Last year, a manufacturer produced 950,000 DVD players. Of these, approximately $6 \%$ were defective. Assume that a simple random sample of $n = 160$ players is drawn. Use the Poisson approximation to the binomial distribution to compute the standard deviation of the number of DVD players that were defective.
A) 3.1
B) 0.1
C) 9.6
D) 6

Question

Quizplus · Accepted Answer

The Poisson approximation to the binomial distribution's standard deviation is $\sqrt{\lambda}$, where $\lambda$ is the mean number of successes (defective DVD players in this case). The mean $\lambda$ can be calculated as $n \times p$, where $n = 160$ and $p = 0.06$ (the probability of a DVD player being defective). Thus, $\lambda = 160 \times 0.06 = 9.6$. The standard deviation is $\sqrt{9.6} \approx 3.1$.

Last Year, a Manufacturer Produced 950,000 DVD Players 6%6 \%6% Were Defective

Last Year, a Manufacturer Produced 950,000 DVD Players $6 \%$ Were Defective