Statistics Question

Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 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 57 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

The middle 50% would go from the 25th to 75th percentile.

The z-score for 75th percentile for the standard normal distribution is 0.67

z = (x - mean)/SD

for z = 0.67, mean = 265 and SD = 16 we have:

0.67 = (x - 265)/16

0.67 * 16 + 265 = x

x = 275.72

275.72 - 265 = 10.72

Since the normal distribution is symmetric, the value of x corresponding to the 25th percentile is:

265 - 10.72 = 254.28

(254.28, 275.72)

For 2nd part:

z = (x - mean)/(SD/sqrt(sample size)

for z = 0.67, mean = 265, SD = 16 and sample size = 57 we have:

0.67 = (x - 265)/(16/sqrt(57))

0.67 = (x - 265)/2.12

0.67 * 2.12 + 265 = x

x = 266..43

266.43 - 265 = 1.43

Since the normal distribution is symmetric, the value of x corresponding to the 25th percentile is:

265 - 1.43 = 263.57

(263.57,266.43)