chapter 06 random variables and probability distributions · 2016-11-15 · chapter 6 random...

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AP* SOLUTIONS

Chapter 6 Random Variables and Probability Distributions

Section 6.1 Exercise Set 1

6.1: (a) discrete

(b) continuous

(c) discrete

(d) discrete

(e) continuous

6.2: The possible values for x are 1x (the positive integers).

Five possible outcomes, with their corresponding x values, are shown below.

Outcomes x S 1

LS 2 RLS 3 RRS 3

LRLRS 5

6.3: (a) 3, 4, 5, 6, 7

(b) –3, –2, –1, 1, 2, 3

(c) 0, 1, 2

(d) 0, 1


6.4: (a) continuous

(b) continuous

(c) continuous

(d) discrete

(e) continuous

*AP and Advanced Placement Program are registered trademarks of the College Entrance Examination Board, which was not involved in the production of, and does not endorse, this product.


(f) continuous

(g) discrete

6.5: The length of the diagonal of the square is 2 1.4142 . Therefore, 0 2x , and x is a continuous random variable.

6.6: The possible values of y are the even integers, and y is a discrete random variable.

Additional Exercises for Section 6.1

6.7: (a) discrete

(b) continuous

(c) continuous

(d) discrete

6.8: (a) 3.1 3 0.1x

(b) continuous

6.9: (a) 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

(b) –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5

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


6.10: (a) (4) 1 (0.65 0.20 0.10 0.04) 0.01p

(b) (1) 0.20p is interpreted as the probability of randomly selecting a carton of one dozen

eggs and finding exactly 1 broken egg is 0.20.

(c) ( 2) (0) (1) (2) 0.65 0.20 0.10 0.95P y p p p ; The probability that a

randomly selected carton of eggs contains 0, 1, or 2 broken eggs is 0.95.

(d) ( 2) (0) (1) 0.65 0.20 0.85P y p p ; This probability is smaller than the

probability in part (c) because we only want fewer than 2 broken eggs, not two or fewer as was the case in part (c). Therefore, we are only considering 0 or 1 broken egg.

(e) Exactly 10 unbroken eggs is equivalent to exactly 2 broken eggs. (2) 0.10p .


(f) At least 10 unbroken eggs is equivalent to 10, 11, or 12 unbroken eggs, or 2, 1, or 0 broken eggs. From Part (c), ( 2) 0.95P y .

6.11: (a) For 1000 graduates, we would expect to see approximately 450 graduates who contributed nothing (either through declining to make a donation or were not reached by phone), 300 graduates to contribute $10, 200 graduates to contribute $25, and 50 graduates to contribute $50.

(b) The most common value of x in this population is $0.

(c) ( 25) (25) (50) 0.20 0.05 0.25P x p p

(d) ( 0) (10) (25) (50) 0.30 0.20 0.05 0.55P x p p p

6.12: (a) (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)

(b) Because the bottles are randomly selected, each of the outcomes enumerated in Part (a) are equally likely. Therefore, each has probability 1/6.

(c) The x value for each possible outcome is shown in the table below.

Outcome x (1, 2) 0 (1, 3) 1 (1, 4) 1 (2, 3) 1 (2, 4) 1 (3, 4) 2

The probability distribution of x is:

x 0 1 2 p(x) 1/6 4/6 1/6


6.13: (a) The table below shows all 16 possible outcomes of homeowners with earthquake insurance (S), and those without earthquake insurance (F), for four homeowners. Also shown are the computed probabilities, and the value of x, the number among the four who have earthquake insurance.

Outcome Probability xSSSS (0.2)(0.2)(0.2)(0.2) = 0.0016 4SSSF (0.2)(0.2)(0.2)(0.8) = 0.0064 3SSFS (0.2)(0.2)(0.8)(0.2) = 0.0064 3SFSS (0.2)(0.8)(0.2)(0.2) = 0.0064 3FSSS (0.8)(0.2)(0.2)(0.2) = 0.0064 3SSFF (0.2)(0.2)(0.8)(0.8) = 0.0256 2SFSF (0.2)(0.8)(0.2)(0.8) = 0.0256 2SFFS (0.2)(0.8)(0.8)(0.2) = 0.0256 2FFSS (0.8)(0.8)(0.2)(0.2) = 0.0256 2FSSF (0.8)(0.2)(0.2)(0.8) = 0.0256 2FSFS (0.8)(0.2)(0.8)(0.2) = 0.0256 2FFFS (0.8)(0.8)(0.8)(0.2) = 0.1024 1FFSF (0.8)(0.8)(0.2)(0.8) = 0.1024 1FSFF (0.8)(0.2)(0.8)(0.8) = 0.1024 1SFFF (0.2)(0.8)(0.8)(0.8) = 0.1024 1FFFF (0.8)(0.8)(0.8)(0.8) = 0.4096 0

The probability distribution of x is shown in the table below. The values in the probability distribution are found by adding the probabilities for the values of x from the table above.

x 0 1 2 3 4 p(x) 0.4096 0.4096 0.1536 0.0256 0.0016

(b) The most likely outcomes are 0 and 1.

(c) ( 2) (2) (3) (4) 0.1536 0.0256 0.0016 0.1808P x p p p


6.14: (a) ( 4) 0.25P x

(b) ( 4) 0.02 0.03 0.09 0.25 0.39P x

(c) (at most 5 courses) ( 5) 0.02 0.03 0.09 0.25 0.40 0.79P P x

(d) (at least 5 course) ( 5) 0.40 0.16 0.05 0.61P P x


(more than 5 courses) ( 5) 0.16 0.05 0.21P P x

(e) (3 6) (3) (4) (5) (6) 0.09 0.25 0.40 0.16 0.90P x p p p p

(3 6) (4) (5) 0.25 0.40 0.65P x p p

These two probabilities are different because the first probability includes ( 3)P x and

( 6)P x , whereas the second probability does not.

6.15: (a) In the long run, we would expect to see approximately 20% of the single-pizza orders for 12 inch pizza, 25% for 14 inch pizza, 50% for 16 inch pizza, and 5% for 18 inch pizza.

(b) ( 16) 0.20 0.25 0.45P x

(c) ( 16) 0.20 0.25 0.50 0.95P x

6.16:

Outcome Probability x DDD (0.7)(0.7)(0.7) = 0.343 3 DDI (0.7)(0.7)(0.3) = 0.147 2 DID (0.7)(0.3)(0.7) = 0.147 2 IDD (0.3)(0.7)(0.7) = 0.147 2 DII (0.7)(0.3)(0.3) = 0.063 1 IDI (0.3)(0.7)(0.3) = 0.063 1 IID (0.3)(0.3)(0.7) = 0.063 1 III (0.3)(0.3)(0.3) = 0.027 0

The probability distribution is:

x 0 1 2 3 p(x) 0.027 0.189 0.441 0.343


6.17: (a) (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)

(b)

Outcome x (1, 2) 2 (1, 3) 1 (1, 4) 1 (1, 5) 1 (2, 3) 1 (2, 4) 1 (2, 5) 1 (3, 4) 0 (3, 5) 0 (4, 5) 0

The probability distribution is:

x 0 1 2 p(x) 0.3 0.6 0.1


6.18: (a) ( 3) (0) (1) (2) (3) 0.10 0.15 0.20 0.25 0.70P x p p p p

(b) ( 3) (0) (1) (2) 0.10 0.15 0.20 0.45P x p p p

(c) ( 3) (3) (4) (5) (6) 0.25 0.20 0.06 0.04 0.55P x p p p p

(d) (2 5) (2) (3) (4) (5) 0.20 0.25 0.20 0.06 0.71P x p p p p

(e) Note that if 2 lines are not in use, then 4 lines are in use; if 3 lines are not in use, then 3 lines are in use; if 4 lines are not in use, then 2 lines are in use. The desired probability is between 2 and 4 lines (inclusive) are in use.

(2 4) (2) (3) (4) 0.20 0.25 0.20 0.65P x p p p

(f) At least 4 lines not in use corresponds to 4, 5, or 6 lines not in use. Therefore, with 4, 5, or 6 lines not in use, 0, 1, or 2 lines are in use. The desired probability is between 0 and 2 lines (inclusive) in use.

(0 2) (0) (1) (2) 0.10 0.15 0.20 0.45P x p p p .

