Posted by Ariellen on Friday, August 27, 2010 at 7:59pm.
Start with your scores at the mean/median. Spread them out from there, with extreme scores at 3 and 45 (mean/median ± 21).
I found this answer on another site, it is accurate. I had this same problem for my stats class.
The easiest way to do this would be to construct the set based on the conditions. So start from the middle and work out.
The median of the 20 numbers is 24. The median is the middle point and since there is two middle points (the tenth and eleventh number) it's the average of the two. But to make it easy just make both the tenth and eleventh number 24.
Since the average (mean) of the numbers is 24, if you add the same amount to the right as you subtract from the left, you'll always maintain the average of 24. So add or subtract (1) to each side.
on the right side you would have {...24, 25, 26,27,28,29, 30 31, 32, 33}
on the left you would have {15, 16, 17, 18, 19, 20, 21, 22, 23, 24...}
this would give the set {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 24, 25, 26,17,18,19, 30 31, 32, 33}
now, this set fits three of the four condition, but the range (the largest - the smallest) is not 42, its 18 ((33-15=18.)
but that's easy enough to find. We want to find your minimum (min) and your maximum (max) with the following conditions :
1. max - min = 42
2. 24 - x = min
3. 24 + x = max
substituting 3 and 2 back into 1, you get the following equation
(24 + x) - (24 - x) = 42
24 + x - 24 + x = 42
24 - 24 + 2x = 42
2x = 42
x = 21
therefore, your max is (24 +21)= 45 and your min is (24 - 21) = 3
replace those max and min with the max and min from the previous set and you get an answser.
{3,16, 17, 18, 19, 20, 21, 22, 23, 24, 24, 25, 26,27,28,29, 30 31, 32, 45}