10

P( X  1)  1  P( X  1)  1  [ P( X  0)  P( X  1)]

     P( X  0)    240 560 0 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 3-146.

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

3-26

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

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 3-147.

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

3-148.

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)    6 6

6 5

6 4

1

34 0 40 6 34 1 40 6 34 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 3,838,380

and

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

3-149.

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

P( X  1)  1  P( X  0)

    P( X  0)    10 0

38 5

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!

3-27

Applied Statistics and Probability for Engineers, 6th edition

    On the second day, P ( X  0)    6 0

42 5 48 5

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



March 27, 2014

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.

3-150. a) For the first exercise, the finite population correction is 96/99. For the second exercise, the finite population correction is 16/19. Because the finite population correction for the first exercise is closer to one, the binomial approximation to the distribution of X should be better in that exercise. 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

1

4 4

4

3

0

The results from the binomial approximation are close to the probabilities obtained from the hypergeometric distribution. 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 from the hypergeometric distribution as the results obtained in part (b). d) X is approximately binomially distributed with n = 20 and p = 20/140 = 1/7.

P( X  1)  1  P( X  0) 

    

 1  0.0458  0.9542

1 0 6 20 7 7

20 0

The finite population correction is 120/139 = 0.8633 X is approximately binomially distributed with n = 20 and p = 5/140 =1/28

P( X  1)  1  P( X  0) 

    

1 0 27 20 28 28

20 0

 1  0.4832  0.5168

The finite population correction is 120/139 = 0.8633

3-151.

a) 𝑃(𝑋

= 4) =

953−242) (242 4 )( 0 (953) 4

b) 𝑃(𝑋

= 0) =

953−242) (242 0 )( 4 (953) 4

= 0.0041

= 0.3091

c) Probability that all visits are from hospital 1

𝑃(𝑋 = 4) =

(195)(953−195) 4 0 (953) 4

= 0.0017

Probability that all visits are from hospital 2

𝑃(𝑋 = 4) =

953−270) (270 4 )( 0 (953) 4

= 0.0063

Probability that all visits are from hospital 3

𝑃(𝑋 = 4) =

953−246) (246 4 )( 0 (953) 4

= 0.0044

Probability that all visits are from hospital 4

𝑃(𝑋 = 4) =

953−242) (242 4 )( 0 (953) 4

= 0.0041

3-28

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

Probability that all visits are from the same hospital

= .0017 + .0063 + .0044 + .0041 = 0.0165

3-152.

a) 𝑃(𝑋

b)

= 2) =

(3290)(7726−3290) 2 4−2 (7726) 4

= 0.359

𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = 1 −

c) 𝜇

3290

7726−3290) (3290 0 )( 4−0 (7726) 4

= 1 − 0.109 = 0.891

= 𝐸(𝑋) = 𝑛𝑝 = 4 (7726) = 1.703

3-153. a) Let X1 denote the number of wafers in the sample with high contamination. Here X 1 has a hypergeometric distribution with N = 940, K = 358, and n = 10 when the sample size is 10.

 358  940  358     4  10  4   P(X1  4)   0.25  940     10  b) Let X2 denote the number of wafers in the sample with high contamination and from the center of the sputtering tool. Here X2 has a hypergeometric distribution with N = 940, K = 112, and n = 10 when the sample size is 10.

112  940  112     0  10  0   P(X 2  1)  1  P(X 2  0)  1   1  0.28  0.72  940     10  c) Let X3 denote the number of wafers in the sample with high contamination or from the edge of the sputtering tool. Here X3 has a hypergeometric distribution with N = 940, K = 426, and n = 10 when the sample size is 10.

 426  940  426     3  10  3   P(X 3  3)   0.16  940     10  where 68 + 112 + 246 = 426 d) Let X4 denote the number of wafers in the sample with high contamination. Here X 4 has a hypergeometric distribution with N = 940 and K = 358.

