Question

manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasing a batch of 2000 cameras, applies the following sample plan: They randomly select 20 cameras and accept the whole batch if there are more than 17 working cameras. What is the correct probability statement for the company to accept the whole batch

Answer 1

P(X>17) = 0.979

Explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r)
`nCr q^(r) p^(n-r)`

`nCr = (n!)/((n-r)!r!)`

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) =
`20C18 * 0.97^(18) * 0.03^(20-18)`

P(X=18) =
`20C18 * 0.97^(18) * 0.03^(2)`

P(X=18) = 0.0988

P(X=19) =
`20C19 * 0.97^(19) * 0.03^(20-19)`

P(X=19) =
`20C19 * 0.97^(19) * 0.03^(1)`

P(X=19) = 0.3364

P(X=20) =
`20C20 * 0.97^(20) * 0.03^(20-20)`

P(X=20) =
`20C20 * 0.97^(20) * 0.03^(0)`

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch