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Applied Statistics and Probability for Engineers, 5 th edition September 22, 2010 users.encs.concordia.ca/~xzhang/ENGR371solutions/ch08.docx http://users.encs.concordia.ca/~xzhang/ENGR371solutions/ the life hours in a 75-watt light bulb is known to be normally distributed with CHAPTER 8 Section 8-1 8-1 a) The confidence level for ¯ x2.14 σ / nμ≤¯ x+2.14 σ / n is determined by the value of z0 which is 2.14. From Table III, (2.14) = P(Z<2.14) = 0.9838 and the confidence level is 2(0.9838-0.5) = 96.76%. b) The confidence level for ¯ x2.49 σ / nμ≤¯ x+2.49σ / n is determined by the by the value of z0 which is 2.14. From Table III, (2.49) = P(Z<2.49) = 0.9936 and the confidence level is is 2(0.9936-0.5) = 98.72%. c) The confidence level for ¯ x1.85 σ / nμ≤¯ x+1.85σ / n is determined by the by the value of z0 which is 2.14. From Table III, (1.85) = P(Z<1.85) = 0.9678 and the confidence level is 93.56%. 8-2 a) A z = 2.33 would give result in a 98% two-sided confidence interval. b) A z = 1.29 would give result in a 80% two-sided confidence interval. c) A z = 1.15 would give result in a 75% two-sided confidence interval. 8-3 a) A z = 1.29 would give result in a 90% one-sided confidence interval. b) A z = 1.65 would give result in a 95% one-sided confidence interval. c) A z = 2.33 would give result in a 99% one-sided confidence interval. 8-4 a) 95% CI for μ, n=10 =20 ¯ x=1000 ,z=1.96 ¯ x/ nμ≤¯ x +/ n 10001.96( 20/ 10 )≤μ1000+ 1.96 ( 20 / 10) 987.6μ1012.4 b) .95% CI for μ, n=25 =20 ¯ x=1000 ,z=1.96 ¯ x/ nμ≤¯ x +/ n 10001.96( 20/ 25)≤μ1000+ 1.96( 20/ 25) 992.2μ1007.8 c) 99% CI for μ, n=10 =20 ¯ x=1000 ,z=2.58 ¯ x/ nμ≤¯ x +/ n 10002.58( 20/ 10)≤μ1000+ 2.58( 20/ 10) 983.7μ1016.3 d) 99% CI for μ, n=25 =20 ¯ x=1000 ,z=2.58 8-1

Applied Statistics and Probability Ch8 solutions

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Page 1: Applied Statistics and Probability Ch8 solutions

Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

users.encs.concordia.ca/~xzhang/ENGR371solutions/ch08.docxhttp://users.encs.concordia.ca/~xzhang/ENGR371solutions/

the life hours in a 75-watt light bulb is known to be normally distributed withCHAPTER 8

Section 8-1

8-1 a) The confidence level for x̄−2 .14 σ /√n≤μ≤ x̄+2. 14 σ /√n is determined by the value of z0 which is 2.14. From Table III, (2.14) = P(Z<2.14) = 0.9838 and the confidence level is 2(0.9838-0.5) = 96.76%.

b) The confidence level for x̄−2 .49σ /√n≤μ≤ x̄+2 .49σ /√n is determined by the by the value of z0 which is 2.14. From Table III, (2.49) = P(Z<2.49) = 0.9936 and the confidence level is is 2(0.9936-0.5) = 98.72%.

c) The confidence level for x̄−1 .85σ /√n≤μ≤ x̄+1. 85σ /√n is determined by the by the value of z0 which is 2.14. From Table III, (1.85) = P(Z<1.85) = 0.9678 and the confidence level is 93.56%.

8-2 a) A z = 2.33 would give result in a 98% two-sided confidence interval.b) A z = 1.29 would give result in a 80% two-sided confidence interval.c) A z = 1.15 would give result in a 75% two-sided confidence interval.

8-3 a) A z = 1.29 would give result in a 90% one-sided confidence interval.b) A z = 1.65 would give result in a 95% one-sided confidence interval.c) A z = 2.33 would give result in a 99% one-sided confidence interval.

8-4 a) 95% CI for μ , n=10 , σ=20 x̄=1000 , z=1. 96

x̄−zσ /√n≤μ≤ x̄+zσ /√n1000−1. 96(20 /√10 )≤μ≤1000+1.96 (20/√10)987 . 6≤μ≤1012 .4b) .95% CI for μ , n=25 , σ=20 x̄=1000 , z=1. 96

x̄−zσ /√n≤μ≤ x̄+zσ /√n1000−1. 96(20 /√25 )≤μ≤1000+1. 96 (20/√25)992 . 2≤μ≤1007 . 8c) 99% CI for μ , n=10 , σ=20 x̄=1000 , z=2. 58

x̄−zσ /√n≤μ≤ x̄+zσ /√n1000−2.58(20/√10 )≤μ≤1000+2.58(20/√10 )983 .7≤μ≤1016 .3d) 99% CI for μ , n=25 , σ=20 x̄=1000 , z=2. 58

8-1

Page 2: Applied Statistics and Probability Ch8 solutions

Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

x̄−zσ /√n≤μ≤ x̄+zσ /√n1000−2. 58(20/√25 )≤μ≤1000+2. 58(20/√25 )989 . 7≤μ≤1010 .3e) When n is larger, the CI is narrower. The higher the confidence level, the wider the CI.

8-5 a) Sample mean from the first confidence interval = 38.02 + (61.98-38.02)/2 = 50Sample mean from the second confidence interval = 39.95 + (60.05-39.95)/2 = 50b) The 95% CI is (38.02, 61.98) and the 90% CI is (39.95, 60.05). The higher the confidence level, the wider the CI.

8-6 a) Sample mean from the first confidence interval =37.53 + (49.87-37.53)/2 = 43.7Sample mean from the second confidence interval =35.59 + (51.81-35.59)/2 = 43.7b) The 99% CI is (35.59, 51.81) and the 95% CI is (37.53, 49.87). The higher the confidence level, the wider the CI.

8-7 a) Find n for the length of the 95% CI to be 40. Za/2 = 1.96

1/2 length=(1. 96 )(20 )/√n=2039 .2=20√nn=(39 .2

20 )2

=3 . 84

Therefore, n = 4.b) Find n for the length of the 99% CI to be 40. Za/2 = 2.58

1/2 length=(2. 58 )(20)/√n=2051 .6=20√nn=(51 .6

20 )2

=6 .66

Therefore, n = 7.

8-8 Interval (1): 3124 .9≤μ≤3215 .7 and Interval (2):3110. 5≤μ≤3230 .1Interval (1): half-length =90.8/2=45.4 and Interval (2): half-length =119.6/2=59.8

a) x̄1=3124 . 9+45 . 4=3170 .3

x̄2=3110.5+59. 8=3170. 3 The sample means are the same.

b) Interval (1): 3124 .9≤μ≤3215 .7 was calculated with 95% Confidence because it has a smaller half-length, and therefore a smaller confidence interval. The 99% confidence level will make the interval larger.

8-9 a) The 99% CI on the mean calcium concentration would be longer.b) No, that is not the correct interpretation of a confidence interval. The probability that is between 0.49 and 0.82 is either 0 or 1.

8-2

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

c) Yes, this is the correct interpretation of a confidence interval. The upper and lower limits of the confidence limits are random variables.

8-10 95% Two-sided CI on the breaking strength of yarn: where x = 98, = 2 , n=9 and z0.025 = 1.96

x̄−z0 .025 σ /√n≤μ≤ x̄+z0 . 025σ /√n98−1. 96 (2)/√9≤μ≤98+1 . 96(2 )/√996 . 7≤μ≤99 . 3

8-11 95% Two-sided CI on the true mean yield: where x = 90.480, = 3 , n=5 and z0.025 = 1.96

x̄−z0 .025 σ /√n≤μ≤ x̄+z0 . 025σ /√n90 . 480−1.96 (3)/√5≤μ≤90 . 480+1 . 96(3 )/√587 . 85≤μ≤93 . 11

8-12 99% Two-sided CI on the diameter cable harness holes: where x =1.5045 , = 0.01 , n=10 and z0.005 = 2.58

x̄−z0 .005 σ /√n≤μ≤ x̄+z0 . 005σ /√n1 .5045−2. 58(0 . 01)/√10≤μ≤1 .5045+2 . 58(0 . 01)/√101 .4963≤μ≤1.5127

8-13 a) 99% Two-sided CI on the true mean piston ring diameter For = 0.01, z/2 = z0.005 = 2.58 , and x = 74.036, = 0.001, n=15

74.0353 74.0367

b) 99% One-sided CI on the true mean piston ring diameter For = 0.01, z = z0.01 =2.33 and x = 74.036, = 0.001, n=15

x̄−z0 .01σ

√n≤μ

74 . 036−2.33(0. 001

√15 )≤μ

74.0354 The lower bound of the one sided confidence interval is less than the lower bound of the two-sided confidence. This is because the Type I probability of 99% one sided confidence interval (or in the left tail (or in the lower bound) is greater than Type I probability of 99% two-sided confidence interval (or in the left tail.

