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

a) 281 days.

b) 255 days

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 = (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 = (X - \mu)/(\sigma)`

`1.405 = (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 = (X - \mu)/(\sigma)`

`-1.88 = (X - 270)/(8)`

`X - 270 = -1.88*8`

`X = 255`