Let the lengths of pregnancies be X X follows normal distribution with mean 268 and standard deviation 15 days z=(X-269)/15
a. P(X>308) z=(308-269)/15=2.6 thus: P(X>308)=P(z>2.6) =1-0.995 =0.005
b] Given that if the length of pregnancy is in lowest is 44%, then the baby is premature. We need to find the length that separates the premature babies from those who are not premature. P(X<x)=0.44 P(Z<z)=0.44 z=-0.15 thus the value of x will be found as follows: -0.05=(x-269)/15 -0.05(15)=x-269 -0.75=x-269 x=-0.75+269 x=268.78 The length that separates premature babies from those who are not premature is 268.78 days