©2014 Cengag

6.19: (a)

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(b)

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6.28: (a)

(b)

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(d)

6.29: The probability ( 1)P x is the smallest. ( 3)P x and (2 3)P x are equal to each

other, and larger than the other two probabilities.

6.30: (a) 1

( 5) 5 0.510

P x

(b) 1 1(3 5) 5 3 2 0.2

10 10P x

6.31: (2 3) (2 3) ( 2) ( 7)P x P x P x P x . The reason for this ordering of the

probabilities is that the smallest two probabilities are both equal to 1/10, the third probability is equal to 2/10, and the fourth probability is equal to 3/10. These values can be determined by shading the appropriate regions beneath the density curve and comparing the areas.


6.32: (a) ( ) 0(0.65) 1(0.20) 2(0.10) 3(0.04) 4(0.01) 0.56y y p y ; This is the mean

value of the number of broken eggs in the population of egg cartons.

(b) The only value of y that is less than 0.56 is 0, therefore ( 0.56) ( 0) 0.65P y P y .

This is not particularly surprising because, in the long run, 65% of egg cartons contain no broken eggs.

(c) This computation of the mean is incorrect because it does not take into account the probabilities with which the number of broken eggs need to be weighted. As written, the computation indicates that the number of broken eggs (0, 1, 2, 3, or 4) are all equally likely.


6.33: (a) ( ) 13.5(0.2) 15.9(0.5) 19.1(0.3) 16.38x x p x and

22 2 2 2

2 2 2

( ) (13.5 16.38) (0.2) (15.9 16.38) (0.5) (19.1 16.38) (0.3)

( 2.88) (0.2) ( 0.48) (0.5) (2.72) (0.3) 3.9936

x xx p x

Therefore, 2 3.9936 1.9984x x

(b) The mean, 16.38x cubic feet represents the long run average storage space of

freezers sold by this particular appliance dealer. The standard deviation, 1.9984x cubic

feet, represents a typical deviation in how much the storage space in freezers purchased deviates from the mean.

6.34: Answers may vary. Two possible probability distributions are shown below.

Probability Distribution 1 ( 3x and 1.265x ):

x 1 2 3 4 5 p(x) 0.15 0.20 0.30 0.20 0.15

Probability Distribution 2 ( 3x and 1.643x ):

x 1 2 3 4 5 p(x) 0.30 0.15 0.10 0.15 0.30

6.35: (a) The mean value of the total score is 1 2

1 110 (40) 0

4 4y x x .

This is not surprising. Consider a student who guesses at random the answer to each question. In that case, we would expect 1/5 of the questions to be answered correctly, and 4/5 of the questions to be answered incorrectly. Using this scoring model, the student loses 1/4 point for each incorrect answer, and earns one point for each correct answer. For the 50 question test, we would expect the student to answer 10 questions correctly and 40 questions incorrectly, resulting in a total score of 0.

(b) The formulas for computing variances and standard deviations given in this section

require that the random variables 1x and 2x be independent. These variables are not

independent because 1 2 50x x , which tells us that if we know the number of correct

answers ( 1x ), we also know the number of incorrect answers ( 2x ).


6.36: (a) ( ) 0(0.05) 1(0.10) 5(0.10) 2.8x x p x

2 2 2 22 ( ) 0 2.8 0.5 1 2.8 0.10 5 2.8 0.10 1.66x xx p x

Therefore, 2 1.66 1.28841x x

(b) ( ) 0(0.50) 1(0.30) 2(0.20) 0.7y y p y

2 2 2 22 ( ) 0 0.7 0.50 1 0.7 0.30 2 0.7 0.20 0.61y yy p y

Therefore, 2 0.61 0.781025y y

(c) The total amount of money collected from cars is $3(number of cars) = 3x. Therefore,

3 3 3(2.8) 8.4x x and 2 2 23 3 9(1.66) 14.94x x .

(d) The total amount of money collected from buses is $10(number of buses) = 10x.

Therefore, 10 10 10(0.7) 7y y and 2 2 210 10 100(0.61) 61y y .

(e) If z is the total number of vehicles, then z x y . Therefore,

2.8 0.7 3.5z x y x y . The variance of z is

2 2 2 2 1.66 0.61 2.27z x y x y .

(f) If w is the total amount of money collected in tolls, then 3 10w x y . Therefore,

3 10 3 10 8.4 7 15.4w x y x y . The variance of w is

2 2 2 23 10 3 10 14.94 61 75.94w x y x y .


6.37: The mean length of commercials appearing on this station is

( ) 15(0.1) 30(0.3) 60(0.6) 46.5x x p x seconds

6.38: (a)

( )

1(0.05) 2(0.10) 3(0.12) 4(0.30) 5(0.30) 6(0.11) 7(0.01) 8(0.01)

4.12

x x p x


(b)

22

2 2 2 2

2 2 2 2

( )

1 4.12 (0.05) 2 4.12 (0.10) 3 4.12 (0.12) 4 4.12 (0.30)

5 4.12 (0.30) 6 4.12 (0.11) 7 4.12 (0.01) 8 4.12 (0.01)

1.9456

x xx p x

Therefore, 2 1.9456 1.3948x x

Variance: The mean squared deviation from the mean number of systems sold in one month is 1.9456.

Standard deviation: A typical deviation from the mean number of systems sold in one month is 1.3948

(c) One standard deviation below the mean is 4.12 1.3948 2.7252x x , and one

standard deviation above the mean is 4.12 1.3948 5.5148x x , so

(2.7252 5.5148) (3 5) (3) (4) (5) 0.12 0.30 0.30 0.72P x P x p p p .

(d) Two standard deviations below the mean is 2 4.12 2(1.3948) 1.3304x x , and

two standard deviations above the mean is 2 4.12 2(1.3948) 6.9096x x , so

( 1.3304) ( 6.9096) ( 1) ( 7) 0.05 0.01 0.01 0.07P x P x P x P x .

6.39: Answers may vary. Two possible probability distributions are shown below.

Probability Distribution 1 ( 2.7x and 1.187x ):

x 1 2 3 4 5 p(x) 0.1 0.5 0.1 0.2 0.1

Probability Distribution 2 ( 3.3x and 1.187x ):

x 1 2 3 4 5 p(x) 0.1 0.2 0.1 0.5 0.1


6.40: (a) The sign of y (positive or negative) determines whether or not the peg will fit in the hole. Positive values for y indicate that the peg will fit in the hole.

(b) 2 1 2 1

0.253 0.25 0.003y x x x x .

(c) 2 1 2 1

2 22 2 0.002 0.006 0.00632y x x x x .

(d) Yes, it is reasonable to think that 1x and 2x are independent because both the peg and

the hole are being selected at random. There is no reason to believe that a particular peg is being selected for a particular hole.

(e) Given that the mean of y, 0.003y , is roughly half the standard deviation of y,

0.00632y , we know that zero is less than one-half a standard deviation away from the

mean. Therefore, it is reasonably likely that the value of y will be negative, and result in a peg that is too large to fit in the hole.

6.41: (a) 1 1 1

( ) (1) (2) (6) 3.56 6 6Rx R Rx p x

22 2 21 1 35( ) (1 3.5) (6 3.5) 2.9167

6 6 12R Rx R x Rx p x

2 351.7078

12R Rx x

(b) 3.5Bx , 2 35

12Bx , and 1.7078Bx

(c) If 1 1 7 7R Bw y x x , then 1

7 3.5 3.5 7 0R Bw x x . The variance of

1w is 1

2 2 2 35 35 355.833

12 12 6R Bw x x , so the standard deviation of 1w is

1 1

2 352.4152

6w w .

(d) If 2 23 3( ) 3 3R B R Bw y x x x x , then 2

3 3 3(3.5) 3(3.5) 0R Bw x x . The

variance of 2w is 2 2 2 2 2 35 35 1053 3 (9) (9) 52.5

12 12 2w R Bw x x , so the standard

deviation of 2w is 2 2

2 52.5 7.2457w w .


(e) The answer will vary depending on how willing the player is to take risks. Both games have a mean of 0. Game 2 has a larger standard deviation than game 1, which means that the player who chooses game 2 has the opportunity to win more money, but there is also the risk of losing more money. Game 1 has a smaller standard deviation, so there is less risk involved, but there is also less potential for a larger payout of money.


