Applications of Normal Distribution Assessment

Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 267 and a standard deviation of 15.

What percentage of pregnancies last beyond 246 days?

P(X > 246 days) =

Answer 1

To standardize a random normal variable, we use the following equation to find z-score

Z = x - mean/st. Dev

Z = 246 - 267/15

Z = -1.4

This means we are -1.4 standard deviations below the mean. The question asks what percentage of pregnancies last more than 246 days. This is everything above a standard deviation if -1.4. If we look up probability on our Z-table for +1.4, we get the probability 0.9192. This is a probability for +1.4, but the standard normal curve is symmetrical. Since our Z-score is to the left of the mean, the probability of 0.9192 (or percentage of 91.92%) represents the percentage of pregnancies that last beyond 246 days.