Question

Suppose that 61.5% of brides are younger than their grooms. Suppose one were to consider simple random samples of size 40 of brides.
What is the probability that the proportion of brides in a sample of size 40 who are younger than their grooms exceeds 0.625?

Answer 1

Answer on Question #55369 – Math – Statistics and Probability

Question

Suppose that 61.5% of brides are younger than their grooms. Suppose one were to consider simple random samples of size 40 of brides. What is the probability that the proportion of brides in a sample of size 40 who are younger than their grooms exceeds 0.625?

Solution

First of all, we find $40 \cdot 0.625 = 25$. Let $\xi$ be a number of brides who are younger than their grooms. We must find the next probability:

$$P\{25 < \xi \leq 40\}.$$

Notice that $\xi$ has the binomial distribution with the parameters $p = 0.615, n = 40$.

We shall use the normal approximation (i.e. the integral theorem of Moivre-Laplace):

$$\begin{array}{l} P\{25 < \xi \leq 40\} = P\left\{\frac{25 - 40 \cdot 0.615}{\sqrt{40 \cdot 0.615 \cdot 0.385}} < \frac{\xi - np}{\sqrt{npq}} \leq \frac{40 - 40 \cdot 0.615}{\sqrt{40 \cdot 0.615 \cdot 0.385}}\right\} = P\left\{0.13 < \frac{\xi - np}{\sqrt{npq}} \leq 5\right\} \approx \\ \approx \Phi(5) - \Phi(0.13) = 0.5 - 0.05172 = 0.44828. \end{array}$$

Here $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{t^2}{2}} dt$ is a tabulated function of Laplace and we found $\Phi(5), \Phi(0.13)$ using the table of this function.

Answer: 0.44828.

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