Question

In a survey of men aged 20-29 in a certain country, the mean height is 73.4 inches with a standard deviation of 2.7 inches. Find the minimum height in the top 15% of heights. Use a TI-83, TI-83 plus, or TI-84 calculator, and round your answer to one decimal place.

Answer 1

Answer:

The minimum height in the top 15% of heights is 76.2 inches.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 73.4, \sigma = 2.7$

Find the minimum height in the top 15% of heights.

This is the value of X when Z has a pvalue of 0.85. So it is X when Z = 1.04.

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

$1.04 = \frac{X - 73.4}{2.7}$

$X - 73.4 = 1.04*2.7$

$X = 76.2$

The minimum height in the top 15% of heights is 76.2 inches.