Question

2. Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be re- worked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the approximate) probability that X is a) At most 30? [2] b) Between 15 and 25 (both inclusive)? [2] c) Assume that the probability of at most x shafts being nonconforming is 0.9906. Find x. [3]

Answer 1

10 X be norcan förming shafts in the sample of 200. paobabiliyy of nonconforming shaft chaft = 10% & p. be the de p = 0.10 Then X followe Binomial diataibution with n = 200 f p = Oilo Then F(x) = l = nap = 200 X 0, L = 20 n(x) = 6 = np (-p) = 200 x 0ilxong = = 18. Als comple size Rubat 200 ir laage, then diatribution of X'ju approximate to Normal. XN Normal (M = 20 5*= 18). f We know 2= X-l A Normal (0,0) 1 Standao Normal Pashability that X is at most 30 PIX & 30 pl X-u < 30-20 o 4.2426 Pat236707 Then from nammal pagbability calculador ox P[X30 P12 ( PLX 430 0.991 we 09:3570

Paobability 115 x x 25 PLX 15 = PI X ( 25 PLX-1 < 25-20 4.2426 + 15-20 4.2426 P(x-4 Eple 41.1985] -P[24 - 1985) Then from nowmal probability calculatoro. 0.881 0,762 0.119 o We have to find the point a such that, PIX<a 0.9906 Then from normal paobability calculator, we get. X = 29.968 je P L x < 29.968 1 = 0.9906 Since 2 should be integer. Lapproximately H = 30