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Unformatted text preview: probability of passing the car is 95.02%. Therefore, agency claim should not be believed. c) Probability that exactly one patrol car will pass through the neighborhood on each of four consecutive nights = P (one patrol car, 4 consecutive nights)^independence = (P(one patrol car per night))^4 = ((3^1)*(e^3))/1! = .1494^4 = .000498 = .0005 5.38 Solution Mean = 200*.5% = 1 a) Probability of four or more car batteries in a random sample of 200 to be defective = P(X>=4) = 1P(X<4) = 1 (P(0)+P(1)+P(2)+P(3)) = 1 (.3679+.3679+.1839+.0613) = 1.9810 = .0190 b) No 5.42 Solution N =10, r =4 and n =3 a) Probability distribution of X(success) = (4CX*((104)C(3X)))/10C3 b) Mean = 3*(4/10) = 3*.4 = 1.2 Variance = 3*(4/10)*(1(4/10))*((103)/(101)) = 1.2*.6*.7777 = .5600 5.44 Solution N =10, r =9 and n =3 P(X=3) =( 9C3*(1C0))/10C3 = (84*1)/120 = .7000...
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This note was uploaded on 01/16/2012 for the course MBA 101 taught by Professor Wormer during the Spring '08 term at Indian Institute Of Management, Ahmedabad.
 Spring '08
 WORMER

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