Chapter B.2, Problem 41E

To determine

Compare the Central Tendency for the given data

• 5,12,7,14,8,9,7,12
• 5,12,7,24,8,9,7,12

What will be sensitive to extreme values?

Answer to Problem 41E

• Mean will vary and is sensitive to extreme values.

Explanation of Solution

Calculation:

• Arrange the data in ascending order

5

7

7

8

9

12

12

14

  $\begin{array}{l}\text{Mean = }\frac{\text{Sum of Observations}}{\text{Number of Observations}}\\ {\text{Mean}}_{(\mathrm{i})}=\frac{5+7+7+8+9+12+12+14}{8}\\ {\text{Mean}}_{(\mathrm{i})}=\frac{74}{8}=\text{ 9}\text{.25}\end{array}$

  $\begin{array}{l}\text{Median for Even number of observation}=\frac{{\text{Sum of 4}}^{\mathrm{t}\mathrm{h}}{\text{ \& 5}}^{\mathrm{t}\mathrm{h}}\text{ observation}}{2}\\ \text{Median = }\frac{8+9}{2}=\frac{17}{2}=8.5\end{array}$

• Arrange the data in ascending order

5

7

7

8

9

12

12

24

  $\begin{array}{l}{\text{Mean}}_{(\mathrm{i}\mathrm{i})}=\frac{5+7+7+8+9+12+12+24}{8}\\ {\text{Mean}}_{(\mathrm{i}\mathrm{i})}=\frac{84}{8}=\text{ 10}\text{.5}\end{array}$

  $\begin{array}{l}\text{Median for Even number of observation}=\frac{{\text{Sum of 4}}^{\mathrm{t}\mathrm{h}}{\text{ \& 5}}^{\mathrm{t}\mathrm{h}}\text{ observation}}{2}\\ \text{Median = }\frac{8+9}{2}=\frac{17}{2}=8.5\end{array}$

Since,

Median>Mean for both the cases, they both are Left-Skewed distribution.

Extreme Values have much more effect on Mean Value than the Median Value. The mean has shifted towards the right side on the number line as the extreme right value has changed/increased.

Conclusion:

Thus, Mean will vary and is sensitive to extreme values.