Question

A production process produces 2.5% defective parts. A sample of five parts from the production process is selected. What is the probability that the sample contains exactly two defective parts?

Answer 1

Using the binomial distribution, it is found that there is a 0.0058 = 0.58% probability that the sample contains exactly two defective parts.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem, we want P(X = 2) with p = 0.025, n = 5, hence:

$P(X = 2) = C_{5,2}.(0.025)^{2}.(0.975)^{3} = 0.0058$

There is a 0.0058 = 0.58% probability that the sample contains exactly two defective parts.

More can be learned about the binomial distribution at https://brainly.com/question/24863377