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

308 days or longer. b. If the length of pregnancy is in the lowest 3%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

Answer) (a)Here we need to find p(X 308)

To find the required probability, first, we need to find the z score

z = (X-mean)/(standard deviation)

z=308-267/15 (as standard deviation is 15 and mean is 267)

z = 2.73

so, p(X 308) = p(Z2.73)

p(Z2.73) = 1-p(Z2.73)

p(Z2.73) = 1-0.9968 (from z table)

Therefore, required probability is 0.0032

(b) Here first we need to find the z score for a left tail of 3%

3% means 0.03 which is equivalent to 0.0301 in z table which corresponds to -1.88 z score

Therefore z = -1.88, standard deviation = 15 and mean is 267

z = (X-mean)/(standard deviation)

We need to find x here

x=z*standard deviation + mesan

X = -1.88*15 + 267

X = 238.8

So, therefore the length that separates premature babies from those who are not premature is 238.8 days