The lengths of pregnancies in a small rural village

Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 263 days and a standard deviation of 16 days.

In what range would you expect to find the middle 50% of most pregnancies? 

If you were to draw samples of size 45 from this population, in what range would you expect to find the middle 50% of most averages for the lengths of pregnancies in the sample? 

Answer 1

Let X be length of pregnancies.

The middle 50% of most averages corresponds to the area between the 25th percentile and the 75th percentile.

To find the values corresponding to the 25th percentile and 75th percentile, we first need to find the z-score corresponding to the 75th percentile, which is 0.675.

We then equate that z score to the standardized value of X to find X at the 75th percentile:

0.675 = (X - 263)/16

0.675 * 16 + 263 = X

X = 273.8, which is 10.8 above the mean

Since the middle 50% is symmetric about the mean, the X value for the 25th percentile is 10.8 below the mean, or 263-10.8 = 252.2

The range to find the middle 50% is 252.2, 273.8