Question

Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the probability that at most 10 of the next 150 items produced are unacceptable.

Answer 1

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Explanation:

Let X = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = p = 0.05.

The sample of items selected is of size, n = 150.

The random variable X follows a Binomial distribution with parameters n = 150 and p = 0.05.

According to the Central limit theorem, if a sample of large size (n > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is:
`\mu_(\hat p)= p=0.05`

The standard deviation of this sampling distribution is:
`\sigma_(\hat p)=\sqrt{( p(1-p))/(n)}=\sqrt{(0.05(1-.0.05))/(150) }=0.0178`

If 10 of the 150 items produced are unacceptable then the probability of this event is:

`\hat p=(10)/(150)=0.067`

Compute the value of
`P(\hat p\leq 0.067)` as follows:

`P(\hat p\leq 0.067)=P((\hat p-\mu_(p))/(\sigma_(p)) \leq(0.067-0.05)/(0.0178))=P(Z\leq 0.96)=0.8315`

*Use a z-table for the probability.

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.