heights of young adult u.s women are approximatelynormal with mean 64 inches and standard deviation 2.7 inches. What proportion of all u.s. young adult women are taller than 6 feet?
Z = (score - mean)/SD
Z = [(6*12)-64]/2.7
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
0.15%
Where'd the 12 come from? PLEASE HELP!
12 inches in one foot
To find the proportion of all U.S. young adult women who are taller than 6 feet, we need to convert 6 feet into inches. Since there are 12 inches in a foot, 6 feet is equal to 6 * 12 = 72 inches.
We now have the mean (μ) of the height distribution, which is 64 inches, and the standard deviation (σ), which is 2.7 inches. We can use these values to calculate the proportion.
First, we need to find the z-score, which measures the number of standard deviations an individual is from the mean. We can calculate the z-score using the formula:
z = (X - μ) / σ
where X is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case, we want to find the proportion of women taller than 72 inches, so X = 72.
z = (72 - 64) / 2.7
z = 8 / 2.7
z ≈ 2.963
Once we have the z-score, we can use a standard normal distribution table or a calculator to find the proportion of women taller than 72 inches.
Using the standard normal distribution table, we can look up the z-score of 2.963 and find the corresponding proportion, which is approximately 0.9974.
So, approximately 0.9974 or 99.74% of all U.S. young adult women are taller than 6 feet.