8-14 a) 95% Two-sided CI on the true mean life of a 75-watt light bulb For = 0.05, z/2 = z0.025 = 1.96 , and x = 1014, =25 , n=20

x̄−z0 .025 ( σ√n )≤μ≤ x̄+z0 . 025( σ√n )

8-3

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

1014−1 .96 (25

√20 )≤μ≤1014+1 . 96(25

√20 )1003≤μ≤1025

b) 95% One-sided CI on the true mean piston ring diameter For = 0.05, z = z0.05 =1.65 and x = 1014, =25 , n=20

x̄−z0 .05σ

√n≤μ

1014−1 .65(25

√20 )≤μ

1005≤μ

The lower bound of the one sided confidence interval is lower than the lower bound of the two-sided confidence interval even though the level of significance is the same. This is because all of the Type I probability (or is in the left tail (or in the lower bound).

8-15 a) 95% two sided CI on the mean compressive strengthz/2 = z0.025 = 1.96, and x = 3250, 2 = 1000, n=12

3250−1. 96(31. 62

√12 )≤μ≤3250+1. 96 (31. 62

√12 )3232 .11 ≤μ≤3267 .89

b) 99% Two-sided CI on the true mean compressive strength z/2 = z0.005 = 2.58

x̄−z0 .005 ( σ√n )≤μ≤ x̄+z0 . 005( σ√n )

3250−2. 58(31 . 62

√12 )≤μ≤3250+2. 58(31 .62

√12 )3226 .4 ≤μ≤3273 .6

The 99% CI is wider than the 95% CI

8-16 95% Confident that the error of estimating the true mean life of a 75-watt light bulb is less than 5 hours. For = 0.05, z/2 = z0.025 = 1.96 , and =25 , E=5

8-4

Page 5: Applied Statistics and Probability Ch8 solutions

Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

n=( za/2σE )2

=( 1.96 (25)5 )

2

=96 .04

Always round up to the next number, therefore n = 97

8-17 Set the width to 6 hours with = 25, z0.025 = 1.96 solve for n.

1/2 width=(1 .96)(25 )/√n=349=3√nn=(49

3 )2

=266 .78

Therefore, n = 267.

8-18 99% Confident that the error of estimating the true compressive strength is less than 15 psi For = 0.01, z/2 = z0.005 = 2.58 , and =31.62 , E=15

n=( za/2σE )2

=( 2 .58(31 .62)15 )

2

=29 .6≃30

Therefore, n=30

8-19 To decrease the length of the CI by one half, the sample size must be increased by 4 times (22).

zα /2σ /√n=0 .5 l

Now, to decrease by half, divide both sides by 2.( zα /2σ /√n )/2=( l /2 )/2( zα /2σ /2√n )=l /4

( zα /2σ /√22n )=l / 4Therefore, the sample size must be increased by 22.

8-20 If n is doubled in Eq 8-7:

x̄−zα /2 ( σ√n )≤μ≤ x̄+zα /2 ( σ√n )zα /2σ

√2n=

zα /2σ

1. 414 √n=

zα /2 σ

1 . 414√n= 1

1. 414 ( zα /2 σ

√n )The interval is reduced by 0.293 29.3%

If n is increased by a factor of 4 Eq 8-7:

zα /2σ

√4 n=zα /2σ

2√n=zα /2σ

2√n=1

2 ( zα /2 σ

√n )The interval is reduced by 0.5 or ½.

8-21 a) 99% two sided CI on the mean temperature

z/2 = z0.005 = 2.57, and x̄ = 13.77, = 0.5, n=11

x̄−z0 .005 ( σ√n )≤μ≤ x̄+z0 . 005( σ√n )

8-5

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

13 .77−2 .57 (0 . 5

√11 )≤μ≤13 .77+2. 57(0 .5

√11 )13 .383 ≤μ≤14 . 157

b) 95% lower-confidence bound on the mean temperature

For = 0.05, z = z0.05 =1.65 and x̄ = 13.77, = 0.5, n =11

x̄−z0 .05σ

√n≤μ

13 .77−1 .65 (0 .5

√11 )≤μ

13 .521≤μ

c) 95% confidence that the error of estimating the mean temperature for wheat grown is less than 2 degrees Celsius.

For = 0.05, z/2 = z0.025 = 1.96, and = 0.5, E = 2

n=( za/2σE )2

=( 1.96 (0 .5 )2 )

2

=0 .2401

Always round up to the next number, therefore n = 1.

d) Set the width to 1.5 degrees Celsius with = 0.5, z0.025 = 1.96 solve for n.

1/2 width=(1 .96)(0 . 5 )/√n=0 .750 .98=0. 75√nn=(0. 98

0. 75 )2

=1 .707

Therefore, n = 2.

Section 8-2

8-22t0.025 ,15=2.131 t0.05 ,10=1 . 812 t0.10 ,20=1 .325

t0.005 ,25=2.787 t0.001 ,30=3 .385

8-23 a) t0.025 ,12=2 .179

b) t0.025 ,24=2.064

c) t0.005 ,13=3 .012

d) t0.0005 ,15=4 . 073

8-24 a) t0.05 ,14=1 . 761

b) t0.01 ,19=2 .539

c) t0.001 ,24=3. 467

8-25 a) Mean= sum

N=251.848

10=25. 1848

Variance ==(stDev )2=1.6052=2 .5760

8-6

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

b) 95% confidence interval on mean

n=10 x̄=25. 1848 s=1 .605 t0.025 ,9=2 . 262

x̄−t0.025 ,9(s√n )≤μ≤ x̄+t0.025 ,9(s√n )25 .1848−2 .262(1. 605

√10 )≤μ≤25 .1848+2 .262(1 .605

√10 )24 . 037≤μ≤26 . 333

8-26 SE Mean = stDev

√N=6 . 11

√N=1 .58

, therefore N = 15

Mean= sum

N=751 .40

15=50 .0933

Variance =(stDev )2=6 .112=37 . 3321b) 95% confidence interval on mean

n=15 x̄=50. 0933 s=6 .11 t0.025 ,14=2.145

x̄−t0.025 ,14 (s√n )≤μ≤ x̄+t0 .025 ,14(s√n )50 . 0933−2 .145 (6 .11

√15 )≤μ≤50. 0933+2 . 145(6 . 11

√15 )46 . 709≤μ≤53 . 477

8-27 95% confidence interval on mean tire life

n=16 x̄=60 ,139.7 s=3645 . 94 t0.025 ,15=2.131

x̄−t0.025 ,15(s√n )≤μ≤ x̄+t0.025 ,15(s√n )60139 .7−2 .131(3645 .94

√16 )≤μ≤60139. 7+2. 131(3645 . 94

√16 )58197 . 33≤μ≤62082.07

8-28 99% lower confidence bound on mean Izod impact strength

n=20 x̄=1. 25 s=0 .25 t0.01 ,19=2 .539

x̄−t0.01 ,19(s√n )≤μ

1 .25−2 .539(0. 25

√20 )≤μ

1 .108≤μ

8-29 = 1.10 s = 0.015 n = 25

8-7

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

95% CI on the mean volume of syrup dispensedFor = 0.05 and n = 25, t/2,n-1 = t0.025,24 = 2.064

x̄−t0.025 ,24(s√n )≤μ≤ x̄+t0 .025 ,24(s√n )1 .10−2. 064 (0 . 015

√25 )≤μ≤1. 10+2 .064 (0 .015

√25 )1 .094 ≤μ≤1 .106

8-30 95% confidence interval on mean peak power

n=7 x̄=315 s=16 t0.025 ,6=2 . 447

x̄−t0.025 ,6(s√n )≤μ≤ x̄+t0.025 ,6(s√n )315−2.447 (16

√7 )≤μ≤315+2 . 447(315

√7 )300 . 202≤μ≤329 .798

8-31 99% upper confidence interval on mean SBP

n=14 x̄=118.3 s=9 . 9 t0.01 ,13=2 .650

μ≤ x̄+t0 .005 ,13(s√n )μ≤118 .3+2.650(9. 9

√14 )μ≤125 . 312

8-32 90% CI on the mean frequency of a beam subjected to loads

x̄=231 . 67, s= 1 . 53, n=5, tα /2,n-1=t. 05 ,4=2. 132

x̄−t0.05 ,4(s√n )≤μ≤ x̄+t0.05 ,4 (s√n )231 .67−2 .132(1.53

√5 )≤μ≤231. 67−2. 132(1 .53

√5 )230 . 2 ≤μ≤233 . 1

8-8

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

By examining the normal probability plot, it appears that the data are normally distributed. There does not appear to be enough evidence to reject the hypothesis that the frequencies are normally distributed.