 1)  0.9  358  940  358     0  n  0   P(X 4  1)  1  P(X 4  0)  1   0.9  940     n 

Find the minimum n that satisfies the condition P(X 4

Through trial of values for n, the minimum n is 5. 3-154. Let X denote the number of patients in the sample that adhere. Here X has a hypergeometric distribution with N = 500, K = 50 and n = 20 when the sample size is 20.

3-29

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

 50  500  50     2  20  2   a) P(X  2)   0.291  500     20  b) P(X  2)  P(X  0)  P(X  1)  50  500  50   50  500  50        0  20  0   1  20  1      0.116  0.270  0.386  500   500       20   20 

P(X  2)  1  P(X  0)  P(X  1)  P(X  2)  1  0.116  0.270  0.291  0.323 K 50  20 2 d) E X   np  n N 500 c)

 N n  480  Var ( X )  np (1  p )   20(0.1)(0.9)   1.73  N 1   499  3-155. Let X = the number of sites with lesions in the sample. Here X has hypergeometric distribution with N = 50, K = 5 and n = 8 when the sample size is 8.

a)

 5  50  5     0  8  0   P(X  1)  1  P(X  0)  1   0.599  50    8

 5  50  5   5  50  5        0  8  0   1  8  1   b) P(X  2)  1  [P(X  0)  P(X  1)]  1    0.176  50   50      8 8 c) We need to find the minimum n that satisfies the condition P(X  1)  0.90 Equivalently, 1  P(X  0)  0.90 or P(X  0)  0.10  5  50  5      0  n  0   0.10  50    n

From trials of n values

 5  50  5   5  50  5         0  18  0   0.1   0 17  0  .  50   50       18   17 

The smallest sample size that satisfies the condition is n = 18 3-156.

3-30

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

Let X = the number of major customers that accept the plan in the sample. Here X has hypergeometric distribution with N = 50, K = 15, and n = 10 when the sample size is 10.

a)

15  50  15     2  10  2   P(X  2)   0.241  50     10 

15  50  15     0  10  0   b) P(X  1)  1  P(X  0)  1   0.982  50     10  c) We need to find the minimum K that satisfies the condition P(X  1)  0.95 Equivalently, 1  P(X  0)  0.95 and P(X  0)  0.05 . This requires  K  50  K      0  10  0   0.05  50     10 

We have

 4  50  4   3  50  3         0  10  0   0.05   0 10  0  .  50   50       10   10 

The minimum number of major customers that would need to accept the plan to meet the given objective is 4.

Section 3-9 3-157.

e 4 4 0  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 

e 4 4 4  01954 . 4! e 4 48 d) P( X  8)   0.0298 8! c) P( X  4) 

3-158.

a) P( X  0)  e 0.4  0.6703 e 0.4 (0.4) e 0.4 (0.4) 2   0.9921 1! 2! e 0.4 (0.4) 4 c) P( X  4)   0.000715 4! e 0.4 (0.4)8 d) P( X  8)   109 .  108 8!

b) P( X  2)  e 0.4 

3-31

Applied Statistics and Probability for Engineers, 6th edition

3-159.

3-160.

3-161.

March 27, 2014

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 10 105 P( X  5)   0.0378 . 5! e 10 10 e 10 102 e 10 103 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 e 20 2015 E(Y) = 20. P(Y  15)   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 E(W) = 5. P(W  5)   01755 . 5! λ=1, Poisson distribution. f(x) = e- λ λx/x! a) P(X ≥ 2) = 0.264 b) In order that P(X ≥ 1) = 1 - P(X=0) = 1-e- λ exceeds 0.95, we need λ = 3. Therefore 3(16) = 48 cubic light years of space must be studied.

3-162.

a)  = 14.4, P(X = 0) = 6E10-7 b)  = 14.4/5 = 2.88, P(X = 0) = 0.056 c)  = 14.4(7)(28.35)/225 = 12.7, P(X ≥ 1) = 0.999997 d) P(X ≥ 28.8) = 1 - P(X ≤ 28) = 0.00046. Unusual.

3-163.

a) λ = 0.61 and P(X ≥ 1) = 0.4566 b)  = 0.61(5) = 3.05, P(X = 0) = 0.047

3-164. 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 E(Y) = 1.

