Question

The lengths of pregnancies are normally distributed with a mean of 266 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.

Answer 1

Solution :

Given that ,

a) P(x 309 ) = 1 - P(x   309)

= 1 - P[(x - ) / (309 - 266) / 15]

= 1 -  P(z 2.87 )

= 1 - 0.9979

= 0.0021

b) Using standard normal table,

P(Z < z) = 4%

= P(Z < z ) = 0.04

= P(Z < -1.75 ) = 0.04

z = -1.75

Using z-score formula,

x = z * +

x = -1.75 * 15 + 266

x = 239.75

x = 240 days