Question

The lengths of pregnancies are normally distributed with a mean of 269 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 2​%, 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 309 days or longer is ?.

​(Round to four decimal places as​ needed.)

b. Babies who are born on or before ? days are considered premature.

​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given ,

mean = = 269

standard deviation = = 15

P(x >309 ) = 1 - P(x<309 )

= 1 - P[(x -) / < (309-269) /15 ]

= 1 - P(z <2.67 )

Using z table

= 1 - 0.9962

=0.0038

probability= 0.0038

(b)

Using standard normal table,

P(Z < z) = 2%

=(Z < z) = 0.02  

= P(Z < -2.05 ) = 0.02  

z = -2.05   

Using z-score formula  

x = z +

x = -2.05 *15+269

x = 238.25

x=238