Question

Use the Probability Calculator to answer the following problem.

According to the U.S. National Health Statistics, the mean height of an adult female in the United States is 63.8 inches tall and
standard deviation 2.1 inches. Enter solution as a percentage rounded to two decimal places, l.e.: 0.1885 to 18.85%,
Singer Carrie Underwood is 5 feet, 3 inches tall. What percentage of U.S. adult females are shorter than Underwood?
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Answer 1

To find the percentage of U.S. adult females who are shorter than Carrie Underwood, calculate the z-score using the formula z = (x - μ) / σ, where x is the height of Carrie Underwood, μ is the mean height of adult females in the U.S., and σ is the standard deviation of adult female heights. Then, use the z-score in the standard normal distribution table to find the percentage.

Step-by-step explanation:

To find the percentage of U.S. adult females who are shorter than Carrie Underwood, we need to calculate the z-score and use the z-score in the standard normal distribution table to find the percentage. The z-score formula is:
z = (x - μ) / σ
where x is the height of Carrie Underwood, μ is the mean height of adult females in the U.S., and σ is the standard deviation of adult female heights.

First, convert Carrie Underwood's height from feet and inches to inches:

5 feet = 5 * 12 inches = 60 inches

So, Carrie Underwood's height is 60 + 3 = 63 inches.

Next, calculate the z-score:

z = (63 - 63.8) / 2.1

Finally, use the z-score in the standard normal distribution table to find the percentage of U.S. adult females who are shorter than Carrie Underwood.