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

Answer:

Q3 = 65.98 inches.

Step-by-step 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:

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

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.

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.

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

$0.675 = \frac{X - 64.9}{1.6}$

$X - 64.9 = 1.6*0.675$

$X = 65.98$

Q3 = 65.98 inches.