Question

Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If the top 5 percent and bottom 15 percent are excluded for an experiment, what is the bottom cutoff heights to be eligible for this experiment? Round your answers to one decimal place.

Answer 1

Answer:

The bottom cutoff heights to be eligible for this experiment is 66.1 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.

Mean of 69.0 inches and a standard deviation of 2.8 inches.

This means that $\mu = 69, \sigma = 2.8$

What is the bottom cutoff heights to be eligible for this experiment?

The bottom 15% are excluded, so the bottom cutoff is the 15th percentile, which is X when Z has a pvalue of 0.15. So X when Z = -1.037.

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

$-1.037 = \frac{X - 69}{2.8}$

$X - 69 = -1.037*2.8$

$X = 66.1$

The bottom cutoff heights to be eligible for this experiment is 66.1 inches.