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.

Question

a. What is the probability that a given shipment is acceptable?
b. What is the probability that more than 285 out of 300 shipments are acceptable?
c. What proportion of bearings must meet the specification in order that 99% of the shipments are acceptable?

Answer 1

(a) Let Xrepresent the number of non-defective bearings in a shipment

Then X ~Bin (500, 0.90), so X is approximately normal with mean and standard deviation .

To find (X? 440), use the continuity correction and find the z-score of 439.5.

The z-score of 439.5 is (439.5 ? 450)?6.7802 = ?1.57.

The area to the right of z= ?1.57 is 1 ? 0.0582 = 0.9418.

(X? 440) = 0.9418.

(b) LetY represent the number of shipments out of 300 that are acceptable.

From part (a) the probability that a shipment is acceptable is 0.9418, so Y~ Bin(300, 0.9418).

It follows that Y is approximately normal with mean and standard deviation

To find P(Y>285), use the continuity correction and find the z-score of 285.5.

The z-score of 285.5 is (285.5 ? 282.54)?4.0551 = 0.73.

The area to the right of z= 0.73 is 1 ? 0.7673 = 0.2327.

(X>285) = 0.2327.

(c) Let p be the required proportion of defective bearings, and let X represent the number of defective bearings in a shipment.

Then X~ Bin (500, p), so Xis approximately normal with mean and standard deviation .

The probability that a shipment is acceptable is (X? 440) = 0.99.

Using the continuity correction, 439.5 is the 1st percentile of the distribution of X.

The z-score of the 1st percentile is approximately z= ?2.33.

The z-score can be expressed in terms of p by

This equation can be rewritten as 252, 714.45p2 ? 442, 214.45p+ 193160.25 = 0.

Solving for p yields p= 0.909. (0.841 is a spurious root.)