Question

Suppose that over the long run a manufacturing process produces 1% defective items. Use the Poisson approximation to the binomial distribution to calculate the probability of getting two or more defectives in a sample of 200 items.

Answer 1

Answer:

The probability of getting two or more defectives in a sample of 200 items

0.3233

Step-by-step explanation:

step1:-

The probability of getting 1 percentage defective items is

P(D) = $\frac{1}{100} =0.01$

Given total number of items is n=200

use the Poisson approximation to the binomial distribution that is

λ=np=200(0.01)=2

by using Poisson distribution P(X=r)=$e^-λ (λ^ r)$

Step2:-

The probability of getting two or more defectives items

$P(x\geq 2)=1-{P(x\leq 2)$

$P(x\geq 2)=1-{P(x=0)+P(x=1)+P(x=2)]$

using Poisson distribution

$P(x\geq 2)=1-[{e^{-2} \frac{2^{0} }{0!} } +e^{-2} \frac{2^{1} }{1!} +e^{-2} \frac{2^{2} }{2!} ]$

on simplification, we get

$P(x\geq 2)=1-({e^{-2})(1+2+2)$

$P(x\geq 2)=1-({e^{-2})5$

by using calculator e^-2 value is 0.13533

$P(x\geq 2)=1-({0.13533)X5$

after simplification

$P(x\geq 2)=1-0.676676$

$P(x\geq 2)= 0.3233$

Final answer:-

The probability of getting two or more defectives in a sample of 200 items

0.3233