Question

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A shipment is acceptable if at least 440 of the 500 bearings meet the specification. Assume that each shipment contains a random sample of bearings.

What is the probability that a given shipment is acceptable?

Answer 1

What I've tried so far is a normal approximation to the binomial X ~ Bin(500 , 0.9)

Which goes to x ~ N(450 , 6.71^2)

Then I found P(x>440) = 1 - p(x<440) = 1 - Phi(-10/6.71) = Phi(1.49)
Which from the z-table is .9319

According to the back of the book, the answer is supposed to be 0.9418. What did I do wrong?

Answer 2

In this question the continuous Normal distribution is being used to approximate a Discrete distribution. Therefore the Correction for Continuity must be applied, giving:
\[\large P(X \ge 440)=1-\phi(\frac{-10.5}{6.71})=you\ can\ calculate\]