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.

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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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = 69, \sigma = 2.8[/tex]

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.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.037 = \frac{X - 69}{2.8}[/tex]

[tex]X - 69 = -1.037*2.8[/tex]

[tex]X = 66.1[/tex]

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