# Thread: Help with Application of a Normal Disribution Problem

1. ## Help with Application of a Normal Disribution Problem

The story goes something like this:
Women have a mean height of 63.8 inches with a standard deviation of 2.6.
Men have a mean height of 69.5 inches with a standard deviation of 2.4

The Blah-Blah company requires its drivers to be between 64 inches and 77 inches tall.

What percent of women meet this height requirement? Got this answer
What percent of men meet this height requirement? Got this answer

If the Blah-Blah company changed height requirements to exclude the tallest 3% of men and the shortest 3% of women, what are the new height requirements?

I thought for shortest: (x-63.8)/2.6 = z score from table for .03 or .5120 which would change the minimum to 65".

I thought for tallest: (x-69.5)/2.4 = z score from table for .97 or .8340 which would change the maximum to 71.5".

Now, I might buy that the minimum went from 64 to 65 but no way am I going to believe the maximum went from 77 to 71.5. What am I doing wrong?

2. ## Re: Help with Application of a Normal Disribution Problem

Hey philenaf.

You will need to calculate a probability - can you please show us your working out? [You are correct on getting the z-score but for an interval you will need to get two and then find the probability from that].

3. ## Re: Help with Application of a Normal Disribution Problem

I set up an equation (x-69.5)/2.4 = .9699 where .9699 is the area to the left of 97%
the other equation was (x-63.8)/2.6 = .0301 where .0301 is the area to the left of 3%

4. ## Re: Help with Application of a Normal Disribution Problem

There is no probability - only the new height amounts