Math Ms Sue can you help
posted by Liz .
Here is the height, in inches, of 10 randomly selected members of the girls' dance team and 10 randomly selected of the girls' volleyball team
Based on the interquartile ranges of the two sets of data, which is a reasonable conclusion concerning heights of the players on the two teams?
Volley ball team:67,63,70,67,68,69.70,68,72,68
Dance team:62,67,59,63,67,60,66,60,66,63
A. there is a greater variability of heights for the middle 50% of the dance team than for the middle50% of the volleyball team members
B. There is a greater variability of heights for the middle 50% of the volleyball team than for the middle50% of the dance team members
c. The variability for the 50% was the same for both team members
D. the average height, in inches, was higher for the members of the dance than for the members of the volleyball team.
I had B. since 68 is the middle and the is 5 numbers before it to get to the lowest number,63, and 4 to get to the highest number,72.
In dance 63, 59 is the lowest nuber, being 4 away and the highest is 67, being 4 away., so volleyball team has more variability.
Is this correct. I don't know if I did this correctly, can you help me; thanks

Is there any reason you did not compute the standard deviation for each set of data? Standard deviation indicates variability.

I didn't do the standard deviation for each one because I didn't know how, can you help?, if not do you have something ( a link to forward me to) so I can review standard deviations.
I do appreciate your help. 
Do you have a statistical calculator? http://easycalculation.com/statistics/standarddeviation.php
Your text will tell you how to do it manually, if you want to groan through it, all students need to do that once. 
so the volleyball team had a standard deviation 2.39 and a mean of 68.2
( I messed up when giving you the numbers, there should have been a comma between 69 &70 not a periodgiving 10 numbers)
The dance team I got 3.05 standard deviation and a mean of 63
So the should the answer be A.