The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.
A researcher suggests using 500 births instead of only 5. In this case, which distribution is a reasonable approximation of X?

Question

Quizplus · Accepted Answer

The approximation for a binomial distribution with a large number of trials (n) and a probability of success (p) is a normal distribution with mean $np$ and variance $np(1-p)$. Here, $n=500$ and $p=1-\frac{1}{33}$ since we're looking at the probability of not having a defect. The mean is $500 \times \left(1-\frac{1}{33}\right) = 500 \times \frac{32}{33} = 484.85$. The variance is $500 \times \frac{32}{33} \times \frac{1}{33} = 484.85 \times \frac{1}{33} \approx 14.69$, and the standard deviation (square root of variance) is approximately $\sqrt{14.69} \approx 3.83$. Therefore, the appropriate normal distribution approximation is $N(484.85, 3.83^2)$, which matches option D in terms of mean and standard deviation (not variance directly, but the square of the standard deviation is implied).

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