Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 14 days. A distribution of values is normal with a mean of 267 and a standard deviation of 14.

What percentage of pregnancies last beyond 282 days?
P(X > 282 days) =
%

Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

To find the percentage of pregnancies that last beyond 282 days, we first need to calculate the z-score for 282 days using the formula:

\[
z = \frac{X - \mu}{\sigma}
\]

where \(X\) is the value we are interested in (282 days), \(\mu\) is the mean (267 days), and \(\sigma\) is the standard deviation (14 days).

Substituting in the values:

\[
z = \frac{282 - 267}{14} = \frac{15}{14} \approx 1.0714
\]

Now we round the z-score to three decimal places:

\[
z \approx 1.071
\]

Next, we can use the standard normal distribution table (or a calculator) to find the probability \(P(Z \leq z)\) for \(z = 1.071\).

Looking up the z-score of 1.071 in the standard normal distribution table, we find:

\[
P(Z \leq 1.071) \approx 0.8577
\]

This value represents the percentage of pregnancies that last **less than** 282 days. To find the percentage of pregnancies that last **more than** 282 days, we need to subtract this value from 1:

\[
P(X > 282) = 1 - P(Z \leq 1.071) \approx 1 - 0.8577 \approx 0.1423
\]

To convert this to a percentage, we multiply by 100:

\[
0.1423 \times 100 \approx 14.23
\]

Rounding to one decimal place, the percentage of pregnancies that last beyond 282 days is:

\[
\boxed{14.2}
\]