8-33 The data appear to be normally distributed based on examination of the normal probability plot below.

Therefore, there is evidence to support that the annual rainfall is normally distributed.

95% confidence interval on mean annual rainfall

n=20 x̄=485 . 8 s=90 . 34 t0.025 ,19=2.093

x̄−t0.025 ,19(s√n )≤μ≤ x̄+t0.025 ,19(s√n )485 . 8−2. 093(90 .34

√20 )≤μ≤485 . 8+2. 093(90 .34

√20 )443 . 520≤μ≤528 . 080

8-34 The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the solar energy is normally distributed.

8-9

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

95% confidence interval on mean solar energy consumed

n=16 x̄=65 .58 s=4 . 225 t0.025 ,15=2.131

x̄−t0.025 ,15(s√n )≤μ≤ x̄+t0.025 ,15(s√n )65 . 58−2 .131(4 . 225

√16 )≤μ≤65 .58+2. 131(4 .225

√16 )63 . 329≤μ≤67 .831

8-35 99% confidence interval on mean current requiredAssume that the data are a random sample from a normal distribution.

n=10 x̄=317 .2 s=15 . 7 t0.005 ,9=3 . 250

x̄−t0.005 ,9(s√n )≤μ≤ x̄+t0.005 ,9(s√n )317 . 2−3 . 250(15 .7

√10 )≤μ≤317 .2+3 .250(15 .7

√10 )301 .06≤μ≤333. 34

8-36 a) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the level of polyunsaturated fatty acid is normally distributed.

8-10

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

b) 99% CI on the mean level of polyunsaturated fatty acid.For = 0.01, t/2,n-1 = t0.005,5 = 4.032

x̄−t0.005 ,5 ( s

√n )≤μ≤ x̄+t0 .005 ,5 ( s√n )

16 . 98−4 . 032(0 . 319

√6 )≤μ≤16 . 98+4 . 032(0 . 319

√6 )16 . 455≤μ≤17 . 505

The 99% confidence for the mean polyunsaturated fat is (16.455, 17.505). There is high confidence that the true mean is in this interval

8-37 a) The data appear to be normally distributed based on examination of the normal probability plot below.

b) 95% two-sided confidence interval on mean comprehensive strength

8-11

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

n=12 x̄=2259 . 9 s=35 .6 t0.025 ,11=2 .201

x̄−t0.025 ,11 (s√n )≤μ≤ x̄+ t0 .025 ,11(s√n )2259 .9−2 .201(35 .6

√12 )≤μ≤2259 . 9+2 .201(35 .6

√12 )2237 .3≤μ≤2282.5

c) 95% lower-confidence bound on mean strength

x̄−t0.05 ,11(s√n )≤μ

2259 .9−1 .796 (35 .6

√12 )≤μ

2241 .4≤μ

8-38 a) According to the normal probability plot there does not seem to be a severe deviation from normality for this data. This is due to the fact that the data appears to fall along a straight line.

b) 95% two-sided confidence interval on mean rod diameter For = 0.05 and n = 15, t/2,n-1 = t0.025,14 = 2.145

x̄−t0.025 ,14 ( s√n )≤μ≤ x̄+t0 .025 ,14( s

√n )

8 .23−2 .145 ( 0. 025

√15 )≤μ≤8 .23+2 . 145( 0 .025

√15 )

8-12

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

8.216 8.244c) 95% upper confidence bound on mean rod diameter t0.05,14 = 1.761

μ≤ x̄+t0 .025 ,14(s√n )μ≤8.23+1 . 761(0 .025

√15 )μ≤8.241

8-39 a) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the speed-up of CNN is normally distributed.

Speed

Perc

ent

6.05.55.04.54.03.53.0

99

95

90

80

70

60504030

20

10

5

1

Mean

0.745

4.313StDev 0.4328N 13AD 0.233P-Value

Probabi l i ty P lot of SpeedNormal - 95% CI

b) 95% confidence interval on mean speed-up

n=13 x̄=4 .313 s=0 . 4328 t0.025 ,12=2 .179

x̄−t0.025 ,12(s√n )≤μ≤ x̄+t0.025 ,12(s√n )4 .313−2 .179(0 . 4328

√13 )≤μ≤4 .313+2 .179(0 . 4328

√13 )4 .051≤μ≤4 . 575

c) 95% lower confidence bound on mean speed-up

n=13 x̄=4 .313 s=0 . 4328 t0.05 ,12=1 .782

8-13

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th edition September 22, 2010

x̄−t0.05 ,12(s√n )≤μ

4 .313−1. 782(0 . 4328

√13 )≤μ

4 .099≤μ

8-40 95% lower bound confidence for the mean wall thickness given = 4.05 s = 0.08 n = 25

t,n-1 = t0.05,24 = 1.711

x̄−t0.05 ,24( s

√n )≤μ

4 .05−1 .711( 0. 08

√25 )≤μ

4.023 There is high confidence that the true mean wall thickness is greater than 4.023 mm.

8-41 a) The data appear to be normally distributed. There is not strong evidence that the percentage of enrichment deviates from normality.

b) 99% two-sided confidence interval on mean percentage enrichment

For = 0.01 and n = 12, t/2,n-1 = t0.005,11 = 3.106, x̄= 2 .9017 s=0 .0993

8-14

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th edition September 22, 2010

x̄−t0.005 ,11 (s√n )≤μ≤ x̄+ t0 .005 ,11(s√n )2 .902−3 .106 (0 . 0993

√12 )≤μ≤2.902+3 .106 (0 .0993

√12 )2 .813≤μ≤2. 991

8-15

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Applied Statistics and Probability for Engineers, 5

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Section 8-3

8-42χ0 . 05 ,10

2 =18 .31 χ0 . 025 ,152 =27 .49 χ0 . 01 ,12

2 =26. 22

χ0 . 95 ,202 =10 .85 χ0 . 99 ,18

2 =7 .01 χ0 . 995 ,162 =5 . 14

χ0 . 005 ,252 =46 .93

8-43 a) 95% upper CI and df = 24 χ1−α ,df2 = χ0 . 95 ,24

2 = 13.85

b) 99% lower CI and df = 9 χ α ,df2 = χ0 . 01 ,9

2 = 21.67c) 90% CI and df = 19

χ α /2, df2 = χ0.05 ,19

2 =30 . 14 and χ1−α /2, df2 = χ0 .95 ,19

2 =10. 12

8-44 99% lower confidence bound for 2

For = 0.01 and n = 15, χ α ,n−12 = χ0 . 01 ,14

2 =29.14

14 (0 .008 )2

29 .14≤σ2

0 .00003075≤σ2

8-45 99% lower confidence bound for from the previous exercise is

0 .00003075≤σ2

0 .005545≤σOne may take the square root of the variance bound to obtain the confidence bound for the standard deviation.

8-46 95% two sided confidence interval for n=10 s=4 .8

χ α /2, n−12 = χ0 .025 ,9

2 =19 .02 and

χ1−α /2, n−12 = χ 0. 975 ,9

2 =2 . 70

9( 4 . 8)2

19 . 02≤σ2≤

9(4 .8 )2

2. 7010 . 90≤σ2≤76. 803 .30<σ<8 . 76

8-47 95% confidence interval for : given n = 51, s = 0.37 First find the confidence interval for 2 :

For = 0.05 and n = 51, 71.42 and 32.36

50(0 .37 )2

71 . 42≤σ2≤

50(0 .37 )2

32.36 0.096 2 0.2115 Taking the square root of the endpoints of this interval we obtain,

0.31 < < 0.46

8-16

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

8-48 95% confidence interval for n=17 s=0 . 09

χ α /2, n−12 = χ0 .025 ,16

2 =28 . 85 and

χ1−α /2, n−12 = χ 0. 975 ,16

2 =6 .91

16(0 .09 )2

28 . 85≤σ2≤

16(0 . 09)2

6 . 910 .0045≤σ2≤0 .01880 .067<σ<0 .137

8-49 The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the mean temperature is normally distributed.