6.42:

22

2 2 2 2 2

( )

(0 1.2) (0.54) (1 1.2) (0.16) (2 1.2) (0.06) (3 1.2) (0.04) (4 1.2) (0.20)

2.52

x xx p x

and 2 2.52 1.58745x x

6.43: (a) The mean is ( ) 1(0.2) 2(0.4) 3(0.3) 4(0.1) 2.3x x p x and represents the

long run average number of lots ordered per customer.

(b) The variance is

2 2 2 2 22 ( ) 1 2.3 (0.2) 2 2.3 (0.4) 3 2.3 (0.3) 4 2.3 (0.1) 0.81x xx p x square lots, and the standard deviation is 2 0.81 0.9x x lots. The variance

measures how much, on average, the squared number of lots ordered per customer deviates from the mean. The standard deviation measures how much, on average, the number of lots ordered per customer deviates from the mean.

6.44: (a)

( ) 0(0.10) 1(0.15) 2(0.20) 3(0.25) 4(0.20) 5(0.06) 7(0.04) 2.64x x p x

22

2 2 2 2

2 2 2

( )

(0 2.64) (0.10) (1 2.64) (0.15) (2 2.64) (0.20) (3 2.64) (0.25)

(4 2.64) (0.20) (5 2.64) (0.06) (6 2.64) (0.04)

2.3704

x xx p x

Therefore, the standard deviation is 2 2.3704 1.5396x x

(b) Three standard deviation below the mean is 3 2.64 3(1.5396) 1.9788x x and

three standard deviations above the mean is 3 2.64 3(1.5396) 7.2588x x . All


possible values of x are within three standard deviations of the mean, so no values are farther than three standard deviations from the mean; the probability is 0.


6.45: (a) x can take on the values 0, 1, 2, 3, 4, or 5.

(b) The probability for each value of x is computed using the formula 5

! 5! 1 3( ) (1 )

!( )! !(5 )! 4 4

x xx n xn

p x p px n x x x

. For example, two of the six

probabilities are calculated as follows: 0 5 0

55! 1 3(0) (1)(1)(0.75) 0.2373

0!(5 0)! 4 4p

and

2 5 22 35! 1 3

(2) (10)(0.25) (0.75) 0.26372!(5 2)! 4 4

p

. The remaining probabilities

are calculated similarly and the results are shown in the table below.

x 0 1 2 3 4 5 p(x) 0.2373 0.3955 0.2637 0.0879 0.0146 0.0010

(c)

543210

0.4

0.3

0.2

0.1

0.0

x

p(x)


6.46: (a) 2 24!( 2) (2) (0.9) (0.1) 0.0486

2!(4 2)!P x p

; The probability that exactly two of

the four randomly selected households have cable TV is 0.0486.

(b) 4 04!( 4) (4) (0.9) (0.1) 0.6561

4!(4 4)!P x p

(c) ( 3) 1 ( 4) 1 0.6561 0.3439P x P x

6.47: (a) 2 35!( 2) (2) (0.25) (0.75) 0.2637

2!(5 2)!P x p

(b)

0 5 1 45! 5!

( 1) (0) (1) (0.25) (0.75) (0.25) (0.75)0!(5 0)! 1!(5 1)!

0.2373 0.3955 0.6328

P x p p

(c) (2 ) 1 ( 1) 1 0.6328 0.3672P x P x

(d) ( 2) 1 ( 2) 1 0.2637 0.7363P x P x

6.48: (a) Binomial distribution with n = 20 and p = 0.05. Accept if shipped lot contains 0 or 1 defective part. Therefore,

0 20 1 1920! 20!(0) (1) (0.05) (0.95) (0.05) (0.95)

0!(20 0)! 1!(20 1)!

0.3585 0.3774 0.7359

p p

(b) Binomial distribution with n = 20 and p = 0.10. Accept if shipped lot contains 0 or 1 defective part. Therefore,

0 20 1 1920! 20!(0) (1) (0.10) (0.90) (0.10) (0.90)

0!(20 0)! 1!(20 1)!

0.1216 0.2702 0.3918

p p

(c) Binomial distribution with n = 20 and p = 0.20. Accept if shipped lot contains 0 or 1 defective part. Therefore,

0 20 1 1920! 20!(0) (1) (0.20) (0.80) (0.20) (0.80)

0!(20 0)! 1!(20 1)!

0.0115 0.0576 0.0691

p p


6.49: (a) The probability distribution of x is not binomial because the question asks about songs being played until a song by this particular artist is played. There is not a fixed number of trials, which is required for the binomial distribution. This setting is geometric.

(b) (i) 3(4) (1 0.08) (0.08) 0.0623p

(ii)

0 1 2 3

( 4) (1) (2) (3) (4)

(0.92) (0.08) (0.92) (0.08) (0.92) (0.08) (0.92) (0.08)

0.2836

P x p p p p

(iii) ( 4) 1 ( 4) 1 0.2836 0.7164P x P x

(iv) ( 4) (4) ( 4) 0.0623 0.7164 0.7787P x p P x

(c) The differences between the four probabilities are shown in bold font.

(i) The probability that it takes exactly four songs until the first song by the particular artist is played is 0.0623.

(ii) The probability that it takes at most four songs until the first song by the particular artist is played is 0.2836.

(iii) The probability that it takes more than four songs until the first song by the particular artist is played is 0.7164.

(iv) The probability that it takes at least four songs until the first song by the particular artist is played is 0.7787.

6.50: (a) The probability distribution of x is geometric, with success probability p = 0.15.

(b) 2( 3) (1 0.15) (0.15) 0.1084P x

(c) 2( 4) (1) (2) (3) 0.15 (0.85)(0.15) (0.85) (0.15) 0.3859P x p p p

(d)

2

( 3) 1 ( 3)

1 (1) (2) (3)

1 0.15 (0.85)(0.15) (0.85) (0.15) 0.6141

P x P x

p p p



6.51: (a)

16 4 20 0

( 15) (16) (17) (18) (19) (20)

20! 20!0.8 1 0.8 0.8 1 0.8

16!(20 16)! 20!(20 20)!

0.6296

P x p p p p p

(b)

16 4 20 0

( 15) (16) (17) (18) (19) (20)

20! 20!0.4 1 0.4 0.4 1 0.4

16!(20 16)! 20!(20 20)!

0.000317

P x p p p p p

(c) If the true proportion of computer owners who have a firewall installed is 0.80p ,

then 14 620!( 14) (14) 0.80 1 0.80 0.1091

14!(20 14)!P x p

.

If the true proportion of computer owners who have a firewall installed is 0.40p ,

then 14 620!( 14) (14) 0.40 1 0.40 0.00485

14!(20 14)!P x p

.

Because the probability that 14 of the 20 randomly selected computer owners have a firewall installed is greater when 0.80p than when 0.40p , then the true proportion is

more likely to be 0.80p .

6.52: (a) 4 26!(4) ( 4) (0.8) (1 0.8) 0.2458

4!(6 4)!p P x

. In the long run, 24.58% of

random samples of 6 passengers will contain exactly 4 passengers who prefer to rest or sleep.

(b) 6 06!(6) ( 6) (0.8) (1 0.8) 0.2621

6!(6 6)!p P x

(c) ( 4) (4) (5) (6)P x p p p . From parts (a) and (b), (4) 0.2458p and

(6) 0.2621p . 5 16!(5) 0.8 1 0.8 0.3932

5!(6 5)!p

. Therefore,

( 4) 0.2458 0.2621 0.3932 0.9011P x .


6.53: (a) (8) 0.302p

(b) Using the complement rule to simplify the probability calculation,

( 7) 1 ( 8) 1 (8) (9) (10)

1 0.302 0.268 0.107 0.323

P x P x p p p

(c)

( 5) (6) (7) (8) (9) (10)

0.088 0.201 0.302 0.268 0.107 0.966

P x p p p p p

6.54: Let x be the number of cars failing inspection. The probabilities in parts (a) and (b) were calculated using appendix table 9 with n = 15 and p = 0.3.

(a)

( 5) (0) (1) (2) (3) (4) (5)

0.005 0.031 0.092 0.170 0.219 0.206 0.723

P x p p p p p p

(b)

(5 10) (5) (6) (7) (8) (9) (10)

0.206 0.147 0.081 0.035 0.012 0.003 0.484

P x p p p p p p

(c) Let y be the number of cars that pass inspection, so p = 0.7. The mean is 25(0.7) 17.5y np , and the standard deviation is

(1 ) 25(0.7)(1 0.7) 2.2913y np p .

