STA4032S12Solumidterm

Replication is equivalent to the event that at most 5

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replication is equivalent to the event that at most 5 of the seleced cells in the sample are able to replicate, i.e. { X 5 } . So, we have P ( X 5) = 1 - P ( X = 6) = 1 - 0 . 069 = 0 . 931 . (d) E ( X ) = n K N = 6 · 24 36 = 4 and σ 2 = V ( X ) = 6 · 24 36 1 - 24 36 30 35 = 1 . 1428 . 2. A factory produces its products with three machines. Machine I, II, and III produces 50%, 30%, and 20% of the products, but 4%, 2%, and 4% of their products are defective, respectively. (10) (a) What is the probability that a randomly selected product is defective? (5) (b) If a randomly selected product was found to be defective, what is the probability that this product was produced by machine I? (5) (c) If a randomly selected product was found to be good, what is the probability that this product was produced by machine I? Solution . (a) Let I , II , and III denote the events that the selected product is produced by machine I, II, and III, respectively. Let D be the event that the selected product is defective. Then, P ( I ) = 0 . 5 , P ( II ) = 0 . 3 , P ( III ) = 0 . 2 , P ( D | I ) = 0 . 04 , P ( D | II ) = 0 . 02 , P ( D | III ) = 0 . 04. So, by the total probability rule, we have P ( D ) = P ( D | I ) P ( I ) + P ( D | II ) P ( II ) + P ( D | III ) P ( III ) = (0 . 04)(0 . 50) + (0 . 02)(0 . 30) + (0 . 04)(0 . 20) = 0 . 034 .

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(b) By Bayes’ theorem, we find P ( I | D ) = P ( D | I ) P ( I ) P ( D ) = (0 . 04)(0 . 50) 0 . 034 = 0 . 5882 . (c) By Bayes’ theorem, we find P ( I | D 0 ) = P ( D 0 | I ) P ( I ) P ( D 0 ) = (1 - 0 . 04)(0 . 50) 1 - 0 . 034 = 0 . 4969 . 3. Let the probability mass function of X be given by f ( x ) = 2 x - 1 16 , x = 1 , 2 , 3 , 4 . (10) (a) Find the cumulative distribution function F ( x ) of X .
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