Question

The lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 4​%, then the baby is premature. Find the length that separates premature babies from those who are not premature. a. The probability that a pregnancy will last 309 days or longer is..... ​(Round to four decimal places as​ needed.) b. Babies who are born on or before...…..days are considered premature. ​(Round to the nearest integer as​ needed.)

Answer 1

Given The lengths of pregnancies are normally distributed with a mean of = 267 days and a standard deviation of = 15 days.

Since the distribution is normal hence Z statistic is applicable for probability calculation.

a). The probability of a pregnancy lasting X = 309 days or longer P(X>= 309) is calculated by finding the Z score at X = 309 which is calculated as:

Thus the probability P(X> = 309) is P(Z>2.8) which is calculated using the excel formula for normal distribution which is =1-NORM.S.DIST(2.8, TRUE), Thus the probability P(X>=309) = 0.0026

The probability that a pregnancy will last 309 days or longer is 0.0026.

b)  If the length of pregnancy is in the lowest 4​%, then the baby is premature, then the length that separates premature babies from those who are not premature is calculated by finding the Z score for the lowest 4% which is calculated by the excel formula which is =NORM.S.INV(0.04), thus the Z is computed as -1.751.

Now using the Z score formula the length of pregnancy is calculated as:

Hence Babies who are born on or before 241 days are considered premature.