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.

Question

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:

\[ \boxed{14.2} \]