Question

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
a. find the probability of a pregnancy lasting 308 days or longer.
b. if the length of pregnancy is in the lowest 3​%, then the baby is premature. find the length that separates premature babies from those who are not premature.

Answer 1

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean $\mu =266$

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find $P(X\geq 308)$

Using normal distribution,

$z=\frac{X-\mu}{\sigma}$

$z=\frac{308-266}{15}$

$=\frac{42}{15}$

Thus $z=2.8$

So $P\left (X\geq 308 \right )=1-P(X<308)$

$=1-P(z<2.8)$

$=1-Table\: value\: of\: 2.8$

$=1-0.99744$

$=0.00256$

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3​%.

The z-value for the lowest of 3% is -1.8808

Then $X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu$

This implies $X=-1.8808*15+266=237.788$

Thus the babies who are born on or before 238 days are considered to be premature.