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 2%, 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 z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 300$, $\mu=287$, and $\sigma = 15$. So $z=\frac{300 - 287}{15}=\frac{13}{15}\approx0.87$.

Step2: Find the probability using the standard normal table

We want $P(X\geq300)$. Since the total area under the standard - normal curve is 1, and $P(X\geq300)=1 - P(X < 300)$. Looking up the z - score of 0.87 in the standard normal table, $P(Z < 0.87)=0.8078$. So $P(X\geq300)=1 - 0.8078 = 0.1922$.

Answer:

0.1922