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 understand finding the z score but I'm not sure how to find the percent part. Help?
Are you using tables, charts, or software?
I assume charts. It's an online class and I'm basically teaching myself.
z = (x -mean)/ sd
z = (72-68)/2
z= 4/2= 2
1-.9772 = .0228
2.28%
I saw this example (I found the same question with different numbers) and it made sense to me until the 1-.9772 = .0228
Where is that coming from?
To find the percent of men who are over 6 feet tall, we first need to convert the height of 6 feet into inches since the given mean and standard deviation are in inches.
Since 1 foot is equal to 12 inches, 6 feet is equal to 6 x 12 = 72 inches.
Next, we need to calculate the z-score of the height of 72 inches. The z-score formula is:
z = (x - μ) / σ
Where:
x = the value we want to convert to a z-score (72 inches in this case)
μ = the mean of the distribution (69.0 inches in this case)
σ = the standard deviation of the distribution (2.8 inches in this case)
So, plugging in the values, we have:
z = (72 - 69.0) / 2.8 = 1.071
Now, we need to find the area under the normal distribution curve to the right of the z-score of 1.071.
Using a standard normal distribution table or a calculator, we find that the area to the right of 1.071 is approximately 0.1419.
To find the percentage, we multiply the decimal value by 100:
0.1419 * 100 = 14.19%
Therefore, approximately 14.2% of men are over 6 feet tall.