The lengths of pregnancies in a small rural village are normally distributed with a mean of 269 days and a standard deviation of 16 days. What is the probability that pregnancy will last fewer than 311 days? Round your answer to at least three decimals.
Solution
Step 1 ：The lengths of pregnancies in a small rural village are normally distributed with a mean of 269 days and a standard deviation of 16 days.
Step 2 ：We are asked to find the probability that a pregnancy will last fewer than 311 days.
Step 3 ：To solve this, we first calculate the z-score, which is given by the formula \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 4 ：Substituting the given values, we get \(z = \frac{311 - 269}{16} = 2.625\).
Step 5 ：We then use the cumulative distribution function to calculate the probability corresponding to this z-score.
Step 6 ：The probability that a pregnancy will last fewer than 311 days is approximately 0.996. This means that in this small rural village, about 99.6% of pregnancies are expected to last fewer than 311 days.
Step 7 ：Final Answer: \(\boxed{0.996}\)
