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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?

Sagot :

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:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

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:

[tex]P(X = 2) = C_{5,2}.(0.025)^{2}.(0.975)^{3} = 0.0058[/tex]

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

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