P(Y  1) 

e111  e1  0.3679 1!

c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable with E(W) = 2.

P(W  0)  e2  0.1353 1 1 d) P(Y  2)  1  P(Y  1)  1  P(Y  0)  P(Y  1)  1  e  e  0.2642 3-165.

a)

E ( X )  0.2 errors per test area

b) P( X

 2)  e 0.2 

e 0.2 0.2 e 0.2 (0.2) 2   0.9989 1! 2!

99.89% of test areas

3-166.

a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with E(X) = 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 E(Y) = 1.

P(Y  1)  1  P(Y  0)  1  e1  0.6321

3-32

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

c) The assumptions of a Poisson process require that the probability of an 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 heavily and lightly loaded sections of the highway. 3-167.

a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random variable with E(X) = 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 has a binomial distribution with n = 10 and p = 0.3935

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 3-168.

a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with E(X) = 0.16.

P( X  0)  e0.16  0.8521 b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with E(Y) = 0.48.

P(Y  1)  1  P(Y  0)  1  e48  0.3812 3-169.

≥ 2) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)] = 1 − [ b) 𝜆 = 0.25(5) = 1.25 per five days 𝑃(𝑋 = 0) = 𝑒 −1.25 = 0.287 c) 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) a) 𝑃(𝑋

𝑒 −1.25 1.25

= 𝑒 −1.25 + 3-170.

1!

+

𝑒 −1.25 1.252 2!

𝑒 −0.25 0.250 0!

= 0.868

= 0) = 𝑒 −1.5 = 0.223 b) 𝐸(𝑋) = 1.5(10) = 15 per 10 minutes 𝑃(𝑋 ≤ 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) a) 𝑃(𝑋

= 𝑒 −15 +

𝑒 −15 15 1!

+

𝑒 −15 152 2!

= 0.000039

c) No, if a Poisson distribution is assumed, the intervals need not be consecutive. 3-171.

a) Let X denote the number of cabs that pass your workplace in 10 minutes. Then, X is a Poisson random variable with

P(X  0) 

T  5

10 5  60 6

e 5 / 6 (5 / 6) 0  0.435 0!

b) Let Y denote the number of cabs that pass your workplace in 20 minutes. Then, Y is a Poisson random variable with

T  5

20 5  60 3

3-33

+

𝑒 −0.25 0.251 1!

] = 0.026

Applied Statistics and Probability for Engineers, 6th edition

e 5 / 3 (5 / 3) 0  0.811 0!

P(Y  1)  1  P(Y  0)  1  c) Let

*

March 27, 2014

be the mean number of cabs per hour and T = 1/6 hour. Find

P(X  0) 

e

 / 6 *

that satisfies the following condition:

( / 6)  0.1 0!

e   / 6  0.1 and  *

*

*

* 6

0

 ln( 0.1)

*  13.816 3-172.

a) Let X denote the number of orders that arrive in 5 minutes. Then, X is a Poisson random variable with

P(X  0)  b)

T  12

5  1. 60

e 1 (1) 0  0.368 0!

P(X  3)  1  P(X  0)  P(X  1)  P(X  2)

 e 1 (1) 0 e 1 (1)1 e 1 (1) 2  P(X  3)  1       0.080 1! 2!   0! c) Let Y denote the number of orders arriving in the length of time T (in hours) that satisfies the condition. The mean of the random variable Y is 12T and T satisfies the following condition:

e 12T (12T ) 0  0.001 0!  0.001 and  12T  ln( 0.001)

P(Y  0) 

e 12T

Therefore, T = 0.57565 hours = 34.54 minutes. 3-173.

a) Let X denote the number of visits in a day. Then, X is a Poisson random variable with T

 1.8 e (1.8) n P(X  5)  1   P(X  n)  1    0.010 n! n 0 n 0 5

1.8

5

b) Let Y denote the number of visits in a week. Then, Y is a Poisson random variable with T 4

4

n 0

n 0

P(X  5)   P(X  n)  

e

12.6

 1.8(7)  12.6 .

