Question

It is known that screws produced by a certain company will b defective with probability .01 independently of each other. the company sells the screws in packages of 25 and offers a money back guarantee that at most 1 of the 25 is defective, using Poisson approximation for binomial distribution the probability that the company must replace a package is approximatelya).01b).1947c).7788d).0264e).2211

Answer 1

Answer: d).0264

Explanation:

Given : It is known that screws produced by a certain company will b defective with probability .01 independently of each other.

The company sells the screws in packages of 25 and offers a money back guarantee that at most 1 of the 25 is defective.

Let x be the binomial variable that represents the defective screws having parameter p= 0.01 and n= 25

For using Poisson approximation for binomial distribution

Mean =
`\lambda=np=25*0.01=0.25`

Poisson distribution formula :
`P(X=x)=(e^(-\lambda)\lambda^x)/(x!)`

Now, the probability that the company must replace a package is

=Probability that package has more than 1 defectives

=
`P(x>1)=1-(P(x\leq1))\\\\=1-[P(x=0)+P(x=1)]\\\\=1-[e^(-0.25)\cdot(0.25^0)/(0!)+e^(-0.25)\cdot(0.25^1)/(1!)]\\\\=1-0.9736=0.0264`

Hence, the required probability is 0.0264