Question

Assume that the heights of women are normally distributed with a mean of 64.9 inches and a standard deviation of 1.6 inches. Find Q3, the third quartile that separates the bottom 75% from the top 25%. *

Answer 1

Q3 = 65.98 inches.

Explanation:

Normal Probability Distribution:

Problems of normal distributions 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.

Normally distributed with a mean of 64.9 inches and a standard deviation of 1.6 inches.

This means that
`\mu = 64.9, \sigma = 1.6`

Find Q3, the third quartile that separates the bottom 75% from the top 25%.

This is X when Z has a pvalue of 0.75. So X when Z = 0.675.

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

`0.675 = (X - 64.9)/(1.6)`

`X - 64.9 = 1.6*0.675`

`X = 65.98`

Q3 = 65.98 inches.