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?

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.