Let [tex]X[/tex] be the random variable for the length of a pregnancy. [tex]X[/tex] follows a normal distribution, so we can transform it to a random variable [tex]Z[/tex] with the standard normal distribution via the rule
[tex]Z=\dfrac{X-\mu_X}{\sigma_X}[/tex]
a.
[tex]\mathbb P(X\ge309)=1-\mathbb P(X<309)[/tex]
[tex]=1-\mathbb P\left(\dfrac{X-269}{15}<\dfrac{309-269}{15}\right)[/tex]
[tex]\approx1-\mathbb P(Z<2.67)\approx0.0038[/tex]
b. A baby is premature if the duration of the pregnancy is up to [tex]x[/tex], where
[tex]\mathbb P(X\le x)=0.04[/tex]
This percentage corresponds to a [tex]z[/tex] value of about -1.75; that is,
[tex]\mathbb P(Z\le-1.75)\approx0.04[/tex]
To find the corresponding value of [tex]x[/tex], we solve:
[tex]\dfrac{x-269}{15}=-1.75\implies x\approx242.75[/tex]
meaning a pregnancy has to exceed 242.75 days in duration in order for the baby to not be deemed premature.
