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 307 days or longer. b. If the length of pregnancy is in the lowest 44​%,then the baby is premature. Find the length that separates premature babies from those who are not premature.

a. The probability that a pregnancy will last 307 days or longer is [ ]. (Round to four decimal places)

b. Babies who are born on or before [ ] days are considered premature. (Round to the nearest integer as needed)

Answer 1

Solution :

Given that ,

a) P(x 307 ) = 1 - P(x   307)

= 1 - P[(x - ) / (307 - 266) / 15]

= 1 -  P(z 2.73)

= 1 - 0.9968

= 0.0032

The probability that a pregnancy will last 307 days or longer is 0.0032

b) Using standard normal table,

P(Z < z) = 44%

= P(Z < z ) = 0.44

= P(Z < -0.15 ) = 0.44  

z = -0.15

Using z-score formula,

x = z * +

x = -0.15 * 15 + 266

x = 263.75

Babies who are born on or before 264 days are considered premature