Question

Suppose that 63.6% of brides are younger than their grooms. Suppose one were to consider simple random samples of size 36 of brides.
What is the probability that the proportion of brides in a sample of size 36 who are younger than their grooms exceeds 0.646?

Answer 1

Answer on Question #62744 – Math – Statistics and Probability

Question

Suppose that 63.6% of brides are younger than their grooms. Suppose one were to consider simple random samples of size 36 of brides.

What is the probability that the proportion of brides in a sample of size 36 who are younger than their grooms exceeds 0.646?

Solution

Given both $np = 36 \cdot 0.636 = 22.896 > 10$,

$n(1 - p) = 36(1 - 0.636) = 13.104 > 10$, the distribution of sample proportion will be approximately normally distributed with a mean $p = 0.636$ and standard deviation of $SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.636(1 - 0.636)}{36}}$.

The probability that the proportion of brides in a sample of size 36 who are younger than their grooms exceeds 0.646 will be

$$P(\hat{p} > 0.646) = P\left(\frac{\hat{p} - p}{SE} > \frac{0.646 - 0.636}{\sqrt{\frac{0.636(1 - 0.636)}{36}}}\right) \approx$$ $$\approx P\left(Z > \frac{0.646 - 0.636}{\sqrt{\frac{0.636(1 - 0.636)}{36}}}\right) = 1 - P\left(\frac{0.646 - 0.636}{\sqrt{\frac{0.636(1 - 0.636)}{36}}}\right) =$$ $$= 1 - 0.54962 = 0.45038,$$

where $Z$ is normally distributed with a mean 0 and standard deviation of 1.

Answer: 0.45038.

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