(d) One standard deviation below the mean is 17.5 2.2913 15.2087y y and one

standard deviation above the mean is 17.5 2.2913 19.7913y y . Therefore,

(15.2087 19.7913) (16 19) (16) (17) (18) (19)

0.134 0.165 0.171 0.147 0.617

P y P y p p p p

6.55: (a) geometric distribution

(b) ( 2) (0.9)(0.1) 0.09P x

(c)

2

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

1 0.1 (0.9)(0.1) (0.9) (0.1) 0.729

P x P x p p p


6.56: This is the geometric setting with p = 0.05.

(a) ( 2) (1) (2) 0.05 0.95(0.05) 0.0975P x p p

(b) 3( 4) (0.95) (0.05) 0.0429P x

(c)

2 3

( 4) 1 ( 4) 1 (1) (2) (3) (4)

1 0.05 0.95(0.05) 0.95 (0.05) 0.95 (0.05) 0.8145

P x P x p p p p


6.57: The probability of correctly guessing 6 or more out of 10 times is ( 6)P x , where x is the

number of correct identifications. When guessing when there are two outcomes, we use a success probability of p = 0.5. This probability is (calculated using appendix table 9):

( 6) (6) (7) (8) (9) (10)

0.205 0.117 0.044 0.010 0.001 0.377.

P x p p p p p

Since this computed probability of getting 6 or more of the 10 identifications correct is not particularly small, it is not unlikely that the graphologist could get 6 or more identifications merely by guessing. Therefore, this does not indicate that the graphologist has an ability to distinguish the handwriting of psychotics.

6.58: The binomial probability distribution with n = 5 and p = 0.5 is

x 0 1 2 3 4 5 p(x) 0.03125 0.15625 0.31250 0.31250 0.15625 0.03125


6.59: (a) This is a binomial distribution with n = 100 and p = 0.20.

(b) The expected score is 100(0.20) 20x np answers correct.

(c) The variance is 2 (1 ) 100(0.2)(1 0.2) 16x np p and the standard deviation is

2 16 4x x .

(d) A score of 50 is 50 20

7.54

z

standard deviations above the mean. In any

distribution, being 7.5 or more standard deviations above the mean is quite unlikely, which indicates that it is not likely that you would score over 50 on the exam.

6.60: This scenario is sampling without replacement. Since more than 5% of the population is being sampled (the percentage of the population being sampled is 2,000/10,000 = 20%), the binomial distribution will not give a good approximation of the probability distribution of the number of invalid signatures.

6.61: If x denotes the number among the 15 who want a Diet Coke, then x takes on the values between 0 and 15, inclusive. If 15 customers want Diet Coke, then 5 want Coke, and so on. The table below outlines the values of x that correspond to whether or not everyone (those who want Diet Coke as well as those who want Coke) is satisfied (Y represents yes, and N represents no).

Want Diet Coke (x)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Want Coke 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Everyone

Satisfied(Y/N)? N N N N N Y Y Y Y Y Y N N N N N

Therefore, the desired probability is computed using the binomial distribution with n = 15 and p = 0.6. From appendix table 9:

(5 10) (5) (6) (7) (8) (9) (10)

0.024 0.061 0.118 0.177 0.207 0.186 0.773.

P x p p p p p p

6.62: (a) This is a binomial setting, with n = 25 and p = 0.5. The probability that the coin is judged biased is ( 7) ( 18) 0.0216 0.0216 0.0432P x P x .

(b) This is a binomial setting, with n = 25 and p = 0.9. The probability that the coin is judged biased is ( 7) ( 18) 0.000 0.9977 0.9977P x P x . Therefore, the probability

that the coin is judged fair is 1 0.9977 0.0023 .


Changing the scenario to P(H) = 0.1, use the binomial setting with n = 25 and p = 0.1. Therefore, the probability that the coin is judged biased is

( 7) ( 18) 0.9977 0.0000 0.9977P x P x . Therefore, the probability that the coin is

judged fair is 1 0.9977 0.0023 .

(c) With P(H) = 0.6, use the binomial setting with n = 25 and p = 0.6. Therefore, the probability that the coin is biased is ( 7) ( 18) 0.0012 0.1536 0.1548P x P x .

Therefore, the probability that the coin is judged fair is 1 0.1548 0.8452 . With P(H) = 0.4, use the binomial setting with n = 25 and p = 0.4. Therefore, the probability that the coin is judged biased is ( 7) ( 18) 0.1536 0.0012 0.1548P x P x . Therefore, the

probability that the coin is judged fair is 1 0.1548 0.8452 . The probabilities are large compared to the probabilities in Part (b) because the probabilities of getting a head in Part (c) are closer to 0.5 than those in Part (b). Probabilities closer to 0.5 will make it more likely that the coin will be judged fair.

(d) Changing the definition of being judged fair to 7 18x increases the likelihood that the coin will be judged fair, and decreases the probability that the coin is judged to not be fair. Although the new rule is more likely to judge a fair coin as fair, it is also more likely to judge a biased coin as fair.


6.63: (a) Note that since S and F are equally likely (p = 0.5), each probability in the table below

is 4(0.5) 0.0625 .

Outcome y Probability SSSS 4 4(0.5) 0.0625

SSSF 3 4(0.5) 0.0625

SSFS 2 4(0.5) 0.0625

SFSS 2 4(0.5) 0.0625

FSSS 3 4(0.5) 0.0625

SSFF 2 4(0.5) 0.0625

SFFS 1 4(0.5) 0.0625

FFSS 2 4(0.5) 0.0625

SFSF 1 4(0.5) 0.0625

FSFS 1 4(0.5) 0.0625

FSSF 2 4(0.5) 0.0625

FFFS 1 4(0.5) 0.0625

FFSF 1 4(0.5) 0.0625

FSFF 1 4(0.5) 0.0625

SFFF 1 4(0.5) 0.0625

FFFF 0 4(0.5) 0.0625

Therefore, the probability distribution is

y 0 1 2 3 4 p(y) 0.0625 0.4375 0.3125 0.125 0.0625

The mean is

( ) 0(0.0625) 1(0.4375) 2(0.3125) 3(0.125) 4(0.0625) 1.6875y y p y .


(b)

Outcome y Probability SSSS 4 4(0.6) 0.1296

SSSF 3 3(0.6) (0.4) 0.0864

SSFS 2 3(0.6) (0.4) 0.0864

SFSS 2 3(0.6) (0.4) 0.0864

FSSS 3 3(0.6) (0.4) 0.0864

SSFF 2 2 2(0.6) (0.4) 0.0576

SFFS 1 2 2(0.6) (0.4) 0.0576FFSS 2 2 2(0.6) (0.4) 0.0576SFSF 1 2 2(0.6) (0.4) 0.0576FSFS 1 2 2(0.6) (0.4) 0.0576FSSF 2 2 2(0.6) (0.4) 0.0576FFFS 1 3(0.6)(0.4) 0.0384

FFSF 1 3(0.6)(0.4) 0.0384

FSFF 1 3(0.6)(0.4) 0.0384

SFFF 1 3(0.6)(0.4) 0.0384

FFFF 0 4(0.4) 0.0256

Therefore, the probability distribution is

y 0 1 2 3 4 p(y) 0.0256 0.3264 0.3456 0.1728 0.1296

The mean is

( ) 0(0.0256) 1(0.3264) 2(0.3456) 3(0.1728) 4(0.1296) 2.0544y y p y .



