The heights of adult men in America are normally distributed

Question

The heights of adult men in America are normally distributed, with a mean of 69.1 inches and standard deviation of 2.65 inches.

What percentage of adult males in America are over 5 feet 11 inches tall? Round your answer as a percentage to the nearest whole number.

 %

Answer 1

Hello Kelley,

The formula for Z Score is

Z score = ( x – µ ) / σ

Where x is the raw score, µ is the mean and σ is the standard deviation.

with a mean of 69.1 inches and standard deviation of 2.65 inches.

So we already have the mean and standard deviation.

What percentage of adult males in America are over 5 feet 11 inches tall?

5 feet 11 inches is 71 inches. Now finding the z-score is simple matter of substitution and using the z table https://www.ztable.net/ . Your Z will be 0.72. And when Z=0.72, the area to left of X=71 will be 0.7642.

Which means 23.58 or ~24% of adult males in America are over 5 feet 11 inches tall?

Hope this helps,

Samantha

Answer 2

Kelley

What you need to do here is find the Z value when X=5 feet 11 inches, then use this and a table of areas under the standard normal to find the percentage.

5 feet 11 inches = 5 x 12 + 11 inches = 71 inches.

Z = (X-µ)/σ. µ (MEAN) = 69.1 and σ (standard deviation) is 2.65

So when X = 71, Z = (71-69.1) / 2.65 = 0.7169811321

The nearest Z value to this in the area table is Z=0.72. When Z=0.72, area to left of X=71 is 0.7642.

This means the percentage of males less than 71 inches is 76.42, and the percentage more is 100 - 76.42 = 23.58

To the nearest percent that is 24%

Mike