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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{61 - 54}{8}$

$Z = 0.875$

$Z = 0.875$ has a pvalue of 0.8092

X = 38.9

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38.9 - 54}{8}$

$Z = -1.89$

$Z = -1.89$ 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