Mean Temp

Perc

ent

252423222120191817

99

95

90

80

70

60504030

20

10

5

1

Mean

0.367

21.41StDev 0.9463N 8AD 0.352P-Value

Probabi l i ty P lot of Mean TempNormal - 95% CI

95% confidence interval for n=8 s=0 . 9463

χ α /2, n−12 = χ0 .025 ,7

2 =16 . 01 and χ1−α /2, n−12 = χ 0. 975 ,7

2 =1 .69

7( 0. 9463 )2

16 . 01≤σ2≤

7(0 . 9463 )2

1 .690 .392≤σ 2≤3 .7090 .626<σ<1. 926

8-50 95% confidence interval for n=41 s=15 . 99

χ α /2, n−12 = χ0 .025 ,40

2 =59 . 34 and

χ1−α /2, n−12 = χ 0. 975 ,40

2 =24 . 43

8-17

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

40 (15. 99 )2

59 .34≤σ2≤

40(15 .99 )2

24 .43172 .35≤σ2≤418.63313 .13<σ<20 . 46

The data don’t appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is not enough evidence to support that the time of tumor appearance is normally distributed. So the 95% confidence interval for σ is invalid.

TimeOfTumor

Perc

ent

140120100806040

99

95

90

80

70

605040

30

20

10

5

1

Mean

<0.005

88.78StDev 15.99N 41AD 1.631P-Value

Probabi l i ty P lot of TimeOfTumorNormal - 95% CI

8-51 95% confidence interval for n=15 s=0 . 00831

χ α /2, n−12 = χ0 .025 ,14

2 =26 . 12 and χ1−α ,n−12 = χ 0. 95 ,14

2 =6 . 53

σ 2≤14 (0 . 00831)2

6 . 53σ 2≤0 . 000148σ≤0 . 0122

The data do not appear to be normally distributed based on an examination of the normal probability plot below. Therefore, the 95% confidence interval for σ is not valid.

8-18

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

Gauge Cap

Perc

ent

3.503.493.483.473.463.453.44

99

95

90

80

70

605040

30

20

10

5

1

Mean

0.018

3.472StDev 0.008307N 15AD 0.878P-Value

Probabi l i ty P lot of Gauge CapNormal - 95% CI

8-52 a) 99% two-sided confidence interval on 2

n=10 s=1 .913 χ0 . 005 ,9

2 =23 .59 and

χ0 . 995 ,92 =1 .73

9(1 . 913)2

23 .59≤σ2≤

9(1 .913 )2

1. 731 .396≤σ 2≤19 . 038

b) 99% lower confidence bound for 2

For = 0.01 and n = 10, χ α ,n−12 = χ0 . 01 ,9

2 = 21.67

9(1 . 913)2

21 .67≤σ2

1 .5199≤σ2

c) 90% lower confidence bound for 2

For = 0.1 and n = 10, χ α ,n−1

2 = χ0 . 1,92 =

14.68

9(1 . 913)2

14 . 68≤σ2

2 .2436≤σ 2

1 .498≤σ

d) The lower confidence bound of the 99% two-sided interval is less than the one-sided interval. The lower confidence bound for 2 is in part (c) is greater because the confidence is lower.

Section 8-4

8-53 a) 95% Confidence Interval on the fraction defective produced with this tool.

8-19

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

p̂=13300

=0 . 04333 n=300 zα /2=1.96

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .04333−1 . 96√0. 04333 (0. 95667 )300

≤p≤0 . 04333+1.96 √0 . 04333(0 . 95667 )300

0 .02029≤p≤0 .06637

b) 95% upper confidence bound zα=z0. 05=1 .65

p≤ p̂+z α /2√ p̂(1− p̂ )n

p≤0.04333+1 .650√0 .04333(0 .95667 )300

p≤0.06273

8-54 a) 95% Confidence Interval on the proportion of such tears that will heal.

p̂=0. 676 n=37 zα /2=1.96

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .676−1 .96√0 .676 (0.324 )37

≤p≤0 .676+1 .96 √0 .676(0 .324 )37

0 .5245≤p≤0 .827

b) 95% lower confidence bound on the proportion of such tears that will heal.

p̂−zα √ p̂(1− p̂ )n

≤p

0 .676−1 . 64√0 .676 (0 .33 )37

≤p

0 .549≤p

8-55 a) 95% confidence interval for the proportion of college graduates in Ohio that voted for George Bush.

p̂=412768

=0. 536 n=768 zα /2=1.96

8-20

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th edition September 22, 2010

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .536−1 . 96√0.536(0 .464 )768

≤p≤0 .536+1 . 96√0 .536(0 .464 )768

0 .501≤p≤0. 571

b) 95% lower confidence bound on the proportion of college graduates in Ohio that voted for George Bush.

p̂−zα √ p̂(1− p̂ )n

≤p

0 .536−1 . 64√0.536( 0.464 )768

≤p

0 .506≤p

8-56 a) 95% Confidence Interval on the death rate from lung cancer.

p̂=8231000

=0 .823 n=1000 zα /2=1.96

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .823−1 .96√0.823 (0.177 )1000

≤p≤0 .823+1. 96√0 .823(0 .177 )1000

0 .7993≤p≤0 .8467

b) E = 0.03, = 0.05, z/2 = z0.025 = 1.96 and = 0.823 as the initial estimate of p,

n=( zα /2

E )2

p̂ (1− p̂)=( 1 .960 .03 )

2

0 .823(1−0.823 )=621 .79,

n 622.

c) E = 0.03, = 0.05, z/2 = z0.025 = 1.96 at least 95% confident

n=( zα /2

E )2

(0 .25)=( 1. 960 .03 )

2

(0 .25)=1067 .11,

n 1068.

8-57 a) 95% Confidence Interval on the proportion of rats that are under-weight.

p̂=1230

=0 . 4 n=30 zα /2=1.96

8-21

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .4−1 .96 √0 .4 (0 .6 )30

≤p≤0 .4+1.96√0 .4 (0.6 )30

0 .225≤p≤0 .575

b) E = 0.02, = 0.05, z/2 = z0.025 = 1.96 and = 0.4as the initial estimate of p,

n=( zα /2

E )2

p̂ (1− p̂)=( 1 .960 .02 )

2

0 .4 (1−0 .4 )=2304 .96,

n 2305.

c) E = 0.02, = 0.05, z/2 = z0.025 = 1.96 at least 95% confident

n=( zα /2

E )2

(0 .25)=( 1. 960 .02 )

2

(0 .25)=2401.

8-58 a) 95% Confidence Interval on the true proportion of helmets showing damage

p̂=1850

=0 . 36 n=50 zα /2=1.96

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .36−1 . 96√0. 36 (0. 64 )50

≤p≤0 . 36+1 .96 √0 . 36(0 . 64 )50

0 .227≤p≤0 . 493

b)

n=( zα /2

E )2

p (1−p)=( 1 .960 .02 )

2

0 .36(1−0 .36 )=2212 .76

n≃2213

c)

n=( zα /2

E )2

p (1−p)=( 1 .960 .02 )

2

0 .5(1−0 . 5)=2401

8-59 The worst case would be for p = 0.5, thus with E = 0.05 and = 0.01, z/2 = z0.005 = 2.58 we obtain a sample size of:

n=( zα /2

E )2

p (1−p)=( 2 .580 .05 )

2

0 .5(1−0 . 5)=665 .64, n 666

8-22

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

8-60 E = 0.017, = 0.01, z/2 = z0.005 = 2.58

n=( zα /2

E )2

p (1−p)=( 2 .580 .017 )

2

0 . 5(1−0.5 )=5758 .13, n 5759

Section 8-6

8-61 95% prediction interval on the life of the next tire given = 60139.7 s = 3645.94 n = 16 for =0.05 t/2,n-1 = t0.025,15 = 2.131

x̄−t0.025 ,15 s √1+1n≤xn+1≤ x̄+t0 . 025 ,15 s√1+1

n

60139 .7−2 .131 (3645 . 94 )√1+116

≤xn+1≤60139 .7+2.131(3645. 94 )√1+116

52131 .1≤ xn+1≤68148 .3

The prediction interval is considerably wider than the 95% confidence interval (58,197.3 62,082.07). This is expected because the prediction interval needs to include the variability in the parameter estimates as well as the variability in a future observation.

8-62 99% prediction interval on the Izod impact data

n=20 x̄=1. 25 s=0 .25 t0.005 ,19=2.861

x̄−t0.005 ,19 s √1+1n≤xn+1≤ x̄+t0 . 005 ,19 s√1+1

n

1 .25−2 .861(0 . 25 )√1+120

≤xn+1≤1 .25+2 . 861(0 .25 )√1+120

0 .517≤xn+1≤1. 983

The lower bound of the 99% prediction interval is considerably lower than the 99% confidence interval (1.108 ). This is expected because the prediction interval needs to include the variability in the parameter estimates as well as the variability in a future observation.

8-63 95% Prediction Interval on the volume of syrup of the next beverage dispensed= 1.10 s = 0.015 n = 25 t/2,n-1 = t0.025,24 = 2.064

8-23

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

x̄−t0.025 ,24 s√1+1n

≤xn+1≤ x̄+t0. 025 ,24 s √1+1n

1 .10−2. 064 (0 . 015)√1+125

≤xn+1≤1.10−2 . 064( 0. 015 )√1+125

1 .068≤xn+1≤1 . 13

The prediction interval is wider than the confidence interval:1 .094 ≤μ≤1 .106

8-64 90% prediction interval the value of the natural frequency of the next beam of this type that will be tested. given x = 231.67, s =1.53 For = 0.10 and n = 5, t/2,n-1 = t0.05,4 = 2.132

x̄−t0.05 ,4 s √1+1n≤xn+1≤ x̄+t0 . 05 ,4 s√1+1

n

231 .67−2 .132(1 .53 )√1+15

≤xn+1≤231. 67−2. 132(1.53 )√1+15

228 .1≤xn+1≤235 .2The 90% prediction in interval is greater than the 90% CI.

8-65 95% Prediction Interval on the volume of syrup of the next beverage dispensed

n=20 x̄=485 . 8 s=90 . 34 t/2,n-1 = t0.025,19 = 2.093

x̄−t0.025 ,19 s √1+1n≤xn+1≤ x̄+t0 .025 ,19 s√1+1

n

485 .8−2.093( 90.34 )√1+120

≤xn+1≤485 .8−2 .093(90 .34 )√1+120

292 .049≤xn+1≤679 .551The 95% prediction interval is wider than the 95% confidence interval.

8-66 99% prediction interval on the polyunsaturated fat

n=6 x̄=16 .98 s=0 . 319 t0.005 ,5=4 .032

x̄−t0.005 ,5 s √1+1n

≤xn+1≤ x̄+t0.005 ,5 s√1+1n

16 .98−4 .032(0 .319 )√1+16

≤xn+1≤16 .98+4 .032(0 .319)√1+16

15 .59≤xn+1≤18 .37The length of the prediction interval is much longer than the width of the confidence interval

16 . 455≤μ≤17 . 505 .

8-67 Given = 317.2 s = 15.7 n = 10 for =0.05 t/2,n-1 = t0.005,9 = 3.250

8-24

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

x̄−t0.005 ,9s √1+1n

≤xn+1≤ x̄+t0 .005 ,9s √1+1n

317 .2−3 .250(15.7 )√1+110

≤xn+1≤317 .2−3 .250(15 .7 )√1+110

263 .7≤xn+1≤370.7The length of the prediction interval is longer.

8-68 95% prediction interval on the next rod diameter tested

n=15 x̄=8 .23 s=0 .025 t0.025 ,14=2.145

x̄−t0.025 ,14 s√1+1n≤xn+1≤ x̄+t0 .025 ,14 s √1+1

n

8 .23−2 .145 (0 .025 )√1+115

≤xn+1≤8.23−2.145(0 .025)√1+115

8 .17≤xn+1≤8 .29

95% two-sided confidence interval on mean rod diameter is 8.216 8.244

8-69 90% prediction interval on the next specimen of concrete testedgiven = 2260 s = 35.57 n = 12 for = 0.05 and n = 12, t/2,n-1 = t0.05,11 = 1.796

x̄−t0.05 ,11 s√1+1n≤xn+1≤ x̄+ t0 .05 ,11s√1+1

n

2260−1.796(35 .57)√1+112

≤xn+1≤2260+1.796(35 .57)√1+112

2193 .5≤xn+1≤2326 .5

8-70 90% prediction interval on wall thickness on the next bottle tested. given = 4.05 s = 0.08 n = 25 for t/2,n-1 = t0.05,24 = 1.711

x̄−t0.05 ,24 s √1+1n≤xn+1≤ x̄+ t0 .05 ,24 s√1+1

n

4 .05−1 .711(0 .08 )√1+125

≤xn+1≤4 .05−1 .711(0 .08)√1+125

3 .91≤xn+1≤4 .19

8-71 90% prediction interval for enrichment data given = 2.9 s = 0.099 n = 12 for = 0.10 and n = 12, t/2,n-1 = t0.05,11 = 1.796

8-25

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th edition September 22, 2010

x̄−t0.05 ,12 s√1+1n

≤xn+1≤ x̄+t0.05 ,12 s√1+1n

2 .9−1 .796 (0.099 )√1+112

≤xn+1≤2 .9+1.796(0 .099 )√1+112

2 .71≤xn+1≤3 .09The 90% confidence interval is

x̄−t0.05 ,12 s√1n

≤μ≤ x̄+t0. 05 ,12 s√1n

2 .9−1 .796 (0. 099 )√112

≤μ≤2 . 9−1 .796 (0 .099 )√112

2 .85≤μ≤2. 95

The prediction interval is wider than the CI on the population mean with the same confidence.

The 99% confidence interval is

x̄−t0.005 ,12 s √1n

≤μ≤ x̄+t0 . 005 ,12 s√1n

2 .9−3 .106 (0 .099 )√112

≤μ≤2 . 9+3 .106 (0 .099 )√112

2 .81≤μ≤2 . 99The prediction interval is even wider than the CI on the population mean with greater confidence.

8-72 To obtain a one sided prediction interval, use t,n-1 instead of t/2,n-1 Since we want a 95% one sided prediction interval, t/2,n-1 = t0.05,24 = 1.711and = 4.05 s = 0.08 n = 25

x̄−t0.05 ,24 s √1+1n≤xn+1

4 .05−1 .711(0 .08 )√1+125

≤xn+1

3 .91≤xn+1

The prediction interval bound is much lower than the confidence interval bound of 4.023 mm

8-73 95% tolerance interval on the life of the tires that has a 95% CLgiven = 60139.7 s = 3645.94 n = 16 we find k=2.903

8-26

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

x̄−ks , x̄+ks60139 .7−2 . 903 (3645. 94 ) , 60139.7+2 . 903 (3645.94 )

(49555 .54 , 70723.86 )95% confidence interval (58,197.3 62,082.07) is shorter than the 95%tolerance interval.

8-74 99% tolerance interval on the Izod impact strength PVC pipe that has a 90% CLgiven x=1.25, s=0.25 and n=20 we find k=3.368

x̄−ks , x̄+ks1 .25−3 .368 (0 . 25 ) , 1. 25+3 .368 ( 0. 25 )

(0 .408 , 2 .092 )

The 99% tolerance interval is much wider than the 99% confidence interval on the population mean (1.090 1.410).

8-75 95% tolerance interval on the syrup volume that has 90% confidence level= 1.10 s = 0.015 n = 25 and k=2.474

x̄−ks , x̄+ks1 .10−2. 474 (0 .015 ) , 1 .10+2.474 (0 .015 )

(1.06 , 1 .14 )

8-76 99% tolerance interval on the polyunsaturated fatty acid in this type of margarine that has a confidence level of 95% x = 16.98 s = 0.319 n=6 and k = 5.775

x̄−ks , x̄+ks16 . 98−5 . 775 (0 .319 ) , 16 . 98+5 .775 (0 .319 )

(15 .14 , 18 . 82)The 99% tolerance interval is much wider than the 99% confidence interval on the population mean (16.46 17.51).

8-77 95% tolerance interval on the rainfall that has a confidence level of 95%

n=20 x̄=485 . 8 s=90 . 34 k=2 .752

x̄−ks , x̄+ks485 . 8−2. 752 (90 .34 ) , 485 .8+2. 752 (90 .34 )

(237 . 184 , 734 .416 )The 95% tolerance interval is much wider than the 95% confidence interval on the population mean (

443 . 52≤μ≤528 .08 ).

8-78 95% tolerance interval on the diameter of the rods in exercise 8-27 that has a 90% confidence levelx = 8.23 s = 0.0.25 n=15 and k=2.713

8-27

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x̄−ks , x̄+ks8 .23−2 .713 (0 . 025 ) , 8 . 23+2. 713 (0 .025 )

(8 . 16 , 8 .30 )The 95% tolerance interval is wider than the 95% confidence interval on the population mean (8.216 8.244).