(12.6) n  0.005 n!

c) Let Z denote the number of visits in T days that satisfies the given condition. The mean of the random variable Z is 1.8T.

e 1.8T (1.8T ) 0  0.99 0!  0.01 and  1.8T  ln( 0.01) . As a result, T  2.56

P(Z  1)  1  P(Z  0)  1 

e 1.8T

d) With T=1, determine  such that Solving the equation gives  3-174.

5

days.

e  n  0.1 n! n 0 5

P(X  5)  1   P(X  n)  1  

 3.15

n 0

a) Let X denote the number of inclusions in cast iron with a volume of cubic millimeter. Then, X is a Poisson random variable with   2.5 and T = 1

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Applied Statistics and Probability for Engineers, 6th edition

P(X  1)  1  P(X  0)  1 

March 27, 2014

e 2.5 (2.5) 0  0.918 0!

b) Let Y denote the number of inclusions in cast iron with a volume of 5.0 cubic millimeters. Then, Y is a Poisson random variable with T  2.5(5)  12.5

e 12.5 (12.5) n P(Y  5)  1   P(X  n)  1    0.995 n! n 0 n 0 4

4

c) Let Z denote the number of inclusions in a volume of V cubic millimeters that satisfies the

P(Z  1)  0.99

condition

The mean of the random variable Z is 2.5V.

P(Z  1)  1  P(Z  0)  1 

e 2.5V (2.5V ) 0  0.99 . 0!

As a result, V=1.84 cubic millimeters. d) With T = 1, determine  that satisfies

P(X  1)  1  P(Z  0)  1  As a result,

e   ( ) 0  0.95 0!

  3.00 inclusions per cubic millimeter.

Supplemental Exercises

3-175.

E( X ) 

1 1 1 1 3 1 1        , 8  3 4  3 8  3 4 2

3-176.

2

2

2

1 1  1  1 3 1  1  V ( X )                    0.0104 8 3  4 3 8 3  4 1000  1 999 a) P( X  1)    1 0.001 (0.999)  0.3681   1000  0.0010 0.999999  0.6319 b) P( X  1)  1  P( X  0)  1    0  1000  1000  1000  1000 999 0.0010 0.999   0.0011 0.999   0.0012 0.999998 c) P( X  2)    0   1   2   0.9198 d) E ( X )  1000(0.001)  1 V ( X )  1000(0.001)(0.999)  0.999

3-177.

a) n = 50, p = 5/50 = 0.1, because E(X) = 5 = np

 50   50   50  50 49 48  2)   0.10 0.9   0.11 0.9   0.12 0.9  0.112 0 1 2  50  49  50  50 1 0  48 c) P( X  49)    49 0.1 0.9   50 0.1 0.9  4.51  10     b) P( X

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Applied Statistics and Probability for Engineers, 6th edition

3-178.

March 27, 2014

a) Binomial distribution with p = 0.01, n = 12 b) P(X > 1) = 1 - P(X ≤ 1)= 1 -

12  0 12    p (1  p)12 -   p 1 (1  p)14 0  1 

= 0.0062

c) µ = E(X) = np =12(0.01) = 0.12 V(X) = np(1 - p) = 0.1188 3-179.

3-180.

σ=

V (X ) = 0.3447

a)

(0.5)12  0.000244

b)

C126 (0.5)6 (0.5)6 = 0.2256

c)

C512 (0.5) 5 (0.5) 7  C612 (0.5) 6 (0.5) 6  0.4189

a) Binomial distribution with n =100, p = 0.01 b) P(X ≥ 1) = 0.634 c) P(X ≥ 2)= 0.264 d) µ = E(X) = np =100(0.01) = 1 V(X) = np(1 - p) = 0.99 and σ=

V (X ) = 0.995

e) Let pd= P(X≥2)= 0.264, Y = number of messages that require two or more packets be resent. Y is binomial distributed with n = 10, pm = pd(1/10) = 0.0264 P(Y ≥ 1) = 0.235 3-181.

Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random variable with p = 0.20. a) P(X = 4) = (1-0.2)30.2 = 0.1024 b) By independence, (0.8)10 = 0.1074. (Also, P(X > 10) = 0.1074)

3-182.

Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X is a geometric random variable with p = 0.6. P(X  3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.4 2(0.6) = 0.936.

3-183.

Let X denote the number of fills needed to detect three underweight packages. Then, X is a negative binomial random variable with p = 0.001 and r = 3. a) E(X) = 3/0.001 = 3000 b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, X = 1731.18

3-184.

Geometric random variable with p = 0.1 a) f(x) = (1-p)x-1p = 0.9(x-1)0.1 b) P(X=5) = 0.94(0.1) = 0.0656 c) µ = E(X) = 1/p=10 d) P(X ≤ 10) = 0.651

3-185.

a) E(X) = 6(0.5) = 3 P(X = 0) = 0.0498 b) P(X ≥ 3) = 0.5768 c) P(X ≤ x) ≥ 0.9, and by trial x = 5 d) σ2 = λ = 6. Not appropriate.

3-186.

Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a hypergeometric distribution with N = 15, n = 3, and K = 2.

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

 2 13     0 3 13!12! P( X  1)  1  P( X  0)  1      1   0.3714 15  10!15!   3 3-187.

Let X denote the number of calls that are answered in 30 seconds or less. Then, X is a binomial random variable with p = 0.75. a) P(X = 9) =

10   (0.75)9 (0.25)1  0.1877 9

b) P(X  16) = P(X = 16) +P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

 20   20   20    (0.75) 16 (0.25) 4   (0.75) 17 (0.25) 3   (0.75) 18 (0.25) 2  16   17   18   20   20    (0.75) 19 (0.25) 1   (0.75) 20 (0.25) 0  0.4148  19   20  c) E(X) = 20(0.75) = 15 3-188.

Let Y denote the number of calls needed to obtain an answer in less than 30 seconds. a) P(Y  4)  (1  0.75) b) E(Y) = 1/p = 1/0.75 = 4/3

3-189.

3-190.

3

0.75  0.253 0.75  0.0117

Let W denote the number of calls needed to obtain two answers in less than 30 seconds. Then, W has a negative binomial distribution with p = 0.75.  5 a) P(W = 6) =   (0.25)4 (0.75)2  0.0110  1 b) E(W) = r/p = 2/0.75 = 8/3 a) Let X denote the number of messages sent in one hour.

P( X  5) 

e 5 55  0.1755 5!

b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with E(Y) = 7.5.

e 7.5 (7.5)10 P(Y  10)   0.0858 10! c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with E(W) = 2.5

P(W  2)  P(W  0)  P(W  1)  0.2873

3-191.

X is a negative binomial with r=4 and p=0.0001 E ( X )  r / p  4 / 0.0001  40000 requests

3-192.

X  Poisson with E(X) = 0.01(100) = 1

P(Y  3)  e 1 

e 1 (1)1 e 1 (1) 2 e 1 (1) 3    0.9810 1! 2! 3!

3-193.

Let X denote the number of individuals that recover in one week. Assume the individuals are independent. Then, X is a binomial random variable with n = 20 and p = 0.1. P(X  4) = 1  P(X  3) = 1  0.8670 = 0.1330.

3-194.

a) P(X = 1) = 0, P(X = 2) = 0.0025, P(X = 3) = 0.01, P(X = 4) = 0.03, P(X = 5) = 0.065 P(X = 6) = 0.13, P(X = 7) = 0.18, P(X = 8) = 0.2225, P(X = 9) = 0.2, P(X = 10) = 0.16

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Applied Statistics and Probability for Engineers, 6th edition

b) P(X = 1) = 0.0025, P(X=1.5) = 0.01, P(X = 2) = 0.03, P(X = 2.5) = 0.065, P(X = 3) = 0.13 P(X = 3.5) = 0.18, P(X = 4) = 0.2225, P(X = 4.5) = 0.2, P(X = 5) = 0.16 3-195.

Let X denote the number of assemblies needed to obtain 5 defectives. Then, X is a negative binomial random variable with p = 0.01 and r=5. a) E(X) = r/p = 500 b) V(X) = 5(0.99)/0.012 = 49500 and  = 222.49

3-196.

