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.

probability of a pregnancy lasting 309 days is 0.0032

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

Step-by-step 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