Question

"2. The lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 33%, then the baby is premature. Find the length that separates premature babies from those who are not premature."

Answer 1

Answer:

a) 0.26% probability of a pregnancy lasting 309 days or longer.

b) 260.4 days

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 267, \sigma = 15$

a. Find the probability of a pregnancy lasting 309 days or longer.

This is 1 subtracted by the pvalue of Z when X = 309. So

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

$Z = \frac{309 - 267}{15}$

$Z = 2.8$

$Z = 2.8$ has a pvalue of 0.9974

1 - 0.9974 = 0.0026

0.26% probability of a pregnancy lasting 309 days or longer.

b. If the length of pregnancy is in the lowest 33%, then the baby is premature. Find the length that separates premature babies from those who are not premature."

This is the value of X when Z has a pvalue of 0.33. So it is X when Z = -0.44.

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

$-0.44 = \frac{X - 267}{15}$

$X - 267 = -0.44*15$

$X = 260.4$