Question

The lengths of pregnancies are normally distributed with a mean of 269269 days and a standard deviation of 1515 days. a. Find the probability of a pregnancy lasting 309309 days or longer. b. If the length of pregnancy is in the lowest 44​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

a. 0.38%

b. 266.75 days

Explanation:

We have the following data, mean (m) 269 and standard deviation (sd) 15, therefore:

a. The first thing is to calculate the number z:

z (x) = (x - m) / sd

z (309) = (309-269) / 15 = 2.67

When looking in the normal distribution table (attached), we have that at this value of z, the probability is:

P (z> 2.67), that is to say we must look in the table -2.67 and this value corresponds to 0.0038, that is to say 0.38%

b. Find the z-value with a left tail of 44%, i.e. 0.44. We look in the table for this value and what value of z corresponds.

invNorm (0.44) = -0.15

Find the corresponding number of days:

x = z * sd + m

we replace

d = -0.15 * 15 + 269 = 266.75 days