Question

The lengths of pregnancies are normally distributed with a mean of 270 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 2%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

We have the lengths of pregnancies modeled as a random variable normally distributed with mean 270 days and standard deviation of 15 days.

a. We have to find the probability of a pregnancy lasting 309 days or longer.

We can calculate the z-score for this value X = 309 and this distribution, and then look for the probability in the standard normal ditribution for that z-score.

We can calculate the z-score as:

`z=(X-\mu)/(\sigma)=(309-270)/(15)=(39)/(15)=2.6`

We now can calculate the probability as:

`P(X>309)=P(z>2.6)=0.0047`

b. We now know that the probability is 2% for the baby to be premature.

This probability correspond to a z-value as:

`P(z<-2.05375)=0.02`

We can now calculate the value of X that correspond to this z-score as:

`\begin{gathered} X=\mu+z\cdot\sigma \\ X=270+(-2.05375)\cdot15 \\ X=270-30.80625 \\ X\approx239 \end{gathered}`

Then, if a pregnancy lasts less than 239 days, the baby is considered to be premature.

Answer

a) P(x>309) = 0.0047

b) 239 days