Here n assemblies are checked. Let X denote the number of defective assemblies. If P(X  1)  0.95, then P(X = 0)  0.05. Now,

n  (0.01) 0 (0.99) n  99 n 0 n(ln( 0.99))  ln( 0.05) ln( 0.05) n  298.07 ln( 0.95)

P(X = 0) =

and 0.99n  0.05. Therefore,

Therefore, n = 299

3-197.

Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1.

3-198.

Let X denote the number of products that fail during the warranty period. Assume the units are independent. Then, X is a binomial random variable with n = 500 and p = 0.02. a) P(X = 0) =

 500   (0.02) 0 (0.98) 500  4.1 x 10-5 0  

b) E(X) = 500(0.02) = 10 c) P(X >2) = 1  P(X  2) = 0.9995 3-199.

3-200.

f X (0)  (0.1)(0.7)  (0.3)(0.3)  0.16 f X (1)  (0.1)(0.7)  (0.4)(0.3)  0.19 f X (2)  (0.2)(0.7)  (0.2)(0.3)  0.20 f X (3)  (0.4)(0.7)  (0.1)(0.3)  0.31 f X (4)  (0.2)(0.7)  (0)(0.3)  0.14 a) P(X  3) = 0.2 + 0.4 = 0.6 b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8 c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7 d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9 e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1)  (3.9)2 = 3.09

3-201. x f(x)

3-202.

2 0.2

5.7 0.3

6.5 0.3

8.5 0.2

Let X and Y denote the number of bolts in the sample from supplier 1 and 2, respectively. Then, X is a hypergeometric random variable with N = 100, n = 4, and K = 30. Also, Y is a hypergeometric random variable with N = 100, n = 4, and K = 70. a) P(X = 4 or Y=4) = P(X = 4) + P(Y = 4)

3-38

March 27, 2014

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

 30  70   30  70        4 0 0 4         100  100       4   4   0.2408

b) P[(X = 3 and Y = 1) or (Y = 3 and X = 1)]=

 30  70   30  70         3 1 1 3         0.4913 100     4 

3-203.

Let X denote the number of errors in a sector. Then, X is a Poisson random variable with E(X) = 0.32768. a) P(X > 1) = 1  P(X  1) = 1  e–0.32768  e–0.32768(0.32768) = 0.0433 b) Let Y denote the number of sectors until an error is found. Then, Y is a geometric random variable and P = P(X  1) = 1  P(X = 0) = 1  e–0.32768 = 0.2794 E(Y) = 1/p = 3.58

3-204.

Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson random variable with E(X) = 0.25(8) = 2. a) P(X  3) = 1  P(X  2) = 1  [e-2 + e-2(2) + (e-222)/2!] = 1  0.6767 = 0.3233. b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with E(Y) = 4, and P(Y > 2) =1- P(Y  2) = e-4 + (e-441)/1!+ (e-442)/2! =1 - [0.01832 + 0.07326 + 0.1465] = 0.7619.

3-205.

a) Hypergeometric random variable with N = 500, n = 5, and K = 125

125  375     0  5  6.0164 E10  f X (0)    0.2357 2.5524 E11  500     5  125  375     1  4  125(8.10855E8)  f X (1)    0.3971 2.5525E11  500     5  125  375     2  3  7750(8718875)  f X (2)    0.2647 2.5524 E11  500     5  125  375     3  2  317750(70125)  f X (3)    0.0873 2.5524 E11  500     5  125  375     4  1  9691375(375)  f X (4)    0.01424 2.5524 E11  500     5 

3-39

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

125  375     5  0  2.3453E8  f X (5)    0.00092 2.5524 E11  500     5  b)

x f(x)

3-206.

0 1 2 3 4 5 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 5 6 7 8 9 10 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000

Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p. If P(X  1) = 0.9, then P(X = 0) = 0.1. Then

 30  0  ( p) (1  p)30  0.1 , and 30[ln(1p)] = ln(0.1), 0

and p = 0.0739

3-207.

