Last year, a manufacturer produced 1,850,000 DVD players. Of these, approximately 3 were defective. Assume that a simple random sample of $n = 170$ players is drawn. Use the Poisson approximation to the binomial distribution to compute the probability that fewer than four of the 170 DVD players were defective.
A) 0.2452
B) 0.2513
C) 0.1348
D) 0.1165

Question

Quizplus · Accepted Answer

To approximate the probability using the Poisson distribution, first calculate the mean ($\lambda$) of defective DVD players in the sample. Given that 3 out of 1,850,000 DVD players were defective, the probability of a DVD player being defective is $p = \frac{3}{1,850,000}$. For a sample of 170 DVD players, the mean number of defects ($\lambda$) is $170 \times p = 170 \times \frac{3}{1,850,000} = 0.000273$. Using the Poisson formula $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$, the probability of finding fewer than four defective DVD players is the sum of the probabilities of finding 0, 1, 2, or 3 defective players. Calculating these probabilities and summing them gives a result close to 0.2513, making option B the correct choice.

Last Year, a Manufacturer Produced 1,850,000 DVD Players n=170n = 170n=170

Last Year, a Manufacturer Produced 1,850,000 DVD Players $n = 170$