Question

The lengths of pregnancies are normally distributed with a mean of 268 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

probability of a pregnancy lasting 309 days is 0.0032

the length of pregnancy is in the lowest 4​% is 241.74 days

Explanation:

Given data

mean = 268

standard deviation = 15

to find out

probability of a pregnancy lasting 309 days or longer and the length of pregnancy is in the lowest 4​%

solution

we know mean (M) is 268 and standard deviation (SD) is 15

so probability will be in 1st part where pregnancy lasting 309 days

P(X > 309 ) = P( X− mean > 309 − mean )

we know

Z = (309 − mean) / standard deviation ,

it will be Z = 309−268/15 = 2.73

we can say these both are equal

P(X > 309 ) = P( Z > 2.73)

now we use the standard normal z-table i.e.

P( Z > 2.73 ) = 0.0032

so P( X > 309 ) is 0.0032

probability of a pregnancy lasting 309 days is 0.0032

and in 2nd part z value with 4% .i.e 0.04

z = -1.7507 from the standard table

so days = z × standard deviation + mean

days = -1.7507 × 15 + 268

days = 241.74

the length of pregnancy is in the lowest 4​% is 241.74 days