Question

The lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 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. Click to view page 1 of the table.LOADING... Click to view page 2 of the table.LOADING... a. The probability that a pregnancy will last 308 days or longer is nothing. ​(Round to four decimal places as​ needed.) b. Babies who are born on or before nothing days are considered premature. ​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 267

standard deviation = = 15

a) P(x > 308) = 1 - p( x< 308)

=1- p P[(x - ) / < (308 - 267) / 15]

=1- P(z < 2.73 )

Using z table,

= 1 - 0.9968

= 0.0032

b) Using standard normal table,

P(Z < z) = 2%

= P(Z < z ) = 0.02

= P(Z < -2.05 ) = 0.02  

z = -2.05

Using z-score formula,

x = z * +

x = -2.05 * 15 + 267

x = 236 days