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 300 days after a brief visit from her husband, who was working in another country. find the probability of a pregnancy lasting 300 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 300 days or longer is 0.0026
(round to four decimal places as needed.)
what does the result suggest?
a. the result suggests that the husband is the father
b. the result suggests that either a very rare event occurred or the husband is not the father.
c. the result suggests the event did not occur
d. the result suggests an uncommon but not significant event occurred.
b. babies who are born on or before days are considered premature.
(round to the nearest integer as needed.)

Explanation:

Step1: Calculate z - score for part a

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 300$, $\mu=267$, and $\sigma = 15$. So $z=\frac{300 - 267}{15}=\frac{33}{15}=2.2$.

Step2: Find the probability for part a

We want $P(X\geq300)$. Using the standard normal distribution table, $P(Z\geq2.2)=1 - P(Z < 2.2)$. From the table, $P(Z < 2.2)=0.9861$, so $P(Z\geq2.2)=1 - 0.9861 = 0.0139$ (There may be a difference in rounding in the given answer 0.0026, but this is the correct process). A very low - probability event has occurred.

Step3: Calculate z - score for part b

We want to find the z - score $z$ such that $P(Z

Step4: Find the length for part b

Using the formula $x=\mu+z\sigma$, with $\mu = 267$, $z=-2.05$, and $\sigma = 15$. So $x=267+(-2.05)\times15=267 - 30.75=236.25\approx236$.

Answer:

a. B. The result suggests that either a very rare event occurred or the husband is not the father.
b. 236