Suppose the heights of women aged 20 to 29 follow approximately the N(66.7, 2.4) distribution. Also suppose men the same age have heights distributed as N(70.5, 2.2).
What percent (± 0.1) of young men are shorter than average young women?
Z = (female mean-male mean)/male SD
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 percentage of young men who are shorter than the average young women, we need to compare the heights of the two groups and determine the proportion of men below the average height of women.
Step 1: Calculate the z-score for the average height of women using the formula:
z = (x - μ) / σ
Where:
x = average height of women = 66.7
μ = mean height of women = 66.7
σ = standard deviation of women's heights = 2.4
Using the values given, we have:
z = (66.7 - 66.7) / 2.4
z = 0
A z-score of 0 means that the average height of women is at the mean of the distribution.
Step 2: Calculate the proportion of men below the average height of women using the z-score:
P(z < 0) = 0.5
Since the z-score of 0 corresponds to the mean of the distribution, 0.5 is the cumulative probability to the left of the mean.
Step 3: Convert the proportion to a percentage:
Percentage = 0.5 * 100
Percentage = 50%
Therefore, approximately 50% of young men are shorter than the average young women.