Question

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 270270 days and a standard deviation of 1111 days. ​(a) What is the minimum pregnancy length that can be in the top 88​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom 33​% of pregnancy​ lengths? ​(a) The minimum pregnancy length is nothing days. ​(Round to one decimal place as​ needed.) ​(b) The maximum pregnancy length is nothing days.

Answer 1

Answer:

a) The minimum pregnancy length is 27.1 days.

b) The maximum pregnancy length is 265.2 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:

$\mu = 270, \sigma = 11$

(a) What is the minimum pregnancy length that can be in the top 88​% of pregnancy​ lengths? ​

This is the value of X when Z has a pvalue of 1-0.88 = 0.12. So it is X when Z = -1.175.

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

$-1.175 = \frac{X - 270}{11}$

$X - 270 = -1.175*11$

$X = 257.1$

The minimum pregnancy length is 27.1 days.

(b) What is the maximum pregnancy length that can be in the bottom 33​% of pregnancy​ lengths?

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

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

$-0.44 = \frac{X - 270}{11}$

$X - 270 = -0.44*11$

$X = 265.2$

The maximum pregnancy length is 265.2 days.