Question

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 307 days or longer. b. If the length of pregnancy is in the lowest 2​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

Answer:

Step-by-step explanation:

Let X be the length of pregnancy.

X is N(268, 15)

a) Prob of pregnancy lasting 307 days or longer

= P(X>307) = $P(Z>\frac{307-268}{15} )\\=P(Z>2.6)$

=0.5-0.4953

=0.0047

b) Lowest 2% is 2nd percentile

Z=-2.55

X score = 268-2.55(15)

= 229.75 days

If length of days >230 days then it is not premature.