The rate of defects among CD players of a certain brand is 1.1%.Use the Poisson approximation to the binomial distribution to find the probability that among 270 such CD players received by a store,there is at most one defective.
A)0.1524
B)0.8476
C)0.2037
D)0.9487
E)0.7963

Question

Quizplus · Accepted Answer

We are given that the rate of defects is 1.1%, which means the probability of a CD player being defective is 0.011. Let X be the number of defective CD players among 270. We want to find P(X ≤ 1).

Since the probability of a CD player being defective is small (0.011), we can approximate the binomial distribution with a Poisson distribution. The Poisson parameter λ is equal to the mean of the binomial distribution, which is n times the probability of success (defect) in one trial. Therefore, λ = 270 × 0.011 = 2.97.

Now we can use the Poisson distribution to find P(X ≤ 1):

P(X ≤ 1) = P(X = 0) + P(X = 1)
          = e^(-λ) (λ^0/0!) + e^(-λ) (λ^1/1!)
          = e^(-2.97) (1 + 2.97)
          = 0.2037

Therefore, the probability that among 270 CD players there is at most one defective is 0.2037, which is closest to choice C.

The Rate of Defects Among CD Players of a Certain