Question

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 270 days and a standard deviation of 8 days. ​(a) What is the minimum pregnancy length that can be in the top 8​% of pregnancy​ lengths? ​(b) What is the maximum pregnancy length that can be in the bottom 3​% of pregnancy​ lengths?

Answer 1

Answer:

a) 281 days.

b) 255 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:

$\mu = 270, \sigma = 8$

​(a) What is the minimum pregnancy length that can be in the top 8​% of pregnancy​ lengths?

100 - 8 = 92th percentile.

X when Z has a pvalue of 0.92. So X when Z = 1.405.

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

$1.405 = \frac{X - 270}{8}$

$X - 270 = 1.405*8$

$X = 281$

(b) What is the maximum pregnancy length that can be in the bottom 3​% of pregnancy​ lengths?

3rd percentile.

X when Z has a pvalue of 0.03. So X when Z = -1.88

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

$-1.88 = \frac{X - 270}{8}$

$X - 270 = -1.88*8$

$X = 255$