Question

The lengths of pregnancies are normally distributed with a mean of 250 days and a standard deviation of 15 days.

a. Find the probability of a pregnancy lasting 308 days or longer?








b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8% (8th percentile), find the length that separates premature babies from those who are not premature.

Answer 1

a) 0.005% probability of a pregnancy lasting 308 days or longer

b) The pregnancy length that separates premature babies from those who are not premature is 229 days.

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 question, we have that:

`\mu = 250, \sigma = 15`

a. Find the probability of a pregnancy lasting 308 days or longer?

This is 1 subtracted by the pvalue of Z when X = 308. So

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

`Z = (308 - 250)/(15)`

`Z = 3.87`

`Z = 3.87` has a pvalue of 0.99995

1 - 0.99995 - 0.00005

0.005% probability of a pregnancy lasting 308 days or longer

b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8% (8th percentile), find the length that separates premature babies from those who are not premature.

The 8th percentile is X when Z has a pvalue of 0.08. So it is X when Z = -1.405.

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

`-1.405 = (X - 250)/(15)`

`X - 250 = -1.405*15`

`X = -1.405*15 + 250`

`X = 229`

The pregnancy length that separates premature babies from those who are not premature is 229 days.