6.64: (a) The area under the z curve to the left of 1.75 is ( 1.75) 0.9599P z

(b) The area under the z curve to the left of -0.68 is ( 0.68) 0.2483P z

(c) The area under the z curve to the right of 1.20 is ( 1.20) 1 ( 1.20) 1 0.8849 0.1151P z P z

(d) The area under the z curve to the right of -2.82 is ( 2.82) 1 ( 2.82) 1 0.0024 0.9976P z P z

(e) The area under the z curve between -2.22 and 0.53 is ( 2.22 0.53) ( 0.53) ( 2.22) 0.7019 0.0132 0.6887P z P z P z

(f) The area under the z curve between -1 and 1 is ( 1 1) ( 1) ( 1) 0.8413 0.1587 0.6826P z P z P z

(g) The area under the z curve between -4 and 4 is ( 4 4) ( 4) ( 4) 0.999968 0.000032 0.999936P z P z P z (using technology)

6.65: (a) ( 2.36) 0.9909P z

(b) ( 2.36) 0.9909P z

(c) ( 1.23) 0.1093P z

(d) (1.14 3.35) ( 3.35) ( 1.14) 0.9996 0.8729 0.1267P z P z P z

(e) ( 0.77 0.55) ( 0.55) ( 0.77) 0.2912 0.2206 0.0706P z P z P z

(f) ( 2) 1 ( 2) 1 0.9772 0.0228P z P z

(g) ( 3.38) 1 ( 3.38) 1 0.0004 0.9996P z P z

(h) ( 4.98) 1P z


6.66: (a) 5 5

( 5) ( 0) 0.50.2

P x P z P z

(b) 5.4 5

( 5.4) ( 2) 0.97720.2

P x P z P z

(c) 5.4 5

( 5.4) ( 2) 0.97720.2

P x P z P z

(d) 4.6 5 5.2 5

(4.6 5.2) ( 2 1) ( 1) ( 2)0.2 0.2

0.8413 0.0228 0.8185

P x P z P z P z P z

(e) 4.5 5

( 4.5) ( 2.5) 1 ( 2.5) 1 0.0062 0.99380.2

P x P z P z P z

(f) 4.0 5.0

( 4.0) 1 ( 4.0) 1 1 ( 5.0) 1 0 10.2

P x P x P z P z

6.67: Let s be the typist’s speed in words per minute.

(a) At most 60 wpm: 60 60

( 60) ( 0) 0.515

P s P z P z

; Less than 60 wpm:

60 60

( 60) ( 0) 0.515

P s P z P z

(b) 90 60 45 60

(45 90) ( 90) ( 45)15 15

( 2) ( 1) 0.9772 0.1587 0.8185

P s P s P s P z P z

P z P z

(c) 105 60

( 105) ( 3) 0.001315

P s P z P z

; It would be surprising to find a

typist in this population whose speed exceeded 105 wpm because the probability of finding such a typist is only 0.13%.

(d) The probability of finding one typist with a typing speed that exceeds 75 wpm is

75 60( 75) ( 1) 0.1587

15P s P z P z

. The probability that both typists have

typing speeds that exceed 75 wpm is (0.1587)(0.1587) = 0.0252 (by the multiplication rule for independent events).


(e) We need the value of z* such that ( *) 0.20P z z . Using technology, * 0.8416z .

Therefore, 60

0.8416 ( 0.8416)(15) 60 47.37615

ss

. Any typist with a typing

speed of 47.376 wpm or less is in the slowest 20% of typists.

6.68: The proportion of corks that are not defective is found by

2.9 3 3.1 3(2.9 3.1) ( 1 1) 0.6827

0.1 0.1P x P z P z

. Therefore, the

proportion of corks produced by this machine that are defective is 1 0.6827 0.3173 .

6.69: The proportion of corks that are not defective is

2.9 3.05 3.1 3.05(2.9 3.1) ( 15 5) 1

0.01 0.01P y P z P z

. Therefore, the

proportion of corks produced by this machine that are defective is approximately 1 1 0 . The second machine produces fewer defective corks.


6.70: (a) The area to the left of 1.28z is ( 1.28) 0.1003P z .

(b) The area to the right of 1.28z is ( 1.28) 1 ( 1.28) 1 0.8997 0.1003P z P z .

(c) The area between 1z and 2z is ( 1 2) ( 2) ( 1) 0.97725 0.158955 0.8186P z P z P z .

(d) The area to the right of 0z is ( 0) 0.5P z (consider the symmetry of the standard

normal curve).

(e) The area to the right of 5z is ( 5) 1 ( 5) 1P z P z .

(f) The area between 1.6z and 2.5z is ( 1.6 2.5) ( 2.5) ( 1.6) 0.99379 0.054799 0.93899P z P z P z .

(g) The area to the left of 0.23z is ( 0.23) 0.59095P z .

6.71: (a) ( 0.10) 0.5398P z

(b) ( 0.10) 0.4602P z

(c) (0.40 0.85) ( 0.85) ( 0.40) 0.80234 0.65542 0.1469P z P z P z

(d) ( 0.85 0.40) ( 0.40) ( 0.85) 0.344578 0.197662 0.1469P z P z P z


(e) ( 1.25) 1 ( 1.25) 1 0.10565 0.8944P z P z

(f) ( 1.50 or 2.50) ( 1.50) ( 2.50) 0.066807 0.00621 0.07302P z z P z P z

6.72: Let x be the carbon monoxide exposure, in parts per million (ppm). Therefore,

20 18.6( 20) ( 0.245614) 0.402991

5.7P x P z P z

and

25 18.6( 25) ( 1.12281) 0.13076

5.7P x P z P z

6.73: (a) 14.8 15.0

( 14.8) ( 2) 0.022750.1

P x P z P z

(b) 14.7 15.0 15.1 15.0

(14.7 15.1) ( 3 1) 0.84000.1 0.1

P x P z P z

(c) Let 1x and 2x represent the actual capacity of the two randomly selected tanks.

Therefore, 1 2 1 2( 15 15) ( 15) ( 15) (0.5)(0.5) 0.25P x x P x P x .

6.74: (a) 4,000 3,500

( 4,000) ( 0.83333) 0.2023600

P x P z P z

;

3,000 3,500 4,000 3,500(3,000 4,000)

600 600

( 0.83333 0.83333) 0.5953

P x P z

P z

(b)

( 2,000 or 5,000) ( 2,000) ( 5,000)

2,000 3,500 5,000 3,500

600 600

( 2.5) ( 2.5) 0.00621 0.00621 0.01242

P x x P x P x

P z P z

P z P z

(c) Seven pounds is equal to (7)(453.59) 3175.13 grams. Therefore,

3,175.13 3,500( 3,175.13) ( 0.54145) 0.7059

600P x P z P z

.

(d) The most extreme 0.1% of all full-term baby birth weights corresponds to the smallest 0.05% and largest 0.05% of all full-term baby birth weights. Therefore, we want to find z* such that ( *) 0.0005P z z . Using technology, this corresponds to * 3.291z , or 3.291

standard deviations below the mean and (by symmetry) 3.291 standard deviations above the mean. We can find the value of the full-term baby weight a specified number of


standard deviations below or above the mean by solving the z-score expression for x, or 3,500

3,500 (600)600

xz x z

, and substitute 3.291z or 3.291z to find x. The

full-term baby birth weight that corresponds to 3.291 standard deviations below the mean is 3,500 (3.291)(600) 1,525.4x grams. The full-term baby birth weight that corresponds

to 3.291 standard deviations above the mean is 3,500 (3.291)(600) 5, 474.6x grams.

Any full-term baby that is less than 1,525.4 grams or greater than 5,474.6 grams is characterized as the most extreme 0.1%.

6.75: We want to find z* such that ( *) 0.10P z z . From the standard normal table or

technology, * 1.282z . Therefore, the task times that qualify individuals for such

training are 120

1.282 120 1.282(20) 94.3620

xx

seconds or less.


6.76: It is not reasonable to think that traffic flow is approximately normal because traffic flow cannot be negative, and 0 traffic flow is 1.58 standard deviations below the mean

0 0.411.58

0.26z

, which indicates that approximately 5.7% of traffic flow values

would be negative if the traffic times actually followed a normal distribution.

6.77: Let x be the diameter of a randomly selected ball bearing. The ball bearing is acceptable if its diameter lies between 0.496 and 0.504 inches. Outside that range, the ball bearing is unacceptable. The probability that the ball bearing is acceptable given the new distribution of diameters is:

0.496 0.499 0.504 0.499(0.496 0.504)

0.002 0.002

( 1.5 2.5) 0.9270

P x P z

P z

Therefore, the probability that the ball bearing is not acceptable is 1 0.9270 0.0730 , so 7.3% of ball bearings are not acceptable.

6.78: Since these values are times, they must all be positive. In the normal distribution with mean 9.9 and standard deviation 6.2, approximately 5.5% of processing times would be less than or equal to zero.


6.79: (a) 650 500

( 650) ( 2) 0.0227575

P x P z P z

(b) 400 500 550 500

(400 550) ( 1.333 0.667) 0.65675 75

P x P z P z

(c) 400 500 550 500

(400 550) ( 1.333 0.667) 0.65675 75

P x P z P z

6.80: Let w represent the weight of mozzarella cheese on a medium pizza.