Let T denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with E(X) = 10T. Then, P(X=0) = 0.9 and e-10T = 0.9, resulting in T = 0.0105 hours = 37.8 seconds

3-208.

a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with E(X) = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679 b) Let Y denote the number of flaws in one panel. P(Y  1) = 1  P(Y = 0) = 1  e-0.02 = 0.0198 Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198. E(W) = 1/0.0198 = 50.51 panels.

c)

P(Y  1)  1  P(Y  0)  1  e 0.02  0.0198

Let V denote the number of panels with 1 or more flaws. Then V is a binomial random variable with n = 50 and p = 0.0198

 50   50  P(V  2)   0.0198 0 (.9802) 50   0.01981 (0.9802) 49 0 1  50    0.0198 2 (0.9802) 48  0.9234 2 3-209.

a) Let X denote the number of cacti per 10,000 square meters.

10,000  2.8 106 e 2.8 (2.8) 0 b) The unit is 10,000 square meters and T = 1. P(X  0)   0.061 0! Then, X is a Poisson random variable with E(X) =

280

c) Let Y denote the number of cacti in a region of area T (in units of 10,000 square meters). The mean of the random variable Y is 2.8T and

P(Y  2)  1  P(Y  0)  P(Y  1)  0.9

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

e 2.8T (2.8T ) 0 e 2.8T (2.8T )1   0.1 0! 1!

This can be solved in computer software to obtain 2.8T  3.8897 Therefore, T = 3.8897/2.8 = 1.39 (10,000 square meters) = 13,900 square meters. 3-210. X = the number of sites with lesions in the sample Then X has hypergeometric distribution with N = 50 and n = 8. We need to find the minimum K that meets the condition P(X  1)  0.95

1  P(X  0)  0.95 and P(X  0)  0.05 Therefore,  K  50  K      0  8  0   0.05  50    8

Equivalently,

From trials of values for K, we have

15  50  15  14  50  14         0  8  0   0.05   0  8  0  .  50   50      8 8 So, the minimum number of sites with lesions that satisfies the given condition is 15. We also need to find the minimum K that meets the condition

P(X  1)  0.99

1  P(X  0)  0.99 and P(X  0)  0.01 Therefore,  K  50  K      0  8  0   0.01  50    8

Equivalently,

From trials of values for K, we have

 21 50  21  20  50  20         0  8  0   0.01   0  8  0  .  50   50      8 8 So, the minimum number of sites with lesions that satisfies the given condition is 21.

Mind Expanding Exercises 3-211.

The binomial distribution P(X = x) =

n! px(1-p)n-x r!n  r !

The probability of the event can be expressed as p = /n and the probability mass function can be written as

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

n! [/n]x[1 – (/n)]n-x x!n  x ! n  (n  1)  (n  2)  (n  3)......  (n  x  1) λ x P(X = x) (1 – (/n))n-x x x! n P(X = x) =

Now we can re-express as: [1 – (/n)]n-x = [1 – (/n)]n[1 – (/n)]-x In the limit as n  

n  (n  1)  (n  2)  (n  3)......  (n  x  1) n

x

1

As n   the limit of [1 – (/n)]-x  1 Also, we know that as n  

1 - λ n n  e  Thus, x

λ λ e x!

P(X = x) =

The distribution of the probability associated with this process is known as the Poisson distribution and we can express the probability mass function as

e  x f(x) = x! 

3-212.

Show that

 (1  p)i 1 p  1 using an infinite sum.

i 1 

To begin,

 (1  p) i 1

i 1



p  p  (1  p)i 1 , i 1

From the results for an infinite sum this equals 

p (1  p)i 1  i 1

p p  1 1  (1  p) p

3-213.

