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 formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 309$ (the value of interest), $\mu=267$ (mean), and $\sigma = 15$ (standard deviation). So, $z=\frac{309 - 267}{15}=\frac{42}{15}=2.8$.

Step2: Find the probability

We want $P(X\geq309)$, which is equivalent to $P(Z\geq2.8)$ using the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z\geq z)=1 - P(Z < z)$. Looking up $P(Z < 2.8)$ in the standard - normal table, we find that $P(Z < 2.8)=0.9974$. So, $P(Z\geq2.8)=1 - 0.9974 = 0.0026$.

Step3: For part b, find the z - score corresponding to the lower 2%

We want to find the z - score $z$ such that $P(Z < z)=0.02$. Looking up in the standard - normal table (or using a calculator with a normal - distribution function), the z - score corresponding to a left - tail area of 0.02 is approximately $z=-2.05$.

Step4: Find the length of pregnancy

Using the z - score formula $z=\frac{x-\mu}{\sigma}$ and solving for $x$, we get $x=\mu+z\sigma$. Substituting $\mu = 267$, $z=-2.05$, and $\sigma = 15$, we have $x=267+(-2.05)\times15=267 - 30.75 = 236.25$.

Answer:

a. 0.0026
b. 236.25 days