Question

The lengths of pregnancies are normally distributed with a mean of 208 days and a standard deviation of 15 days. A wife claimed to have given birth 308 days after a brief visit from her husband who was serving in the Navy. What is the probability of a pregnancy lasting 308 days or longer

Answer 1

0%

Explanation:

We have the following information:

mean (m) = 208 days

standard deviation (s) = 15 days

We have that the z value would be:

z = (x - m) / sd

replacing we have:

z = (308 - 208) / 15

z = 6.66

However,

P (x> 308) = 1 - (x <308)

P (x> 308) = 1 - (z <6.66)

if we look for this value of z in the table we assume that it is 1, since the table reaches 3 and the value is 1. Therefore:

P (x> 308) = 1 - 1

P (x> 308) = 0

Which means that the probability is 0%, therefore it is a very unusual pregnancy.