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

E ( X )  [(a  (a  1)  ...  b](b  a  1) a 1  b  i  i    i 1    i1

(b  a  1)

 (b  a  b  a )    2   2



 b(b  1) (a  1)a    2   2

2

(b  a  1)

(b  a  1)

 (b  a )(b  a  1)   2  

(b  a  1)

(b  a ) 2

b  b 2 (b  a  1)(b  a ) 2    i  (b  a ) i   4 i a i a   i a V (X )   b  a 1 b  a 1 2 b(b  1)(2b  1) (a  1)a(2a  1)  b(b  1)  ( a  1) a  (b  a  1)(b  a)   (b  a )    6 6 2 4   b  a 1 (b  a  1) 2  1  12 b

 i  b2a 

3-214.

2

Let X denote a geometric random variable with parameter p. Let q = 1 – p. 

E ( X )   x(1  p)

x 1

x 1

 p



p  p  xq

x 1

x 1

d  x d  q   q  p    dq x 1 dq  1  q 



d x q x 1 dq

 p

 1(1  q)  q(1)   p (1  q) 2  

 1  1  p 2   p  p 





V ( X )   ( x  1p ) 2 (1  p ) x 1 p   px 2  2 x  x 1

x 1





 p  x 2 q x 1  2 xq x 1  x 1

x 1



 p  x 2 q x 1  x 1

2 p2





1 p

q

1 p

(1  p)

x 1

x 1

1 p2



 p  x 2 q x 1  x 1

1 p2

 p dqd  q  2q 2  3q 3  ... 

1 p2

 p dqd  q (1  2q  3q 2  ...)  

1 p2

 p dqd  (1qq )2   p12  2 pq(1  q ) 3  p (1  q ) 2     2(1  p)  p  1  (1  p)  q  p2 p2 p2

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1 p2

x 1

Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

3-215. Let X = number of passengers with a reserved seat who arrive for the flight, n = number of seat reservations, p = probability that a ticketed passenger arrives for the flight. a) In this part we determine n such that P(X  120)  0.9. By testing several values for n, the minimum value is n = 131. b) In this part we determine n such that P(X > 120)  0.10 which is equivalent to 1 – P(X  120)  0.10 or 0.90  P(X  120). By testing several values for n, the solution is n = 123. c) One possible answer follows. If the airline is most concerned with losing customers due to over-booking, they should only sell 123 tickets for this flight. The probability of over-booking is then at most 10%. If the airline is most concerned with having a full flight, they should sell 131 tickets for this flight. The chance the flight is full is then at least 90%. These calculations assume customers arrive independently and groups of people that arrive (or do not arrive) together for travel make the analysis more complicated. 3-216.

Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with p = 0.01 and n is to be determined. If P ( X  1)  0.90 , then P ( X  0)  0.10 . Now, P(X = 0) =

n 3-217.

 p (1  p) n 0

0

n

 (1  p) n . Consequently, (1  p) n  0.10 , and

ln 0.10  229.11 . Therefore, n = 230 is required. ln( 1  p)

If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming product in a lot that is 20% nonconforming. If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomial approximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size 5000 is nonconforming, then the probability of zero nonconforming products in the sample is approximately 7E-12. Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might be much larger than is needed.

3-218. Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and n is to be determined such that P( X  100)  0.95 n

P ( X  100)

102 0.666 103 0.848 104 0.942 105 0.981 Therefore, 105 components are needed.

3-219. Let X denote the number of rolls produced. Revenue at each demand 1000 2000 0.3x 0.3x 0  x  1000 mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x 0.05x 0.3(1000) + 0.3x 1000  x  2000 0.05(x-1000) mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 2000  x  3000 0.05(x-1000) 0.05(x-2000) 0 0.05x

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3000 0.3x 0.3x

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Applied Statistics and Probability for Engineers, 6th edition

March 27, 2014

mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x 0.05x 0.3(1000) + 0.3(2000) + 0.3(3000)+ 3000  x 0.05(x-1000) 0.05(x-2000) 0.05(x-3000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x3000)]0.2 - 0.1x

0  x  1000 1000  x  2000 2000  x  3000 3000  x

Profit 0.125 x 0.075 x + 50 200 -0.05 x + 350

The bakery can produce anywhere from 2000 to 3000 and earn the same profit.

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Max. profit $ 125 at x = 1000 $ 200 at x = 2000 $200 at x = 3000 $200 at x = 3000