Heights of young adult U.S. women are approximately normal with mean 64" and standard deviation 2.7". What proportion of all U.S. young adult women are taller than 6 feet?
Convert 6' to inches.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To solve this question, we need to convert the height of 6 feet to inches because the given mean and standard deviation are in inches.
1 foot is equal to 12 inches, so 6 feet would be 6 * 12 = 72 inches.
Now, let's calculate the z-score for a height of 72 inches using the formula:
z = (x - μ) / σ
where:
x = height of 6 feet (72 inches)
μ = mean height (64 inches)
σ = standard deviation (2.7 inches)
z = (72 - 64) / 2.7
z ≈ 2.96
Next, we can use a standard normal distribution table or a calculator to find the proportion of the area under the normal curve beyond a z-score of 2.96.
Looking up the z-score of 2.96 in a standard normal distribution table, we find that the proportion of the area to the right of this z-score is approximately 0.0014.
Therefore, approximately 0.14% of all U.S. young adult women are taller than 6 feet.