Question

(13 points)

Answer 1

Answer:

Both distributions are left skewed, so we can use the median and interquartile range.

The median for Team A is $4$

The median for Team B is $8$

The interquartile range for Team A $=Q_{3}-Q_{1}=5-2=3$

The interquartile range for Team B $=Q_{3}-Q_{1}=9-6=3$

Now we can express the difference in the measures of center as a multiple of the measures of variation as:

$\frac{Median_{A}-Median_{B}}{IQR}=\frac{4-8}{3}=-1.33$

So, the difference in the medians is about 1.33 times the Interquartile range.