The lengths of pregnancies are normally distributed with a m

The lengths of pregnancies are normally distributed with a mean of 250 days and a standard deviation of 15 days.

a. Find the probability of a pregnancy lasting 308 days or longer?

b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8%, find the length that separates premature babies from those who are not premature.

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    308      
u = mean =    250      
          
s = standard deviation =    15      
          
Thus,          
          
z = (x - u) / s =    3.866666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.866666667   ) =    5.51665*10^-5 OR 0.0000551665 [ANSWER]

****************

b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.08      
          
Then, using table or technology,          
          
z =    -1.40507156      
          
As x = u + z * s,          
          
where          
          
u = mean =    250      
z = the critical z score =    -1.40507156      
s = standard deviation =    15      
          
Then          
          
x = critical value =    228.9239266   [ANSWER]