Question

A quality control inspector has drawn a sample of 1414 light bulbs from a recent production lot. If the number of defective bulbs is 22 or more, the lot fails inspection. Suppose 20%20% of the bulbs in the lot are defective. What is the probability that the lot will fail inspection? Round your answer to four decimal places.

Answer 1

Answer: 0.8021

Explanation:

The given problem is a binomial distribution problem, where

`n=14,\ p=0.2, q=1-0.2=0.8`

The formula of binomial distribution is :-

`P(X=r)=^(n)C_(r)p^(r)q^(n-r)`

The probability that the lot will fail inspection is given by :_

`P(X\geq2)=1-(P(X\leq1))\\\\=1-(P(0)+P(1))\\\\`

`=1-(^(14)C_(0)(0.2)^(0)(0.8)^(14-0)+^(14)C_(1)(0.2)^(1)(0.8)^(14-1))\\\\=1-((1)(0.8)^(14)+(14)(0.2)(0.8)^(13))\\\\=0.802087907\approx0.8021`

Hence, the required probability = 0.4365