8-79 99% tolerance interval on the brightness of television tubes that has a 95% CLgiven = 317.2 s = 15.7 n = 10 we find k=4.433

x̄−ks , x̄+ks317 .2−4 . 433 (15.7 ) , 317 . 2+4 . 433 (15 .7 )

(247 . 60 , 386 .80)

The 99% tolerance interval is much wider than the 95% confidence interval on the population mean

301 .06≤μ≤333. 34 .

8-80 90% tolerance interval on the comprehensive strength of concrete that has a 90% CLgiven = 2260 s = 35.57 n = 12 we find k=2.404

x̄−ks , x̄+ks2260−2.404 (35. 57 ) , 2260+2 .404 (35 .57 )

(2174 .5 , 2345 .5 )

The 90% tolerance interval is much wider than the 95% confidence interval on the population mean 2237 .3≤μ≤2282.5 .

8-81 99% tolerance interval on rod enrichment data that have a 95% CLgiven = 2.9 s = 0.099 n = 12 we find k=4.150

x̄−ks , x̄+ks2 .9−4 .150 (0 .099 ) , 2.9+4 . 150 (0.099 )

(2 .49 , 3 .31 )

The 99% tolerance interval is much wider than the 95% CI on the population mean (2.84 2.96)

8-82 a) 90% tolerance interval on wall thickness measurements that have a 90% CLgiven = 4.05 s = 0.08 n = 25 we find k=2.077

x̄−ks , x̄+ks4 .05−2 .077 (0.08 ) , 4 .05+2.077 (0 .08 )

(3 .88 , 4 .22)

8-28

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

The lower bound of the 90% tolerance interval is much lower than the lower bound on the 95% confidence interval on the population mean (4.023 )

b) 90% lower tolerance bound on bottle wall thickness that has confidence level 90%. given = 4.05 s = 0.08 n = 25 and k = 1.702

x̄−ks4 .05−1 .702 (0 . 08 )

3. 91The lower tolerance bound is of interest if we want to make sure the wall thickness is at least a certain value so that the bottle will not break.

8-29

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Supplemental Exercises

8-83 Where α 1+α 2=α . Let α=0 . 05

Interval for α 1=α2=α /2=0 . 025

The confidence level for x̄−1 .96σ /√n≤μ≤ x̄+1. 96 σ /√n is determined by the by the value of z0

which is 1.96. From Table III, we find (1.96) = P(Z<1.96) = 0.975 and the confidence level is 95%.

Interval for α 1=0. 01 , α 2=0 . 04

The confidence interval is x̄−2 .33 σ /√n≤μ≤ x̄+1 . 75σ /√n , the confidence level is the same

because α=0 . 05 . The symmetric interval does not affect the level of significance; however, it does affect the length. The symmetric interval is shorter in length.

8-84 = 50 unknowna) n = 16 s = 1.5

to=52−508/√16

=1

The P-value for t0 = 1, degrees of freedom = 15, is between 0.1 and 0.25. Thus we would conclude that the results are not very unusual.

b) n = 30

to=

52−508/√30

=1.37

The P-value for t0 = 1.37, degrees of freedom = 29, is between 0.05 and 0.1. Thus we conclude that the results are somewhat unusual.

c) n = 100 (with n > 30, the standard normal table can be used for this problem)

zo=

52−508/√100

=2. 5

The P-value for z0 = 2.5, is 0.00621. Thus we conclude that the results are very unusual.

d) For constant values of and s, increasing only the sample size, we see that the standard error of decreases and consequently a sample mean value of 52 when the true mean is 50 is more unusual for the larger sample sizes.

8-85 μ=50 , σ 2=5

a) For n=16 findP( s2≥7 . 44 ) or P( s2≤2 .56 )

P( S2≥7 . 44 )=P( χ152 ≥

15(7 . 44 )52 )=0. 05≤P ( χ15

2 ≥22 . 32)≤0. 10

Using MinitabP( S2≥7 . 44 )=0.0997

P( S2≤2 .56 )=P( χ152 ≤

15(2 .56 )5 )=0 . 05≤P ( χ15

2 ≤7 . 68)≤0.10

Using MinitabP( S2≤2 .56 )=0.064

8-30

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th edition September 22, 2010

b) For n=30 findP( S2≥7 . 44 ) or P( S2≤2 .56 )

P( S2≥7 . 44 )=P( χ292 ≥

29(7 . 44 )5 )=0. 025≤P ( χ29

2 ≥43 .15)≤0. 05

Using MinitabP( S2≥7 . 44 )= 0.044

P( S2≤2 .56 )=P( χ292 ≤

29(2 . 56)5 )=0. 01≤P ( χ29

2 ≤14 . 85)≤0 . 025

Using MinitabP( S2≤2 .56 )= 0.014.

c) For n=71 P( s2≥7 . 44 ) or P( s2≤2 .56 )

P( S2≥7 . 44 )=P( χ702 ≥

70(7 . 44 )5 )=0. 005≤P ( χ70

2 ≥104 .16)≤0 .01

Using MinitabP( S2≥7 . 44 )=0.0051

P( S2≤2 .56 )=P( χ702 ≤

70(2 .56 )5 )=P ( χ70

2 ≤35 .84 )≤0 .005

Using MinitabP( S2≤2 .56 )< 0.001

d) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much larger than the population variance will decrease.

e) The probabilities get smaller as n increases. As n increases, the sample variance should approach the population variance; therefore, the likelihood of obtaining a sample variance much smaller than the population variance will decrease.

8-86 a) The data appear to follow a normal distribution based on the normal probability plot since the data fall along a straight line.

b) It is important to check for normality of the distribution underlying the sample data since the confidence intervals to be constructed should have the assumption of normality for the results to be reliable (especially since the sample size is less than 30 and the central limit theorem does not apply).

c) No, with 95% confidence, we can not infer that the true mean could be 14.05 since this value is not contained within the given 95% confidence interval.

d) As with part b, to construct a confidence interval on the variance, the normality assumption must hold for the results to be reliable.

e) Yes, it is reasonable to infer that the variance could be 0.35 since the 95% confidence interval on the variance contains this value.

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f) i) & ii) No, doctors and children would represent two completely different populations not represented by the population of Canadian Olympic hockey players. Because neither doctors nor children were the target of this study or part of the sample taken, the results should not be extended to these groups.

8-87 a) The probability plot shows that the data appear to be normally distributed. Therefore, there is no evidence conclude that the comprehensive strength data are normally distributed.

b) 99% lower confidence bound on the mean x⃗= 25 .12, s=8. 42, n=9 t0.01 ,8=2.896

x̄−t0.01 ,8(s√n )≤μ

25 .12−2. 896 (8 . 42

√9 )≤μ

16 . 99≤μThe lower bound on the 99% confidence interval shows that the mean comprehensive strength is most likely be greater than 16.99 Megapascals.

c) 98% two-sided confidence interval on the mean x⃗= 25 .12, s=8. 42, n=9 t0.01 ,8=2.896

x̄−t0.01 ,8 (s√n )≤μ≤ x̄+t0 . 01,8 (s√n )25 .12−2. 896 (8 . 42

√9 )≤μ≤25. 12+2 .896 (8 . 42

√9 )16 . 99≤μ≤33 .25

The bounds on the 98% two-sided confidence interval shows that the mean comprehensive strength will most likely be greater than 16.99 Megapascals and less than 33.25 Megapascals. The lower bound of the 99% one sided CI is the same as the lower bound of the 98% two-sided CI (this is because of the value of )

d) 99% one-sided upper bound on the confidence interval on 2 comprehensive strength

s=8 . 42, s2=70 . 90 χ0 . 99 ,82 =1. 65

σ 2≤8 (8 . 42)2

1 . 65σ 2≤343 . 74

The upper bound on the 99% confidence interval on the variance shows that the variance of the comprehensive strength is most likely less than 343.74 Megapascals2.

e) 98% two-sided confidence interval on 2 of comprehensive strength

s=8 . 42, s2=70 . 90 χ0 . 01 ,92 =20. 09 χ0 . 99 ,8

2 =1. 65

8( 8. 42 )2

20 . 09≤σ2≤

8(8 . 42 )2

1 .6528 . 23≤σ 2≤343 . 74

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th edition September 22, 2010

The bounds on the 98% two-sided confidence-interval on the variance shows that the variance of the comprehensive strength is most likely less than 343.74 Megapascals2 and greater than 28.23 Megapascals2.

