Question

Assume that the duration of human pregnancies can be described by a Normal distribution with mean 278 days and standard deviation 12 days. About 15% of babies are born prematurely. What is the length of a pregnancy that is the cutoff between being born prematurely and not being born prematurely?

Answer 1

X = 262 days

Explanation:

For Normal Distribution N( 0,1) we have z table and we can find z score for value of 15 % or 0,15. As 0,15 is not in the table we extrapolate then

0,1492 ⇒ 1,4

0,1515 ⇒ 1,3 0,15 ⇒ x 0,1515 - 0,15 = 0,0015

Δ = 0,0023 ⇒ 0,1

By rule of three

If 0,0023 ⇒ 0,1

0,0015 ⇒ x

x = 0,0652

Therefore the value between 1,3 and 1,4 is:

1,4 - 0,0652

z (score ) = 1,3348 and as we are looking for values below the mean ( that is at the left of the bell shape curve we need to change the value to a negative one z = - 1,3348

Now when N ( 278 , 12 ) is normalized to N ( 0 , 1)

z( score ) = ( X - μ ) / σ

- 1,3348 * 12 = X - 278

X = 278 - 16,0176

X = 278 - 16

X = 262 days