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.

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 bearing must meet the specification in order than 99% of the shipments are acceptable?

Step 1 of 4</p>

Here given 90% of the bearings meat a thickness specification

The p=0.9

A shipment contains 500 bearings

Let X represents no. of acceptable bearings

Then XB(500,0.9)

Here X is approximately following the normal distribution

Mean(=np

=500(0.9)

=450

Variance =npq

=500(0.9)(0.1)

=45

Hence =

=6.71

Step 2 of 4</p>

a) Here we have to find the probability of the given shipment is acceptable

So we have to find P(X440)

Use the continuity correction find the Z value for 439.5

Then Z=(

=(439.5-450)/6.71

= -1.56

Now find P(Z1.49) value from the standard normal tables

P(Z-1.56)=1-P(Z1.56)

=1-0.0606

=0.9394

Hence the probability of the given shipment is acceptable is 0.0681