(a) 0.525 0.5 0.550 0.5

(0.525 0.550) (1 2) 0.1360.025 0.025

P w P z P z

(b) ( 2) 1 ( 2) 1 0.9772 0.0228P z P z

(c) Let w1, w2, and w3 represent the weight of the cheese on the three pizzas. The probability that three randomly selected medium pizzas all have at least 0.475 lb of cheese

is 1 2 3( 0.475) ( 0.475) ( 0.475)P w w w . Since the pizzas were selected

randomly, they are independent of each other, so, by applying the multiplication rule for independent events, the probability is computed as

1 2 3( 0.475) ( 0.475) ( 0.475)P w P w P w . These probabilities are all equal to each

other, so

0.475 0.5( 0.475) ( 1) 1 ( 1) 1 0.1587 0.8413

0.025P w P z P z P z

, and

31 2 3( 0.475) ( 0.475) ( 0.475) (0.8413) 0.5955P w P w P w .

6.81: (a) 29 30 31 30

(29 31) ( 0.8333 0.8333) 0.59531.2 1.2

P x P z P z

(b) 25 30

( 25) ( 4.1667) 0.0000151.2

P x P z P z

(computed using

technology); This extremely small probability indicates that it would be surprising to find that the efficiency of a randomly selected car of this model is less than 25 mpg.

(c) Let 1x , 2x , and 3x represent the three randomly selected cars. We are interested in

computing 1 2 3( 32) ( 32) ( 32)P x x x . The cars are randomly selected, so the

three cars are independent, so the multiplication rule for independent events can be applied:

1 2 3 1 2 3( 32) ( 32) ( 32) ( 32) ( 32) ( 32)P x x x P x P x P x . The probability

that any one car has a fuel efficiency exceeding 32 mpg is


32 30( 32) ( 1.6667) 0.04779

1.2P x P z P z

, so the probability that all three

cars have efficiencies exceeding 32 mpg is 30.04779 0.00011 .

(d) We need the value of x* such that ( *) 0.95P x x . This is equivalent to finding

( *) 0.05P x x , or the 5th percentile. From the table of standard normal probabilities or

using technology, we find that the z-score that corresponds to the 5th percentile is -1.645.

Therefore, * 30

1.645 * 30 1.645(1.2) 28.0261.2

xx

mpg is the desired fuel

efficiency.

6.82: (a) 45 60

( 45) ( 1.5) 1 ( 1.5) 1 0.0668 0.933210

P x P z P z P z

(b) We need the value of z* such that ( *) 0.10P z z . This is equivalent to

( *) 0.90P z z , the 90th percentile. Using technology, we find that ( 1.282) 0.90P z .

Therefore, 60

1.282 10(1.282) 60 72.8210

xx

minutes.

6.83: Let 1x and 2x represent the two independently selected batteries that are put into the DVD

player.

(a) To function for at least 4 hours, both batteries must last at least 4 hours. Therefore,

1 2 1 2

2 2

( 4) ( 4) ( 4) ( 4)

4 6 4 6

0.8 0.8

( 2.5) 0.99379 0.9876

P x x P x P x

P z P z

P z

(b) To function for at most 7 hours, first find the probability that both batteries last more than 7 hours (which is the complement of the event that the DVD player functions for at most 7 hours).

1 2 1 2

2 2

( 7) ( 7) ( 7) ( 7)

7 6 7 6

0.8 0.8

( 1.25) 0.10565 0.011162

P x x P x P x

P z P z

P z


Using the complement rule, the probability that the player works is therefore 1 0.011162 0.9888 .

(c) We need the probability that both batteries in the DVD player last longer than x* hours to be 0.05. For this to be the case, the probability that one or the other battery lasts longer

than x* hours must be 0.05 0.22361 . Therefore, we need to find the value of z* such

that ( *) 0.22361P z z . Using the table of standard normal probabilities or technology,

we find that * 0.76z , so * 6

0.76 * 6 0.76(0.8) 6.6080.8

xx

hours.

6.84: Let x represent the amount of vitamin E in each capsule. Therefore,

4.9 5( 4.9) ( 2) 0.0228

0.05P x P z P z

;

5.2 5( 5.2) ( 4) 0

0.05P x P z P z

6.85: (a) Note that 5 feet 7 inches is 67 inches, so the desired probability is

67 66( 67) ( 0.5) 0.691

2P x P z P z

. This is less than 94%, so the claim is not

correct.

(b) Approximately 69.1% of women would be excluded from employment as a result of the height restriction.

6.86: We want to find z* such that ( *) 0.15P z z . This is equivalent to ( *) 0.85P z z , the

85th percentile. Using technology, we find that ( 1.036) 0.85P z . Therefore,

781.036 7(1.036) 78 85.252

7

xx

. Since your score of 89 is higher than 85.252,

you received an A.

6.87: (a) 5.90 6.00 6.15 6.00

(5.90 6.15) ( 1 1.5) 0.77450.10 0.10

P x P z P z

(b) 6.10 6.00

( 6.10) ( 1) 0.15870.10

P x P z P z

(c) 5.95 6.00

( 5.95) ( 0.5) 0.30850.10

P x P z P z

(d) We want to find z* such that ( *) 0.05P z z . Using a table of standard normal

probabilities or technology, we find that ( 1.645) 0.05P z . Therefore,


6.001.645 6.00 1.645(0.10) 6.1645

0.10

xx

. The pH value that will be exceeded by

only 5% of all such pH values is 6.1645.

6.88: We want to find z* such that ( *) 0.20P z z . Using a table of standard normal

probabilities or technology, we find that ( 0.842) 0.20P z . Therefore,

7000.842 50( 0.842) 700 657.9

50

xx

hours. The bulbs should be replaced

every 657.9 hours so that no more than 20% of the bulbs will have already burned out.

6.89: (a) 250 266 300 266

(250 300) ( 1.0 2.125) 0.82516 16

P x P z P z

(b) 240 266

( 240) ( 1.625) 0.05216

P x P z P z

(c) Using the empirical rule, and noting that 16 days is 1 standard deviation, approximately 68% of values lie within 1 standard deviation of the mean. Alternately,

250 266 282 266(250 282) ( 1 1) 0.6827

16 16P x P z P z

.

(d) 310 266

( 310) ( 2.75) 0.0029816

P x P z P z

. This small probability does

make me skeptical of the claim.

(e) Benefits will not be paid if birth occurs within 275 days of the beginning of coverage. We’re interested in the probability that insurance benefits will not be paid if the pregnancy lasts no more than 275 – 14 = 261 days. Therefore, the probability that the insurance company will refuse to pay benefits because of the 275-day requirement is

261 266( 261) ( 0.3125) 0.3773

16P x P z P z

.



6.90: (a) The plot (shown below) does not look linear. The nonlinearity of the normal probability plot supports the author’s assertion that the distribution of fussing times is not normal.

2.01.61.20.80.40.0-0.4-0.8-1.2-1.6-2.0

14

12

10

8

6

4

2

0

Normal Score

Fuss

ing

Tim

e

(b) The computed correlation coefficient is 0.920r . The critical r from Table 6.2 is 0.911. Since the computed r is greater than the critical r, it is reasonable to think that the population distribution is normal.


6.91: (a)

2.01.61.20.80.40.0-0.4-0.8-1.2-1.6-2.0

160

150

140

130

120

110

100

Normal Score

Ris

k B

ehav

ior

Sco

re

(b)

2.01.61.20.80.40.0-0.4-0.8-1.2-1.6-2.0

70

65

60

55

50

45

40

35

Normal Score

PA

NA

S

(c) Because both normal probability plots are approximately linear, it seems reasonable that both risk behavior scores and PANAS scores are approximately normally distributed.

In addition, the correlation coefficients for both of the plots ( 0.985riskr and

0.993PANASr ) are both above the critical r cutoff from Table 6.2 (critical r = 0.911),

which provides further justification for the claim of normality.



6.92:

210-1-2

450

400

350

300

250

200

150

Normal Score

Life

tim

e (h

ours

)

The curvature in the normal probability plot indicates that it does not seem reasonable to think that the distribution of power supply lifetime is approximately normal.

6.93: (a)

1.51.00.50.0-0.5-1.0-1.5

17.5

15.0

12.5

10.0

7.5

5.0

Normal Score

Fat

Con

tent

(gr

ams)

The normal probability plot does look reasonably linear.


