Suppose heights of men are normally distributed with a mean 69.0 inches and standard deviation 2.8 inches. What percent of men are over 6 feet tall? Round to the nearest tenth of a percent.
I understanf finding the z score, but I'm not sure how to find the percent.
Find the z-score
72-69 divided by 2.8 = 1.07 for a z-score.
What percent are over 6 feet tall, so we want the right tail.
You can use a z-table to find the value. Unfortunately, not all z-tables are constructed the same way. If you have a TI-83 or 84 calculator, you can use the DISTR function found above the VARS key.
2nd distr Normalcdf
you enter (1.07, 1000000000000) use a very large number to take you all the way out to the right tail. I got .1423 that means 14.23% would be taller than 6 feet.
If you are using a z-table, compare your answer with what the calculator gives.
Well, to find the percent of men over 6 feet tall, we need to convert 6 feet (or 72 inches) into a Z-score using the given mean and standard deviation. The formula to calculate the Z-score is:
Z = (X - μ) / σ
where X is the value we're interested in, μ is the mean, and σ is the standard deviation.
For 6 feet (72 inches), the Z-score would be:
Z = (72 - 69) / 2.8 ≈ 1.07
Now that we have the Z-score, we can use a Z-table to find the corresponding percentile. The Z-score of 1.07 corresponds to a percentile of approximately 85.75%. This means that about 85.8% of men are shorter than 6 feet tall.
To find the percentage of men over 6 feet tall, we subtract this percentile from 100%:
100% - 85.8% ≈ 14.2%
So, approximately 14.2% of men are over 6 feet tall. Keep in mind that this is an estimation, rounding to the nearest tenth of a percent.
To find the percentage of men who are over 6 feet tall, we need to convert the height of 6 feet into a z-score using the formula:
z = (x - μ) / σ
where:
- x is the value we want to convert to a z-score (in this case, 6 feet = 72 inches),
- μ is the mean of the distribution (69 inches), and
- σ is the standard deviation (2.8 inches).
Calculating the z-score:
z = (72 - 69) / 2.8 = 1.0714
Now, we need to find the percentage of men who are above this z-score. Using a standard normal distribution table or a calculator, we can find that the area to the left of z = 1.0714 is 0.8577.
However, since we want to find the percentage of men who are over 6 feet tall (the area to the right of 6 feet or z > 1.0714), the percentage can be calculated as:
Percentage = (1 - 0.8577) * 100
Percentage ≈ 14.2%
Therefore, approximately 14.2% of men are over 6 feet tall.
To find the percent of men who are over 6 feet tall (72 inches), you first need to convert the given value to a z-score using the formula:
z = (x - μ) / σ
where:
- z is the standard score (z-score),
- x is the given value,
- μ is the population mean (69 inches in this case),
- σ is the standard deviation (2.8 inches in this case).
Substituting the values into the formula:
z = (72 - 69) / 2.8 = 1.071
Once you have the z-score, you can determine the percentage using the standard normal distribution table (also known as the z-table). The z-table provides the area from the left-hand side of the distribution curve, which represents the percentage of data below a specific z-score.
Since we want to find the percentage of men over 6 feet tall, we need to find the area to the right of the z-score. This can be done by subtracting the area to the left of the z-score from 1.
Using the z-table or a calculator, you can find that the area to the left of a z-score of 1.071 is approximately 0.8589. Therefore, the area to the right (which represents the percentage of men over 6 feet tall) is:
1 - 0.8589 = 0.1411
Converting this to a percentage:
0.1411 * 100 ≈ 14.1%
Therefore, approximately 14.1% of men are over 6 feet tall.