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 308 days or longer. b. if the length of pregnancy is in the lowest 3%, then the baby is premature. find the length that separates premature babies from those who are not premature.

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find

Using normal distribution,

Thus

So

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3%.

The z-value for the lowest of 3% is -1.8808

Then

This implies

Thus the babies who are born on or before 238 days are considered to be premature.

there is no points to select.

step-by-step explanation:

can someone me: the midsegment through the legs of an isosceles triangle is 4 in. find the sides of the triangle if its perimeter is 20 in. ,   and point g is the centroid of the right △abc with hypotenuse ab=18 in. find cg.