Question

Which set of four test scores would have the biggest difference between the mean and the median? A. 32, 85, 89, 91, 92, 92 B. 72, 78, 79, 80, 82, 82 C. 65, 70, 75, 80, 85, 90 D. 30, 34, 39, 41, 48, 50

Answer 1

Answer:

A. 32, 85, 89, 91, 92, 92

Step-by-step explanation:

A. 32, 85, 89, 91, 92, 92

Mean

$\bar{x}=\dfrac{32+85+89+91+92+92}{6}\\\Rightarrow \bar{x}=80.167$

Since the number of scores is 6 which is an even number the median will be of the form $\dfrac{x_{n/2}+x_{(n/2)+1}}{2}$

Median

$\dfrac{89+91}{2}=90$

Difference of mean and median = $|80.167-90|=9.83$

B. 72, 78, 79, 80, 82, 82

Mean

$\bar{x}=\dfrac{72+78+79+80+82+82}{6}\\\Rightarrow \bar{x}=78.83$

Median

$\dfrac{79+80}{2}=79.5$

Difference of mean and median = $|78.83-79.5|=0.67$

C. 65, 70, 75, 80, 85, 90

Mean

$\bar{x}=\dfrac{65+70+75+80+85+90}{6}\\\Rightarrow \bar{x}=77.5$

Median

$\dfrac{75+80}{2}=77.5$

Difference of mean and median = $|77.5-77.5|=0$

D. 30, 34, 39, 41, 48, 50

$\bar{x}=\dfrac{30+34+39+41+48+50}{6}\\\Rightarrow \bar{x}=40.33$

Median

$\dfrac{39+41}{2}=40$

Difference of mean and median = $|40.33-40|=0.33$

So, A has the maximum difference of 9.83 between the mean and the median.