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:
[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{309-270}{15}=\dfrac{39}{15}=2.6[/tex]We now can calculate the probability as:
[tex]P(X>309)=P(z>2.6)=0.0047[/tex]b. We now know that the probability is 2% for the baby to be premature.
This probability correspond to a z-value as:
[tex]P(z<-2.05375)=0.02[/tex]We can now calculate the value of X that correspond to this z-score as:
[tex]\begin{gathered} X=\mu+z\cdot\sigma \\ X=270+(-2.05375)\cdot15 \\ X=270-30.80625 \\ X\approx239 \end{gathered}[/tex]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