(b) The correlation coefficient for the (normal score, fat content) pairs is 0.945r . Because the correlation coefficient is greater than the critical r (for n = 10), it is reasonable to think that the fat content distribution is approximately normal.


6.94: (a) Yes, the normal probability plot appears linear.

1.51.00.50.0-0.5-1.0-1.5

31

30

29

28

27

Normal Score

Fuel

Eff

icie

ncy

(mpg

)

(b) 0.994r ; From Table 6.2, we find that the critical r for n = 6 lies between 0.832 (critical r for n = 5) and 0.880 (critical r for n = 10). The computed correlation coefficient is larger than the critical r for n = 6, so it is reasonable to think that the fuel efficiency distribution is approximately normal.


6.95: (a)

210-1-2

400

300

200

100

0

Normal Score

Bea

ring

Loa

d Li

fe

(b) The correlation coefficient for the (normal score, bearing load life) pairs is 0.963r , which is greater than the critical r of 0.929. It is reasonable to think that the distribution of bearing load life is approximately normal.

6.96:

2.52.01.51.00.50.0-0.5-1.0-1.5-2.0-2.5

16.3

16.2

16.1

16.0

15.9

15.8

15.7

15.6

Normal Score

Dis

k D

iam

eter

Based on the linearity of the normal probability plot and the fact that the correlation coefficient for the (normal score, disk diameter) points ( 0.987r ) exceeds the critical r of 0.941 (from Table 6.2), it is reasonable to think that disk diameter is normally distributed.



6.97: (a) 99.5 100 100.5 100

( 100) ( 0.033 0.033) 0.02615 15

P x P z P z

(b) 110.5 100

( 110) ( 0.7) 0.758015

P x P z P z

(c) 109.5 100

( 110) ( 109) ( 0.633) 0.736615

P x P x P z P z

(d) 74.5 100 125.5 100

(75 125) ( 1.7 1.7) 0.910915 15

P x P z P z

6.98: To use the normal approximation in this scenario, first verify that 10np and

(1 ) 10n p , where n = 100 and p = 0.7. (100)(0.7) 70 and 100(1 0.7) 30 , both

values are greater than 10. Second, find the mean and standard deviation of x, where x represents the number of mountain bikes out of 100 randomly selected bikes sold. The

mean is (100)(0.7) 70x np and the standard deviation is

(1 ) 100 0.7 (1 0.7) 21 4.5826x np p .

(a) 75.5 70

( 75) ( 1.200) 0.88494.5826

P x P z P z

(b) 59.5 70 75.5 70

(60 75) ( 2.291 1.200) 0.87394.5826 4.5826

P x P z P z

(c) 80.5 70

( 80) ( 81) ( 2.291) 1 ( 2.291) 0.01104.5826

P x P x P z P z P z

(d) At most 30 are not mountain bikes is the same as at least 70 are mountain bikes.

Therefore, 69.5 70

( 70) ( 0.109) 0.54344.5826

P x P z P z

6.99: Check 10np and (1 ) 10n p to verify that using the normal approximation is

appropriate. Both (225)(0.65) 146.25np and (1 ) (225)(1 0.65) 78.75n p are

greater than 10. Let x represent the number of registered voters out of 225 in a certain area who favor a 7-day waiting period before purchasing a hand gun. Compute the mean and standard deviation to use the normal approximation. The mean is


(225)(0.65) 146.25x np and the standard deviation is

(1 ) 225 0.65 (1 0.65) 7.1545x np p

(a) 149.5 146.25

( 150) ( 0.4543) 0.32487.1545

P x P z P z

(b) 150.5 146.25

( 150) ( 151) ( 0.5940) 0.27637.1545

P x P x P z P z

(c) 124.5 146.25

( 125) ( 124) ( 3.040) 0.00127.1545

P x P x P z P z


6.100: Let y = the number of items produced by an assembly line during an 8-hour shift.

(a) 120.5 150

( 120) ( 2.95) 0.001610

P y P z P z

(b) 124.5 150

( 125) ( 2.55) 0.994610

P y P z P z

(c) 134.5 150 160.5 150

(135 160) ( 1.55 1.05) 0.792610 10

P y P z P y

6.101: Both 100(0.25) 25np and (1 ) 100(1 0.25) 75n p are greater than 10, so using

the normal approximation is appropriate. The mean is 100(0.25) 25x np and the

standard deviation is (1 ) 100(0.25)(1 0.25) 4.330x np p

(a) 19.5 25 30.5 25

(20 30) ( 1.27 1.27) 0.79594.330 4.330

P x P z P z

(b) 20.5 25 29.5 25

(20 30) ( 1.039 1.039) 0.70124.330 4.330

P x P z P z

(c) 34.5 25

( 35) ( 2.194) 0.01414.330

P x P z P z

(d) Two standard deviations below the mean is 2 25 2(4.330) 16.34x x and two

standard deviations above the mean is 2 25 2(4.330) 33.66x x . Therefore, the

probability that x lies within two standard deviations of the mean is:

16.5 25 33.5 25(16.34 33.66) 17 33

4.330 4.330

( 1.963 1.963) 0.9504

P x P x P z

P z


By the complement rule, the probability that x is not within two standard deviations of the mean is 1 0.9504 0.0496 .

6.102: Let x be the number of mufflers out of 400 that are replaced under warranty. Both 400(0.2) 80np and (1 ) 400(1 0.2) 320n p are greater than 10, so using the

normal approximation is appropriate. The mean is 400(0.2) 80x np and the

standard deviation is (1 ) 400(0.2)(1 0.2) 8x np p .

(a) 74.5 80 100.5 80

(75 100) ( 0.6875 2.5625) 0.74898 8

P x P z P z

(b) 70.5 80

( 70) ( 1.1875) 0.11758

P x P z P z

(c) 49.5 80

( 50) ( 3.8125) 0.00006888

P x P z P z

. This small probability

indicates that replacing fewer than 50 mufflers out of 400, given the claim that 20% of mufflers are replaced under warranty, is quite unlikely. Yes, it seems as if the 20% figure should be questioned.


6.103: (a) To verify the appropriate use of the normal approximation, both np and n(1 – p) must be at least 10. Both (60)(0.7) 42np and (1 ) 60(1 0.7) 18n p are at least 10.

(b) Compute mean and standard deviation. (60)(0.7) 42x np and

(1 ) 60 0.7 (1 0.7) 3.5497x np p

(i) 41.5 42 42.5 42

( 42) ( 0.1409 0.1409) 0.11213.5497 3.5497

P x P z P z

(ii) 41.5 42

( 42) ( 41) ( 0.1409) 0.44403.5497

P x P x P z P z

(iii) 42.5 42

( 42) ( 0.1409) 0.55603.5497

P x P z P z

(c) The probability in (i) represents the probability of exactly 42 correct responses on the MENT test for a patient who is trying to fake the test. The probability in (ii) is the probability of less than 42 correct responses on the MENT test for a patient who is trying to fake the test. The probability in (iii) represents the probability of 42 or fewer correct responses on the MENT test for a patient who is trying to fake the test. These are different numbers of correct questions.


(d) The normal approximation is not appropriate here because (1 ) 60(1 0.96) 2.4n p ,

which is less than 10. As such, the binomial distribution must be used.

(e) The small probability in part (d) indicates that it is extremely unlikely for someone who is not faking the test to correctly answer 42 or fewer questions, when compared with the probability that a person who is faking the test to correctly answer 42 or fewer questions (0.556). If someone does score 42 or fewer points on the MENT exam, there is strong evidence (supported by the extremely small probability) that the person must be faking.

6.104: (a) Four conditions for a binomial experiment must be satisfied in order for the random variable to be considered a binomial random variable. The conditions are a fixed number of trials, each trial results in only one of two possible outcomes, outcomes of different trials are independent, and the probability of success is the same for each trial. In this scenario, all four conditions have been satisfied, so it is reasonable to think that the random variable has a binomial distribution with n = 500 and p = 0.031.

(b) Because 500(0.031) 15.5np and (1 ) 500(1 0.031) 484.5n p are both at least

10, it is reasonable to use the normal distribution to approximate probabilities for the random variable x.

(c) Use a mean of 500(0.031) 15.5x np and a standard deviation of

500(0.031)(1 0.031) 3.8755x .

