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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 71e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 71e

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# A geologist has collected 10 specimens of basaltic rock ISBN: 9780321629111 32

## Solution for problem 71E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 71E

A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a.? ?What is the pmf of the number of granite specimens selected for analysis? b. ?What is the probability that all specimens of one of the two types of rock are selected for analysis? c. ?What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

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Answer: Step1: Given, a geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. Here, population size N = 20 Sample size n = 15. The number of granite specimens is M = 10. The number of basaltic specimens is N - M = 10. Step2: a). To find the pmf of the number of granite specimens selected for analysis. Let x denote the number of granite specimens. The possible values of x are 5,6,7,8,9,10. P(X = x) = h(x;n,M,N) M NM (x )( n x = ( ) n ( )( 10 ) P(X = x) = X 15 x , x = 5,6,7,8,9,10. ( 15 Then, 10 10 (X = 5) = h(5;15,10,20) = (5)( 15 5 (15 252 ×1 = 15504 P(X = 5) = 0.0163 10 10 ( 6( 15 6 Similarly, P(X = 6) = h(6;15,10,20) = 20 ( 15 210×10 = 15504 = 0.1354 10 10 ( 7( 15 7 P(X = 7) = h(7;15,10,20) = ( )0 15 = 120×45 15504 = 0.3483 10 10 P(X = 8) = h(8;15,10,20) = (8)( 15 8 (15 45×120 = 15504 = 0.3483 10 10 (9)( 15 9 P(X = 9) = h(9;15,10,20) = ( ) 15 = 10×210 15504 = 0.1354 ( 10 15 10 P(X = 10) = h(10;15,10,20) = 20 (15 = 1×252 15504 = 0.0163 The pmf of the number of granite specimens are tabulated below. x 5 6 7 8 9 10 p(x) 0.0163 0.1354 0.3483 0.3483 0.1354 0.0163 Step3: b). To find the probability that all specimens of one of the two types of rock are selected for analysis. Therefore, P(all specimens of one of the two types) = P(x = 5) + P(x = 10) = 0.0163 + 0.0163 = 0.0326. The probability that all specimens of one of the two types is 0.0326. Step4: c). To find the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value. M Mean E(x) = n.( N ) = 15.( 10) 20 = 7.5 var(x) = = nM(N2)(Nm) N (N1) = 15(10)22015)(2010) 20 (201) = 7500 7600 var(x) = = 0.9868 Standard deviation = vr(x) = 0.9934 within 1 standard deviation of its mean value is ± = 7.5±0.9934 (6.5066,8.4934) Then the required probability is P(6.5066 x 8.4934) = P( 7 x 8) = P(x = 7) + P(x = 8) = 0.3483 + 0.3483 P(x = 7) + P(x = 8) = 0.6966 Therefore, the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value is 0.6966.

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##### ISBN: 9780321629111

This full solution covers the following key subjects: specimens, analysis, selected, granite, Probability. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. The answer to “A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a.? ?What is the pmf of the number of granite specimens selected for analysis? b. ?What is the probability that all specimens of one of the two types of rock are selected for analysis? c. ?What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?” is broken down into a number of easy to follow steps, and 86 words. Since the solution to 71E from 3 chapter was answered, more than 1083 students have viewed the full step-by-step answer. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The full step-by-step solution to problem: 71E from chapter: 3 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.

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