Question

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 4​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

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

b) A pregnancy length of 241 days separates premature babies from those who are not premature.

Explanation:

Problems of normally distributed samples can be 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 = 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 = (X - \mu)/(\sigma)`

`Z = (309 - 267)/(15)`

`Z = 2.8`

`Z = 2.8` has a pvalue of 0.9974

So there is a 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 4​%, 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 0.04. So X when Z = -1.75

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

`-1.75 = (X - 267)/(15)`

`X - 267 = -1.75*15`

`X = 240.75`

A pregnancy length of 241 days separates premature babies from those who are not premature.

Answer 2

We have a normal distribution with a mean of 267 days and a standard deviation of 15 days. To solve this proble we're going to need the help of a calculator.

a. The probability of a pregnancy lasting 309 days or longer is:

P(z>309) = 0.0026 or 0.26%

b. The lowest 4% is separeted by the 240.74 days. The probability of pregnancy lasting 240.74 days is 4%.