Question

The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 267 and a standard deviation of 15. What percentage of pregnancies last beyond 246 days? P(X > 246 days) =

Answer 1

91.92% of pregnancies last beyond 246 days

Explanation:

When the distribution is normal, we use 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 = 267, \sigma = 15`

What percentage of pregnancies last beyond 246 days?

We have to find 1 subtracted by the pvalue of Z when X = 246. So

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

`Z = (246 - 267)/(15)`

`Z = -1.4`

`Z = -1.4` has a pvalue of 0.0808

1 - 0.0808 = 0.9192

91.92% of pregnancies last beyond 246 days