Question

The distribution of heights of adult American men is approximately Normal with mean of $69$inches and a standard deviation of $2.5$inches. Draw a Normal curve on which this mean and standard deviation are correctly located. (Hint: Draw the curve first, locate the points where the curvature changes, and then mark the horizontal axis.

(a) What per cent of men are taller than $74$inches?

(b) Between what heights do the middle $95\%$of men fall?

(c) What per cent of men are between $64$and $66.5$inches tall?

(d) A height of $71.5$inches corresponds to what percentile of adult male American heights?

Short answer

(a) It is declared that $2.5\%$of men are over $74$inches tall.

(b) As a result, $95\%$of men are between the ages of $64$and $74$inches tall.

(c) Therefore $13.5\%$of men are between the ages of $64$and $66.5$inches tall.

(d) $71.5$is the $84$th percentile of adult male heights in the United States.

Step-by-step solution

Part (a) Step 1: Given information

The heights of adult American men are approximately

Mean $=69in$

Standard deviationrole="math" localid="1652780330432" $=2.5in$

Part (a) Step 2: Explanation

The height distribution is roughly $N(69,2.5)$.

The Normal density curve is sketched below, with the necessary points labelled

Approximately $2.5\%$of men are taller than $74$inches, which is $2$standard deviations over the mean, according to the $68-95-99.7$rule

As a result, $2.5\%$of men are over $74$inches tall.

Part (b) Step 1: Given information

The heights of adult American men are approximately

Mean$=69in$

Standard deviation$=2.5in$

Part (b) Step 2: Explanation

According to the $68-95-99.7$rule, $95\%$of males have heights between $64$and $74$inches that are within $2s$of $\mu$.

As a result, $95\%$of men are between the ages of $64$and $74$inches tall.

Part (c) Step 1: Given information

The heights of adult American men are approximately

Mean$=69in$

Standard deviation$=2.5in$

Part (c) Step 2: Explanation

Because $66.5$is one standard deviation below the mean, the around $16\%$of males are shorter than $66.5$inches, according to the $68-95-99.7$rule. Because $64$inches are two standard deviations below the mean, around $2.5\%$are shorter than $64$inches. So roughly $16\%-2.5\%=13.5\%$of men are between the ages of $64$and $66.5$inches tall.

Part (d) Step 1: Given information

The heights of adult American men are approximately

Mean$=69in$

Standard deviation$=2.5in$

Part (d) Step 2: Explanation

The figure $71.5$is one standard deviation above the mean, according to the $68-95-99.7$rule. Thus, $0.68+0.16=0.84$is the area to the left of $71.5$. In other words, $71.5$ is the $84$th percentile of adult male heights in the United States.