It is known that screws produced by a certain company will be 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 screws is defective.Using Poisson approximation for binomial distribution,the probability that the company must replace a package is approximately _________
A) 0.01
B) 0.1947
C) 0.7788
D) 0.0264
E) 0.2211

Question

Quizplus · Accepted Answer

Let X be the number of defective screws in a package of 25. Then X has a binomial distribution with n = 25 and p = 0.01. We want to find the probability that X is greater than 1, since the company guarantees at most 1 defective screw.

Using Poisson approximation, we can approximate X with a Poisson distribution with mean λ = np = 0.25. Then, the probability that X is greater than 1 is:

P(X > 1) ≈ 1 - P(X = 0) - P(X = 1)
          ≈ 1 - e^(-0.25) - 0.25e^(-0.25)
          ≈ 0.0264

Therefore, the probability that the company must replace a package is approximately 0.0264, which is answer choice D.

It Is Known That Screws Produced by a Certain Company