The distribution of heights of a population of adults is approximately normal with mean 66 inches and SD 2.5 inches.

Question

1. Approximately what percent of the adults are over 6 feet tall?
2. Approximately what percent of the adults have heights that are within 1 inch of the average?
3. Approximately what percent of the adults are 70 inches tall, to the nearest inch?
4. Approximately what is the 90th percentile of the heights, in inches?

I REALLY NEED SOME HELP. Thank you in advance!

Answer 1

1. 6 feet = 72 inches

You need to find the z score

(number - mean) divided by SD
72 -66/2.5 = z score

Then use a standard z-table to find the area of under the curve to the right of that number. This number will be your %.

2. Find the z-score for 67 and 65. Find the area between those two scores using the z-table.

3. To the nearest inch. Are you looking for 69.5 to 70.4?

4. 90th percentile means that 90% of the values are below that number or 10% are above. Find the z-score from the table

set the z-score = (height - mean)/SD

Answer 2

Thank you Dr. Jane for giving a succinctly explanatory answer without doing all the work for HELPPPPP, although I do realize the deadline is approaching.
If you don't understand how to do a problem HELPPPPP, ask it on the discussion board and many will be happy to help you. And by the way, this was covered in lecture 5.

Answer 3

Hi guys, thanks for the above. I got all the equations right but my z score to percent was wrong. If I have a random answer z score. which I wont put due to honor code. of pretend 2.6, would my z score from the table be 2 on the left and .06 on the top (to find the intersection) or would I just use 2.6 and 0.0?? once I have that number - pretend its .9456 / would that be 94.56% or would it be 94.56-100 = 5.4%?? your help is soooooooooo appreciated.