Chapter B.2, Problem 13E

i. The Mean of each team member.

To determine

Mean Score.

Answer to Problem 13E

• Mean(Jay) = 169.67
• Mean(Hank) = 199.67
• Mean(Buck) = 209.33

Explanation of Solution

Given:

Team Member	Game 1	Game 2	Game 3
Jay	141	202	166
Hank	199	195	205
Buck	182	231	215

Calculation:

  $\text{Mean/Avg}\text{.=}\frac{\text{Sum of Observations}}{\text{Total Number Of Observations}}$

  $\begin{array}{l}\Rightarrow {\text{Mean}}_{\mathrm{J}\mathrm{a}\mathrm{y}}=\frac{\text{Total Score of 3 Games}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{J}\mathrm{a}\mathrm{y}}=\frac{\text{141+202+166}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{J}\mathrm{a}\mathrm{y}}=\frac{509}{3}\text{ = 169}\text{.67}\mathrm{.........................................................}\left(1\right)\end{array}$

  $\begin{array}{l}\Rightarrow {\text{Mean}}_{\mathrm{H}\mathrm{a}\mathrm{n}\mathrm{k}}=\frac{\text{Total Score of 3 Games}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{H}\mathrm{a}\mathrm{n}\mathrm{k}}=\frac{\text{199+195+205}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{H}\mathrm{a}\mathrm{n}\mathrm{k}}=\frac{599}{3}\text{ = 199}\text{.67}\mathrm{.........................................................}\left(2\right)\end{array}$

  $\begin{array}{l}\Rightarrow {\text{Mean}}_{\mathrm{B}\mathrm{u}\mathrm{c}\mathrm{k}}=\frac{\text{Total Score of 3 Games}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{B}\mathrm{u}\mathrm{c}\mathrm{k}}=\frac{\text{182+231+215}}{3}\\ \Rightarrow {\text{Mean}}_{\mathrm{B}\mathrm{u}\mathrm{c}\mathrm{k}}=\frac{628}{3}\text{ = 209}\text{.33}\mathrm{.........................................................}\left(3\right)\end{array}$

ii. The Mean and Median of all the 9 values.

To determine

Mean and Median.

Answer to Problem 13E

• Mean = 192.89
• Median = 199

Explanation of Solution

Given:

Team Member	Game 1	Game 2	Game 3
Jay	141	202	166
Hank	199	195	205
Buck	182	231	215

Calculation:

• Mean of all 9 Games:

  $\begin{array}{l}{\text{Mean}}_{(\text{9 Games})}\text{ = }\frac{\text{Total Score of 9 Games}}{\text{Total Number Of Games}}\\ \\ \text{Mean = }\frac{141+202+166+199+195+205+182+231+215}{9}\\ \\ \text{Mean = }\frac{1736}{9}=192.89\end{array}$

• Median of all 9 Games:

141	166	182	195	199	202	205	215	231
				Median

  $\begin{array}{l}{\text{Median}}_{\left(\text{9 Games}\right)}\text{= Central Value}\\ {\text{Median}}_{\left(\text{9 Games}\right)}\text{=199}\mathrm{.................................................................................}\left(4\right)\end{array}$

Conclusion:

• Mean = 192.89
• Median = 199

iii. Which measure of central tendency describes the 9 scores?

To determine

Measure of central tendency.

Answer to Problem 13E

Left-Skewed Distribution

Explanation of Solution

Given:

Team Member	Game 1	Game 2	Game 3
Jay	141	202	166
Hank	199	195	205
Buck	182	231	215

Calculation:

• Mean = 192.89
• Median = 199

Since, Mean < Median

The Data above is Left-Skewed Distributed.