Heights of young adult US women are approximately normal with mean 64" and standard deviation 2.7". What proportion of all US young adult women are taller than 6 feet?
Z = (score-mean)/SD
Z = (72-64)/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.
To find the proportion of US young adult women who are taller than 6 feet, we need to convert the height of 6 feet into the same units as the mean and standard deviation. Since the mean is given in inches, we need to convert 6 feet into inches.
First, let's convert 6 feet into inches:
1 foot = 12 inches
6 feet = 6 * 12 inches = 72 inches
Now we know that we need to find the proportion of young adult women who are taller than 72 inches.
To find this proportion, we will use the standard normal distribution and the Z-score formula. The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value we want to find the proportion for
μ is the mean of the distribution
σ is the standard deviation of the distribution
Now, substituting the values we have:
Z = (72 - 64) / 2.7
Calculating this, we get:
Z ≈ 2.96
Using a Z-table or a calculator that can calculate Z-scores, we can find the proportion of values greater than 2.96. Consulting a Z-table, we find that the proportion is approximately 0.0014.
Therefore, approximately 0.14% of all US young adult women are taller than 6 feet.