The upper bound of the 99% one-sided CI is the same as the upper bound of the 98% two-sided CI because value of for the one-sided example is one-half the value for the two-sided example.

f) 98% two-sided confidence interval on the mean x⃗= 23, s=6 . 31, n=9 t0.01 ,8=2.896

x̄−t0.01 ,8(s√n )≤μ≤ x̄+t0 .01,8(s√n )23−2. 896(6 .31

√9 )≤μ≤23+2 . 896(6 .31

√9 )16 . 91≤μ≤29.09

98% two-sided confidence interval on 2 comprehensive strength

s=6 .31 , s2=39. 8 χ0 . 01 ,9

2 =20. 09 χ0 . 99 ,8

2 =1. 658(39 .8 )20 . 09

≤σ 2≤8 (39. 8 )1 . 65

15 . 85≤σ 2≤192 . 97Fixing the mistake decreased the values of the sample mean and the sample standard deviation. Because the sample standard deviation was decreased the widths of the confidence intervals were also decreased.

g) The exercise provides s = 8.41 (instead of the sample variance). A 98% two-sided confidence interval

on the mean x⃗= 25, s=8 . 41 , n=9 t0.01 ,8=2.896

x̄−t0.01 ,8 (s√n )≤μ≤ x̄+t0 . 01,8 (s√n )25−2. 896(8 . 41

√9 )≤μ≤25+2. 896(8 . 41

√9 )16 . 88≤μ≤33 .12

98% two-sided confidence interval on 2 of comprehensive strength

s=8 . 41, s2=70 .73 χ0 . 01 ,9

2 =20. 09 χ0 . 99 ,8

2 =1. 65

8( 8. 41 )2

20 . 09≤σ2≤

8(8 . 41 )2

1 .6528 .16≤σ2≤342. 94

Fixing the mistake did not affect the sample mean or the sample standard deviation. They are very close to the original values. The widths of the confidence intervals are also very similar.

h) When a mistaken value is near the sample mean, the mistake will not affect the sample mean, standard deviation or confidence intervals greatly. However, when the mistake is not near the sample mean, the value can greatly affect the sample mean, standard deviation and confidence intervals. The farther from the mean, the greater is the effect.

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

8-88With = 8, the 95% confidence interval on the mean has length of at most 5; the error is then E = 2.5.

a)

n=( z0. 025

2. 5 )2

82=( 1 .962 .5 )

2

64=39 .34 = 40

b)

n=( z0. 025

2.5 )2

62=( 1 .962 .5 )

2

36=22.13 = 23

As the standard deviation decreases, with all other values held constant, the sample size necessary to maintain the acceptable level of confidence and the length of the interval, decreases.

8-89 x̄=15 . 33 s=0 .62 n=20k=2 .564a) 95% Tolerance Interval of hemoglobin values with 90% confidence

x̄−ks , x̄+ks15 .33−2 .564 (0.62 ) , 15 .33+2 . 564 (0 .62 )

(13 .74 , 16 . 92)

b) 99% Tolerance Interval of hemoglobin values with 90% confidencek=3 .368

x̄−ks , x̄+ks15 .33−3 .368 (0 .62 ) , 15.33+3 . 368 (0.62 )

(13 .24 , 17 .42 )

8-90 95% prediction interval for the next sample of concrete that will be tested.given = 25.12 s = 8.42 n = 9 for = 0.05 and n = 9, t/2,n-1 = t0.025,8 = 2.306

x̄−t0.025 ,8s √1+1n

≤xn+1≤ x̄+t0 .025 ,8s √1+1n

25 .12−2. 306 (8. 42 )√1+19

≤xn+1≤25.12+2 .306 (8. 42)√1+19

4 .65≤xn+1≤45 .59

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8-91 a) There is no evidence to reject the assumption that the data are normally distributed.

b) 95% confidence interval on the mean x⃗= 203 .20, s=7. 5 , n=10 t0.025 ,9=2 . 262

x̄−t0.025 ,9(s√n )≤μ≤ x̄−t0 .025 ,9(s√n )203 . 2−2 .262(7 . 50

√10 )≤μ≤203 .2+2. 262(7 .50

√10 )197 . 84≤μ≤208 . 56

c) 95% prediction interval on a future sample

x̄−t0.025 ,9s √1+1n

≤μ≤ x̄−t0. 025 ,9 s√1+1n

203 . 2−2 .262(7 . 50)√1+110

≤μ≤203 .2+2 . 262(7 .50 )√1+110

185 . 41≤μ≤220 .99d) 95% tolerance interval on foam height with 99% confidencek=4 .265

x̄−ks , x̄+ks203 . 2−4 .265 (7 .5 ) , 203.2+4 .265 (7 . 5 )

(171 .21 , 235 . 19)e) The 95% CI on the population mean is the narrowest interval. For the CI, 95% of such intervals contain the population mean. For the prediction interval, 95% of such intervals will cover a future data value. This interval is quite a bit wider than the CI on the mean. The tolerance interval is the widest interval of all. For the tolerance interval, 99% of such intervals will include 95% of the true distribution of foam height.

8-92a) Normal probability plot for the coefficient of restitution.

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th edition September 22, 2010

There is no evidence to reject the assumption that the data are normally distributed.

b) 99% CI on the true mean coefficient of restitution x = 0.624, s = 0.013, n = 40 ta/2, n-1 = t0.005, 39 = 2.7079

x̄−t0.005 ,39s

√n≤μ≤ x̄+t0. 005 ,39

s

√n0 .624−2. 7079

0 .013

√40≤μ≤0 .624+2 .7079

0 . 013

√400 .618≤μ≤0 .630

c) 99% prediction interval on the coefficient of restitution for the next baseball that will be tested.

x̄−t0.005 ,39 s √1+1n≤xn+1≤ x̄+ t0 .005 ,39 s√1+1

n

0 .624−2.7079(0 .013)√1+140

≤xn+1≤0 .624+2.7079(0 .013)√1+140

0 .588≤xn+1≤0 .660d) 99% tolerance interval on the coefficient of restitution with a 95% level of confidence

( x̄−ks , x̄+ks )(0 .624−3. 213(0 . 013 ), 0 .624+3 .213(0 .013))

(0 .582 , 0 .666 )

e) The confidence interval in part (b) is for the population mean and we may interpret this to imply that 99% of such intervals will cover the true population mean. For the prediction interval, 99% of such intervals will cover a future baseball’s coefficient of restitution. For the tolerance interval, 95% of such intervals will cover 99% of the true distribution.

8-93 95% Confidence Interval on the proportion of baseballs with a coefficient of restitution that exceeds 0.635.

p̂= 840

=0 .2 n=40 zα=1. 65

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

p̂−zα √ p̂(1− p̂ )n

≤p

0 .2−1 .65√0 . 2(0. 8 )40

≤p

0 .0956≤p

8-94 a) The normal probability shows that the data are mostly follow the straight line, however, there are some points that deviate from the line near the middle. It is probably safe to assume that the data are normal.

b) 95% CI on the mean dissolved oxygen concentrationx = 3.265, s = 2.127, n = 20 ta/2, n-1 = t0.025, 19 = 2.093

x̄−t0.025 ,19s

√n≤μ≤ x̄+t0. 025 ,19

s

√n3 .265−2 . 093

2 .127

√20≤μ≤3 .265+2. 093

2 .127

√202 .270≤μ≤4 . 260

c) 95% prediction interval on the oxygen concentration for the next stream in the system that will be tested..

x̄−t0.025 ,19 s √1+1n≤xn+1≤ x̄+t0 .025 ,19 s√1+1

n

3 .265−2 .093 (2.127 )√1+120

≤xn+1≤3 .265+2 .093 (2.127 )√1+120

−1 .297≤xn+1≤7 .827

d) 95% tolerance interval on the values of the dissolved oxygen concentration with a 99% level of confidence

8-37

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

( x̄−ks , x̄+ks )(3 .265−3 .168 (2.127 ), 3 .265+3 .168 (2 .127))

(−3 .473 , 10 .003)

e) The confidence interval in part (b) is for the population mean and we may interpret this to imply that 95% of such intervals will cover the true population mean. For the prediction interval, 95% of such intervals will cover a future oxygen concentration. For the tolerance interval, 99% of such intervals will cover 95% of the true distribution

8-95 a) There is no evidence to support that the data are not normally distributed. The data points appear to fall along the normal probability line.