(i) 9.5 15.5

( 10) ( 1.548) 0.06083.8755

P x P z P z

(ii)

9.5 15.5 25.5 15.5

(10 25)3.8755 3.8755

( 1.548 2.580) 0.9342

P x P z

P z

(iii) 20.5 15.5

( 20) ( 1.290) 0.09853.8755

P x P z P z

(d) (i) In a long run of repeated random samples of size 500 women diagnosed with breast cancer in one breast, approximately 6.08% will contain fewer than 10 women with an undetected tumor in the other breast.

(ii) In a long run of repeated random samples of size 500 women diagnosed with breast cancer in one breast, approximately 93.42% will contain between 10 and 25 (inclusive) women with an undetected tumor in the other breast.


(iii) In a long run of repeated random samples of size 500 women diagnosed with breast cancer in one breast, approximately 9.85% will contain more than 20 women with an undetected tumor in the other breast.

Are You Ready to Move On? Chapter 6 Review Exercises

6.105: The depth, x, can take on any value between 0 and 100 ( 0 100x ), inclusive. x is continuous.

6.106: The table below shows the possible outcomes.

Slip 1 Slip 2 Amount

won $1 $1 $1 $1 $1 $1 $1 $10 $10 $1 $25 $25 $1 $1 $1 $1 $10 $10 $1 $25 $25 $1 $10 $10 $1 $25 $25 $10 $25 $25

Each of the outcomes in the table is equally likely. The probability distribution of w, the amount awarded, is shown below.

w $1 $10 $25 p(w) 0.3 0.3 0.4

6.107: (a) The plane can accommodate 100 passengers, so if 100 or fewer passengers show up for the flight, everyone can be accommodated. Therefore,

( 100) 0.05 0.10 0.12 0.14 0.24 0.17 0.82P x .

(b) This is the complement of all passengers being accommodated, or ( 100) 1 ( 100) 1 0.82 0.18P x P x .

(c) The first person on the standby list hopes that 99 or fewer people show up for the flight. Therefore, ( 99) 0.05 0.10 0.12 0.14 0.24 0.65P x . The third person

on the standby list hopes that 97 or fewer people show up for the flight. Therefore, ( 97) 0.05 0.10 0.12 0.27P x .

©2014 Cengag

6.108: (a1

(b1

6.109: (a

(b

(c

T

6.110: (a

(b

an

P

to

P

tr

o

th

U

pr


a) The proba6 and 17 oun

b) In the long6 and 17 oun

a) ( 10)P x

b) (7P x

c) ( )P x c

Therefore, (0

a)

b) ( 10)P w

nd w = 10. U

(20 2P w

o w = 25. Us

(20 2P w

rapezoidal re

f trapezoids,

he area of the

Using propert

robabilities,


ability that thnces.

g run, 50% onces.

(0.05)(10)

12) (0.05)

0.9 means

0.05) 0.9c ,

) is the area

Using prope

5) is the are

sing properti1

5) (0.052

egion under t

, (25P w

e triangular

ties of triang

in order, are


he weight of

of all random

) 0.5 ; (P x

)(12 7) 0.

that the area

, or 0.9 /c

of the triang

rties of trian

ea of the trap

ies of trapez

0.0375)(5)

the density c1

30) (0.02

region under

gles, ( 3P w

e ( 10)P w


f a randomly

mly selected

15) (0.0

.25

a under the d

0.05 18 m

gular region

ngles, (P w

pezoidal regi

zoids,

) 0.21875 .

curve, betwe

0375 0.025

r the density1

30) (10)(2

( 30)P w


y selected pac

packages of

05)(20 15)

density curve

minutes.

under the de

110) (10)

2

ion under the

. (25P w

een w = 25 a

5)(5) 0.156

y curve betw

(0.025) 0.1

) (25P w


ckage of cer

f cereal will

0.25

e between 0

ensity curve

)(0.025) 0.

e density cur

30) is the

and w = 30.

625 . Finally

ween w = 30

125 . Theref

30) (2w P

whole or in part.

real lies betw

weigh betw

and c is 0.9

between w =

.125 .

rve, from w

e area under t

Using prope

y, ( 30)P w

and w = 40.

fore, the

20 25)w

ween

ween

.

= 0

= 20

the

erties

is

.


6.111: (a) Supplier 1

(b) Supplier 2

(c) I would recommend supplier 1 because the bulbs will last, on average, longer than those from supplier 2. Additionally, the bulbs from supplier 1 have less variability in their lifetimes, so there is more consistency in the bulb lifetimes when compared with supplier 2.

(d) Approximately 1000 hours.

(e) Approximately 100 hours.

6.112: (a) ( ) 0(0.54) 1(0.16) 2(0.06) 3(0.04) 4(0.20) 1.2x x p x

(b) In repeated inspections of cars at this station, the mean number of defective tires is 1.2 tires per car.

(c) ( 1.2) (2) (3) (4) 0.06 0.04 0.20 0.30P x p p p

(d)

2 2 2 2

2 2 2

( ) ( ) (0 1.2) (0.54) (1 1.2) (0.16)

(2 1.2) (0.06) (3 1.2) (0.04) (4 1.2) (0.20)

2.52

x xx p x

Therefore, 2 2.52 1.587x x .


6.113: Let x represent the left atrial diameter.

(a) 24 26.4

( 24) ( 0.571) 0.2844.2

P x P z P z

(b) 32 26.4

( 32) ( 1.333) 1 ( 1.333) 1 0.909 0.0914.2

P x P z P z P z

(c) 25 26.4 30 26.4

(25 30) ( 0.333 0.857) 0.4354.2 4.2

P x P z P z

(d) We need the value of z* such that ( *) 0.20P z z , or ( *) 0.80P z z . Using

technology, * 0.8416z . Therefore we find that, 26.4

0.8416 (0.8416)(4.2) 26.4 29.934.2

xx

. 29.93 mm is the value for which

only about 20% have a larger left atrial diameter.

6.114: Let x be the left atrial diameter in overweight children.

(a) 25 28

( 25) ( 0.638) 0.26174.7

P x P z P z

(b) 32 28

( 32) ( 0.851) 0.19744.7

P x P z P z

(c) 25 28 30 28

(25 30) ( 0.638 0.426) 0.40324.7 4.7

P x P z P z

(d) 26.4 28

( 26.4) ( 0.340) 0.6334.7

P x P z P z

6.115: (a) 50 45

( 50) 1 ( 50) 1 1 ( 1) 1 0.841 0.1595

P x P x P z P z

(b) We need the value of z* such that ( *) 0.90P z z . From the standard normal table

or using technology, * 1.28z . Therefore, 45

1.28 (1.28)(5) 45 51.45

xx

minutes. 51.4 minutes should be allowed for 90% of students taking the test to be able to finish in the allotted time.

(c) We need the value of z* such that ( *) 0.25P z z . From the standard normal table

or using technology, * 0.674z . Therefore,


450.674 ( 0.674)(5) 45 41.6

5

xx

minutes. 41.6 minutes is required for the

fastest 25% of all students to complete the exam.

6.116: Referring back to Example 6.14, we see that both distributions of the number of flaws per panel have the same mean ( 1x y ). In addition, examination of the histograms and

calculation of standard deviations (refer to Example 6.15) reveal that the distribution of

supplier 1 has more variability ( 1x ) than that for supplier 2 ( 0.632y ). Therefore,

supplier 2 is recommended because there is less variability in the number of flaws. Supplier 2 provides glass panels that are more consistent in their quality.

6.117: Yes, the data suggest that the cadmium concentration distribution is not normal because of the curvature of the points apparent in the normal probability plot.

6.118: (a)

210-1-2

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Normal Score

sFas

L

Then normal probability plot appears to have some curvature.

(b) The correlation coefficient is 0.95r , which is greater than the critical r of 0.880 for 10 data points. According to the value of the correlation coefficient it seems reasonable to consider the distribution of sFasL levels to be approximately normal.


6.119: (a) ( ) 13.5(0.2) 15.9(0.5) 19.1(0.3) 16.38x x p x . The variance of x is

2 2 2 22 ( ) 13.5 16.38 0.2 15.9 16.38 0.5 19.1 16.38 0.3 3.9936x xx p x which yields a standard deviation of 2 3.9936 1.9984x x .

(b) 25 8.5 25 8.5 25(16.38) 8.5 401price x x . The mean price paid by the next

customer is $401.

(c) 2 225 (3.9936) 49.96price price . The standard deviation of the price paid by

the next customer is $49.96.

chapter 06 random variables and probability distributions · 2016-11-15 · chapter 6 random...

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