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

Step-by-step 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  = $\lambda=np=25\times0.01=0.25$

Poisson distribution formula : $P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$

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

=Probability that package has more than 1 defectives

=$P(x>1)=1-(P(x\leq1))\\\\=1-[P(x=0)+P(x=1)]\\\\=1-[e^{-0.25}\cdot\frac{0.25^0}{0!}+e^{-0.25}\cdot\frac{0.25^1}{1!}]\\\\=1-0.9736=0.0264$

Hence, the required probability is 0.0264