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Unformatted text preview: 22. (a) P(0.900 – 0.005 < X < 0.900 + 0.005) = P(1.67<Z<1.67) =2 (1.67) 1 0.9050 Φ = Hence 9.5 percent will be defective (that is each will be defective with probability 0.0950) (b) P(0.005/6 < Z < 0.005/6) = 0.005 2 ( ) 1 0.99 σ Φ = When 0.005 ( ) 0.995 Φ = => 0.0019 = 23. With C denoting the life of a chip, and Φ the standard normal distribution function, we have (a) P(C<1.8*10^6) = 6 6 5 1.8*10 1.4*10 ( ) (1.33) 0.9082 3*10Φ = Φ = Thus, if N is the number of the chips whose life is less than 1.8*10^6 then N is a binomial random variable with parameters (100, 0.9082). Hence, P(N>19.5) 19.5 90.82 ( ) 1 ( 24.7) 1 90.82*0.0918≈ Φ = Φ ≈ 27. P(X>5799.5) = P(Z > 799.5/sqrt(2500)) = P(Z>15.99) closed to 0...
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This note was uploaded on 04/15/2010 for the course STAT 418 taught by Professor G.jogeshbabu during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 G.JOGESHBABU
 Probability

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