Answer:
a) The sold games of Jezebel had the greatest spread.
b) The middle 50% of the games sold by Colin has the least spread.
c) Jezebel sold more games than colin.
Explanation:
Range is sued to calculate the spread of the given data.
Range is given by:

So,
Part a)
For Collin:
Range=21-3=18
For Jezebel:
Range=26-5=21
Part b)
Middle 50% of the values will be the values between 1st quartile and 3rd quartile
So, to find quartiles:
For Colin:
=3,9,9,14,15,16,17,20,21
The median is 15.
The lower half is 3,9,9,14
Q1 = (9+9)/2= 18/2 = 9
The upper half is 16,17,20,21
Q3 = (17+20)/2= 37/2= 18.5
The middle 50% is first quartile subtracted from third quartile
So, the spread is:
18.5-9=9.5
For Jezebel:
4,5,8,10,11,14,20,20,26
The median is 11.
The lower half is 4,5,8,10
Q1 = (5+8)/2=13/2=6.5
The upper half is 14,20,20,26
Q3 = (20+20)/2=40/2=20
The middle 50% values' spread is:
20-6.5=13.5
Part c) The answer to part a tells us that Jezebels sold games have more spread than Colin's sold games. Similarly the answer of part b tells us that the spread of middle 50% values of Jezebel's sold games was more than the spread of middle 50% values of Colin's sold games ..