Okay, here is the problem I'm stuck on

The heights of women aged 20 to 29 follow approximately the N(64, 2.7) distribution. Men the same age have heights distributed as N(69.3,2.8). What percent of young men are shorter than the mean height of young women?

I have been trying to figure out where to even begin and I have no clue! If someone can show me how to solve this that would be great!!

You know that the mean height of women in this sample is 64. To determine what percentage of men who are shorter than that, you first o determine what the z-score is for men whse heihgt is 64. Z-score is found from:


z = \frac {x - \mu} \sigma = \frac {64-69.3} {2.8} = -1.9


So now you look up the percentage of the normal distribution that is 1.9 standard deviations or more from the mean. Hope this helps.