Help LuS I I Contact Us I Log Out Montgomery & Runger, Applied Statistics and Pr

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Help LuS I I Contact Us I Log Out Montgomery & Runger, Applied Statistics and Probability for Engin Edition rs, 6th PROBABILITY & STATISTICS (201 HTH) NEXT CES Problem 3.144 A batch contains 35 bacteria cells and 12 of the cells are not capable of cellular replication. Suppose you examine three bacteria cells selected at random, without replacement. Round your answers to three decimal places (e.g. 98.765). (a) What is the probability mass function of the number of cells in the sample, X, that can replicate? f(X=0)- | f(X-1)= f(X=2)= f(X=3)= (b) What is the mean and variance of the number of cells in the sample that can replicate? Mean Variance

Explanation / Answer

Here the given distribution is hypergeometric distribution as we do things without replacement.

N = 35 ; K = 23 ; n = 3 Here K is the number of success in population

P(X) = 23CX12C(3-X)/35C3

(a) P(X = 0) =12C323C0/35C3 = 0.0336

P(X = 1) = 12C223C1/35C3 =  0.2319

P(X = 2) = 12C123C2/35C3 =  0.4639

P(X = 3) = 12C023C3/35C3 =  0.2706

(b) Here mean of the number of cells that can replicate, there are 23 cells which can replicate.

Mean = nK/N = 3 * 23/35 = 1.9714

Variance = nK/N * (N-K)/N * (N-n)/(N-1) = 3 * 23/35 * (35 - 23)/35 * (35 - 3)/(35 -1) = 0.6362

(c) Pr(atleast one out of 3 cells cannot replicate) = 1 - Pr(X = 3) = 1 - 0.2706 = 0.7294

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