Suppose that 10% of all steel shafts produced by a certain process are nonconfor

Question

Suppose that 10% of all steel shafts produced by a certain process are nonconforming, but can be reworked (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 using the normal distribution. Using the normal approximation to the binomial, what is the (approximate) probability that X is at most 20? Using the normal approximation to the binomial, what is the (approximate) probability that X is less than 20? Using the normal approximation to the binomial, what is the (approximate) probability that X is between 10 and 20 inclusive?

Explanation / Answer

here p = 0.1, n = 200 and q = 1- p = 0.9
As per normal approximation to binomial distribution
mean = np = 20
std. dev. = sqrt(npq) = sqrt(200*0.1*0.9) = 4.2426

(A)
P(X<=20) = P(z<(20-20)/4.2426) = 0.5
0.5 is the probability that X is at most 20.

(B)
P(X<20) = P(X<=19) = P(z<(19-20)/4.2426) = 0.2357
0.2357 is the probability that X is less than 20.

(C)
P(10<=X<=20) = P(X<=20) - P(X<10)
= 0.5 - P(z<-2.59) = 0.5 - 0.0048 = 0.4952

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