Answer:
Both distributions are left skewed, so we can use the median and interquartile range.
The median for Team A is [tex]4[/tex]
The median for Team B is [tex]8[/tex]
The interquartile range for Team A [tex]=Q_{3}-Q_{1}=5-2=3[/tex]
The interquartile range for Team B [tex]=Q_{3}-Q_{1}=9-6=3[/tex]
Now we can express the difference in the measures of center as a multiple of the measures of variation as:
[tex]\frac{Median_{A}-Median_{B}}{IQR}=\frac{4-8}{3}=-1.33[/tex]
So, the difference in the medians is about 1.33 times the Interquartile range.
