Question

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 54 inches, and standard deviation of 8 inches. What is the probability that the height of a randomly chosen child is between 38.9 and 61 inches

Answer 1

77.98% probability that the height of a randomly chosen child is between 38.9 and 61 inches

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:

`\mu = 54, \sigma = 8`

What is the probability that the height of a randomly chosen child is between 38.9 and 61 inches

This is the pvalue of Z when X = 61 subtracted by the pvalue of Z when X = 38.9. So

X = 61

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

`Z = (61 - 54)/(8)`

`Z = 0.875`

`Z = 0.875` has a pvalue of 0.8092

X = 38.9

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

`Z = (38.9 - 54)/(8)`

`Z = -1.89`

`Z = -1.89` has a pvalue of 0.0294

0.8092 - 0.0294 = 0.7798

77.98% probability that the height of a randomly chosen child is between 38.9 and 61 inches