Question

the lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15 days.
a. in a letter to an advice column, a wife claimed to have given birth 309 days after a brief visit from her husband, who was working in another country. find the probability of a pregnancy lasting 309 days or longer. what does the result suggest?
b. if the length of pregnancy is in the lowest 4%, then the baby is premature. find the length that separates premature babies from those who are not considered premature.

a. the probability that a pregnancy will last 309 days or longer is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 309\), \(\mu=267\), and \(\sigma = 15\).

$$z=\frac{309 - 267}{15}=\frac{42}{15}=2.8$$

Step2: Find the probability

We want to find \(P(X\geq309)\), which is equivalent to \(P(Z\geq2.8)\).
Since \(P(Z\geq z)=1 - P(Z < z)\), and from the standard normal table \(P(Z < 2.8)=0.9974\)

$$P(Z\geq2.8)=1 - 0.9974=0.0026$$

Answer:

\(0.0026\)