Final answer:
Using the given mean and standard deviation for women's heights, Mary's height, which is taller than 70% of the population, is closest to Option B, 66.4 inches, after finding the corresponding Z-score and applying the formula for a normal distribution.
Step-by-step explanation:
The question asks us to determine the height of Mary, who is taller than 70% of the population of U.S women, based on the given normal distribution of women's heights in the United States. The mean height is 64.5 inches and the standard deviation is 2.3 inches. To find Mary's height, we use the concept of Z-scores in a normal distribution, which can be looked up in a Z-table or calculated using statistical software or a calculator.
The percentage indicated (70%) needs to be converted into a Z-score. The Z-score that corresponds to the 70th percentile (which is the same as being taller than 70% of the population) can be found by using a Z-table, statistical software, or a calculator. For a standard normal distribution, the Z-score for the 70th percentile is approximately 0.52.
Once we have the Z-score, we can use the formula:
X = μ + (Z × σ)
where X is the height we want to find, μ is the mean height, σ is the standard deviation, and Z is the Z-score.
Substituting the values we have:
X = 64.5 + (0.52 × 2.3) = 64.5 + 1.196 = 65.696 inches
Since this is not an available option, we can identify that Mary's height should be closest to the next highest option, which is Option B, 66.4 inches.