The heights of adult men in America are normally distributed, with a mean of 69.2 inches and a standard deviation of 2.65 inches.

Question

The heights of adult men in America are normally distributed, with a mean of 69.2 inches and a standard deviation of 2.65 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.57 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

z = 

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

%

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z = 

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

%

e) Who is relatively taller: a 6'3" American man or a 5'11" American woman? Defend your choice in a meaningful sentence.

Answer 1

To calculate any z-score, you need to use the following formula: z=(x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For parts a & c of the problem, simply plug in the correct numbers that are given in the problem, but remember to make sure everything is in the SAME units (i.e. you're given means & std. dev's in inches, so I would convert the given values for the heights of the male & female to inches before you plug in the numbers and solve. Now you have the value of the z-score for that question.

For parts b & d of the problem, you'll need to use a 'z-score Table' to calculate the % of the population that is above or below some value. A z-score table shows the percentage of values (usually a decimal figure) to the left of a given z-score on a standard normal distribution.

In part b) you'll want to find the values to the left (shorter), but for part d) you'll want to find the values to the right (taller). Since the total area under the bell curve is 1 (as a decimal value which is equivalent to 100%), we subtract the area from the table from 1. For example: if the area to the left of the z-score is 0.8621, then the area to the right of the score is 1-.8621=0.1379.

Watch out: If you have a negative z-score, you'll need to do the following before you look for the %:

if you want the area to the left: use the same table but disregard the negative sign, then subtract the area from the table from 1, and if you want the area to the right: use the same table but disregard the negative sign to find the area above your z-score. Basically, it's the reverse of what you did for a + z-score.

to answer part e) you'll need to look at the answers to the first 4 parts and interpret them to construct a sentence to support your answer.

Answer 2

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

To start with, you might want to visualize the standard deviation as the length of a measuring string that you will use to determine how far a given amount is away from the mean. In this case, that string would be 2.65 inches long, as given above. It is also given above that the mean height of men in America is 69.2 inches. Subtracting that from 75 inches (i.e. 6 feet 3 inches) you are left with 5.8 inches as the amount this man is taller than the mean. However, you do not want to know the inches; you want to know how many lengths of the 2.65 inch measuring string would it take to cover that distance, Two lengths of the string would cover 5.3 inches (i.e. 2 x 2.65 inches) and 3 lengths would cover 7.95 inches, so it we know that our answer should be a little more than 2 lengths of the string.

The z-score is the number of string lengths it would take to cover the distance that an specified value--in this case, the height--is away from the mean, and it is calculated as follows:

.

z = (Specified value - Mean) / Standard Deviation = (75 inches - 69.2 inches) / 2.65 inches = 5.8 / 2.65 = 2.19

z = 2.19

(Note that both the numerator and the denominator are in inches so they cancel out each other leaving z as just a dimensionless number)

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

The percentage can be calculated by hand, but that can be a bit complicated. Fortunately, there are other ways to determine the percentage. Many calculators and programs (like MS Excel) have built in functions to give you this value, but it might be easiest to understand if you use a Z-Score table (also known as a Standard Normal Distribution Table). There is more than one way to display the table; but, for most I have seen, you would find the percentage using the following steps:

With 2.19 as the Z-score:

1. In the first column, find the row that contains the 2.1 (from the 2.19)
2. In the first row, find the column that contains the 0.09 (from hundredth column of the 2.19)
3. Find the value in the cell where the row from step 1 and the column from step 2 intersect.

The value found in step 3 is .4857, which is equivalent to 48.57%. This is the percentage of the total population with heights between the mean (69.2 inches) and the height of the man (75 inches). With a normal distribution, 50% of the population will have heights below the mean height of 69.2 inches. Therefore, the percentage of men shorter than 6 feet 3 inches will be the sum of 50% (the percentage with heights below 69.2 inches) and 48.57% (the percentage with heights between 69.2 inches and 75 inches), which is 98.57%

Note: Instead of using the standard Z-Table, you could avoid the final step of adding 50% to the value by using a Left Z-Table, also known as a Standard Normal Cumulative Probability Table. With it, and using the same step discussed above, you will directly get the value .9857 or 98.57%.

98.6% are shorter than the 6' 3" man.

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

Given:

Mean: 64.8 inches

Std Dev: 2.57 inches

Height: 71 Inches (i.e. 5 feet 11 inches)

z = (71 inches - 64.8 inches) / 2.57 inches = 6.2 inches / 2.57 inches = 2.41

z = 2.41

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

Value from Z-Table: 0.4920

This indicates that 49.20% of women in America have heights between the mean (64.8 inches) and 71 inches; and, since 50% of the women have heights below the 64.8 inch mean, the sum of those percentages will give you the percentage of women in America with heights below 71 inches.

50 % + 49.20% = 99.20% = the percentage of women in America with heights below 71 inches. To determine the percentage that are taller than that, you would subtract that percentage from 100%.

100% - 99.20 % = 0.80%

Therefore, only about 0.8% of women in America are taller than 71 inches.

0.8%

e) Who is relatively taller: a 6' 3" American man or a 5' 11" American woman? Defend your choice in a meaningful sentence.

It was shown above that 98.6% of men in America are shorter than the 6' 3" man, which means that only 1.4% (100% - 98.6%) are taller than that man. However, only 0.8% of women in America are taller than the 5' 11" woman. For this reason, one could argue that, with respect to the gender of each, the woman is relatively taller than the man.