b) 99% CI on the mean tar contentx = 1.529, s = 0.0566, n = 30 ta/2, n-1 = t0.005, 29 = 2.756

x̄−t0.005 ,29s

√n≤μ≤ x̄+t0. 005 ,29

s

√n1 .529−2. 756

0 .0566

√30≤μ≤1 .529+2 . 756

0 .0566

√301 .501≤μ≤1 .557

c) 99% prediction interval on the tar content for the next sample that will be tested..

x̄−t0.005 ,19 s √1+1n≤xn+1≤ x̄+t0 .005 ,19 s√1+1

n

1 .529−2.756(0 .0566 )√1+130

≤xn+1≤1 .529+2.756(0 .0566 )√1+130

1 .370≤xn+1≤1 .688

d) 99% tolerance interval on the values of the tar content with a 95% level of confidence

8-38

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

( x̄−ks , x̄+ks )(1.529−3.350(0 .0566 ) , 1 .529+3.350(0 .0566 ))

(1 .339 , 1 .719 )

e) The confidence interval in part (b) is for the population mean and we may interpret this to imply that 95% of such intervals will cover the true population mean. For the prediction interval, 95% of such intervals will cover a future observed tar content. For the tolerance interval, 99% of such intervals will cover 95% of the true distribution

8-96 a) 95% Confidence Interval on the population proportion

n=1200 x=8 p̂=0. 0067 z/2=z0.025=1.96

p̂−za /2√ p̂(1− p̂ )n

≤p≤ p̂+za /2 √ p̂(1− p̂ )n

0 .0067−1 .96√0 .0067 (1−0 . 0067 )1200

≤p≤0 .0067+1. 96√0 . 0067(1−0 .0067 )1200

0 .0021≤p≤0 .0113b) No, there is not sufficient evidence to support the claim that the fraction of defective units produced is one percent or less at = 0.05. This is because the upper limit of the control limit is greater than 0.01.

8-97 99% Confidence Interval on the population proportion

n=1600 x=8 p̂=0. 005 z/2=z0.005=2.58

p̂−za /2√ p̂(1− p̂ )n

≤p≤ p̂+za /2 √ p̂(1− p̂ )n

0 .005−2 . 58√0 .005(1−0. 005 )1600

≤p≤0 .005+2. 58√0.005 (1−0 . 005)1600

0 .0004505≤p≤0 .009549b) E = 0.008, = 0.01, z/2 = z0.005 = 2.58

n=( zα /2

E )2

p (1−p)=( 2 .580 .008 )

2

0 .005(1−0.005 )=517 .43, n 518

c) E = 0.008, = 0.01, z/2 = z0.005 = 2.58

n=( zα /2

E )2

p (1−p)=( 2 .580 .008 )

2

0 . 5(1−0 .5)=26001. 56, n 26002

d) A bound on the true population proportion reduces the required sample size by a substantial amount. A sample size of 518 is much more reasonable than a sample size of over 26,000.

8-98 p̂=117

484=0 . 242

a) 90% confidence interval; zα /2=1.645

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .210≤p≤0 .274With 90% confidence, the true proportion of new engineering graduates who were planning to continue studying for an advanced degree is between 0.210 and 0.274.

b) 95% confidence interval;

p̂−zα /2 √ p̂(1− p̂ )n

≤p≤ p̂+zα /2√ p̂ (1− p̂ )n

0 .204≤p≤0 . 280

With 95% confidence, we believe the true proportion of new engineering graduates who were planning to continue studying for an advanced degree lies between 0.204 and 0.280.

c) Comparison of parts (a) and (b): The 95% confidence interval is larger than the 90% confidence interval. Higher confidence always yields larger intervals, all other values held constant.

d) Yes, since both intervals contain the value 0.25, thus there in not enough evidence to determine that the true proportion is not actually 0.25.

8-99 a) The data appear to follow a normal distribution based on the normal probability plot since the data fall along a straight line.

b) It is important to check for normality of the distribution underlying the sample data since the confidence intervals to be constructed should have the assumption of normality for the results to be reliable (especially since the sample size is less than 30 and the central limit theorem does not apply).

c) 95% confidence interval for the mean

n=11 x̄=22. 73 s=6 . 33 t0.025 ,10=2.228

x̄−t0.025 ,10(s√n )≤μ≤ x̄+t0.025 ,10(s√n )22 .73−2. 228(6 .33

√11 )≤μ≤22 .73+2. 228(6 .33

√11 )18 . 478≤μ≤26 . 982

d) As with part b, to construct a confidence interval on the variance, the normality assumption must hold for the results to be reliable.

e) 95% confidence interval for variance

n=11 s=6 . 33

χ α /2, n−12 = χ0 .025 ,10

2 =20 . 48 and χ1−α /2, n−12 = χ 0. 975 ,10

2 =3 . 25

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th edition September 22, 2010

10(6 .33)2

20 . 48≤σ2≤

10(6 .33 )2

3. 2519 .565≤σ 2≤123 .289

8-100 a) The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the energy intake is normally distributed.

Energy

Perc

ent

87654

99

95

90

80

70

60504030

20

10

5

1

Mean

0.011

5.884StDev 0.5645N 10AD 0.928P-Value

Probabi l i ty P lot of EnergyNormal - 95% CI

b) 99% upper confidence interval on mean energy (BMR)

n=10 x̄=5. 884 s=0 .5645 t0.005 ,9=3 . 250

x̄−t0.005 ,9(s√n )≤μ≤ x̄+t0.005 ,9(s√n )5 .884−3. 250(0. 5645

√10 )≤μ≤5 .884+3 . 250(0 . 5645

√10 )5 .304≤μ≤6 . 464

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Page 42: Applied Statistics and Probability Ch8 solutions

Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

Mind Expanding Exercises

8-101 a.) P( χ

1− α2,2 r

2 <2λT r< χ α2,2 r

2 )=1−α

=P( χ 1−α2,2 r

2

2T r

< λ<χ α

2,2 r

2

2T r)

Then a confidence interval for μ=1

λ is ( 2T r

χ α2,2 r

2 ,2T r

χ1−

α2,2 r

2 )b) n = 20 , r = 10 , and the observed value of Tr is 199 + 10(29) = 489.

A 95% confidence interval for

1λ is

( 2(489 )34 . 17

,2(489 )9 . 59 )=(28. 62 ,101 . 98 )

8-102

α 1=∫zα 1

∞ 1√2π

e− x2

2 dx=1−∫−∞

zα1

1√2 π

e− x2

2 dx

Therefore, 1−α1=Φ( zα 1

).

To minimize L we need to minimize Φ−1(1−α1 )+Φ(1−α 2) subject to

α 1+α 2=α . Therefore, we

need to minimize Φ−1(1−α1 )+Φ(1−α+α1 ) .

∂∂ α1

Φ−1(1−α 1)=−√2 π e

zα 1

2

2

∂∂ α1

Φ−1(1−α+α1 )=√2π e

zα−α1

2

2

Upon setting the sum of the two derivatives equal to zero, we obtain e

zα−α 1

2

2 =ezα 1

2

2. This is solved by

. Consequently, α 1=α−α1 ,2α 1=α

and α 1=α2=

α2 .

8.103 a) n=1/2+(1.9/.1)(9.4877/4), then n=46

b) (10-.5)/(9.4877/4)=(1+p)/(1-p) p=0.6004 between 10.19 and 10.41.

8-104 a)

P(X i≤~μ )=1/2P(allX i≤

~μ )=(1/2)n

P(allX i≥~μ )=(1/2)n

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Applied Statistics and Probability for Engineers, 5

th edition September 22, 2010

P( A∪B )=P( A )+P(B)−P (A∩B )

=(12 )n

+(12 )n

=2(12 )n

=(12 )

n−1

1−P( A∪B)=P(min (X i )<~μ<max(X i ))=1−( 1

2 )n

b) P(min (X i )<~μ<max( X i))=1−α The confidence interval is min(Xi), max(Xi)

8-105 We would expect that 950 of the confidence intervals would include the value of . This is due to 9the definition of a confidence interval.Let X bet the number of intervals that contain the true mean (). We can use the large sample approximation to determine the probability that P(930 < X < 970).

Letp=950

1000=0 .950

p1=

9301000

=0. 930 and

p2=9701000

=0 .970

The variance is estimated by

p (1−p)n

=0 .950(0 .050)1000

P(0 . 930< p<0 . 970 )=P(Z<(0 . 970−0 .950 )

√0 .950 (0 .050 )1000

)−P(Z<( 0. 930−0 . 950)

√0 . 950(0 . 050)1000

)=P(Z<0 . 02

0 . 006892 )−P(Z<−0 . 020 .006892 )=P(Z<2 . 90)−P(Z<−2 . 90 )=0 . 9963

8-43