Heights of adult men have a mean of 69.0 inches and a standard deviation of 2.8 inches. Approximately what percentage of adult men have a height between 66.2 and 77.4 inches? Must show the number and the empirical rule

You have to find z scores first.

66.2-69.0 then divide by 2.8

77.4 - 69.0 then divide by 2.8

Now, you have to find the area between those two z scores. The empirical rule tells you the % between -3 and 3 standard deviations, between -2 and 2 sds and between -1 and 1 standard deviations.

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To calculate the percentage of adult men with heights between 66.2 and 77.4 inches, we can use the empirical rule, also known as the 68-95-99.7 rule. This rule states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

Given the mean height of adult men as 69.0 inches and the standard deviation as 2.8 inches, we can calculate the number of standard deviations between 66.2 and 77.4 inches by using the formula:

Z = (X - μ) / σ

Where:
Z is the number of standard deviations from the mean
X is the specific value (height in this case)
μ is the mean
σ is the standard deviation

For the lower bound, Z1:

Z1 = (66.2 - 69.0) / 2.8

Z1 ≈ -0.93

For the upper bound, Z2:

Z2 = (77.4 - 69.0) / 2.8

Z2 ≈ 3.00

Now, we can determine the percentage using the empirical rule:

- From the empirical rule, we know that approximately 68% falls within one standard deviation of the mean. This includes the area from -1 to 1 standard deviation.
- Additionally, approximately 95% falls within two standard deviations of the mean. This includes the area from -2 to 2 standard deviations.
- And approximately 99.7% falls within three standard deviations. This includes the area from -3 to 3 standard deviations.

Since the range between -1 and 1 standard deviations covers 68% of the distribution, we can subtract that from the total range to find the percentage for the desired range of 66.2 to 77.4 inches.

Percentage = (1 - (0.68 + 0.95)) * 100

Percentage ≈ (1 - 0.63) * 100

Percentage ≈ 37%

So, approximately 37% of adult men have a height between 66.2 and 77.4 inches.

To find the approximate percentage of adult men who have a height between 66.2 and 77.4 inches using the empirical rule, we can first calculate the Z-scores for both heights.

The Z-score is a measure of how many standard deviations a value is away from the mean. It can be calculated using the formula:

Z = (X - μ) / σ

Where:
- X is the value we want to measure (height in this case)
- μ is the mean of the distribution (69.0 inches)
- σ is the standard deviation of the distribution (2.8 inches)

Calculating the Z-score for the lower height:

Z1 = (66.2 - 69.0) / 2.8
Z1 ≈ -0.9643

Calculating the Z-score for the higher height:

Z2 = (77.4 - 69.0) / 2.8
Z2 ≈ 3.0000

Now, using the empirical rule, we know that approximately:

- 68% of data falls within one standard deviation of the mean.
- 95% falls within two standard deviations.
- 99.7% falls within three standard deviations.

If we consider the interval between -1 and +3 standard deviations, it covers approximately 99.7% of the data. Since the Z-scores for the lower and higher heights fall within this range (-0.9643 is greater than -1, and 3.0000 is less than 3), we can conclude that approximately 99.7% of adult men will have a height between 66.2 and 77.4 inches.

Therefore, the approximate percentage of adult men with heights in this range is 99.7%.