Question

A machine is shut down for repairs if a random sample of 100 items selected from the daily output of the machine reveals at least 15% defectives. Suppose the machine is producing 20% defective items another day, what is the probability that a random sample of 10 items selected from the machine will contain at least two defective items? (Hint use the .5 continuity correction and the binomial distribution directly)

Answer 1

Let X is a random variable shows the number of defective out of 10. Here X has binomial distribution with following parameters n=10 and p=0.20.

Here we cannot use normal distribution because n(1-p) = 10 *0.20 = 2 is less than 5.

The probability that a random sample of 10 items selected from the machine will contain at least two defective items is

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The probability that machine will shut down :

Here X has n=100 and p = 0.20

15% of 100 means 100 * 0.15 = 15

Using continuity correction factor, we need to find the probability

P(X >= 15) = P(X > 15-0.5) = P(X >14.5)