Problem 6 (16 points) Assume that the lengths of pregnancies are normally distri
ID: 3317238 • Letter: P
Question
Problem 6 (16 points)
Assume that the lengths of pregnancies are normally distributed with a mean of 280 days and a standard deviation of 7 days.
1.What is the probability that a randomly selected mother will have a pregnancy lasting longer than 290 days?
2.What is the probability that a randomly selected mother will have a pregnancy lasting shorter than 278 days?
3.What is the probability that a randomly selected mother will have a pregnancy lasting between 271 and 289 days?
4.Find the 90th percentile for the lengths of pregnancies.
Explanation / Answer
Mean = 280 days
Standard deviation = 7 days
P(X < A) = P(Z < (A - mean)/standard deviation)
1) P(X > 290) = 1 - P(X < 290)
= 1 - P(Z < (290 - 280)/7)
= 1 - P(Z < 1.43)
= 1 - 0.9236
= 0.0764
2) P(X < 278) = P(Z < (278-280)/7)
= P(Z < -0.286)
= 0.3874
3) P(271 < X < 289) = P(X < 289) - P(X < 271)
= P(Z < (289 - 280)/7) - P(Z< (271 - 280)/7)
= P(Z < 1.29) - P(Z < -1.29)
= 0.9015 - 0.0985
= 0.8030
4) P(X < A) = 0.9
P(Z < (A - 280)/7) = 0.9
(A - 280)/7 = 1.28
A = 288.96
90th percentile for the lenths of pregnancies = 288.96 days