Chapter 12.3, Problem 19PPE

To determine

To find: The range and mean of the given data sets of each baseball player.

Answer to Problem 19PPE

  $\begin{array}{l}\text{Player 1: Range}=0.62;\text{Mean}=0.300\\ \text{Player 2: Range}=0.29;\text{Mean}=0.301\end{array}$

Explanation of Solution

Given information:

Two data sets are given as below −

  $\begin{array}{l}\text{Player 1 batting averages: 0}\text{.265}0.3270.2940.3160.2810.318\\ \text{Player 2 batting averages: 0}\text{.304}0.2850.3120.2910.3030.314\end{array}$

Calculation:

Given data sets are −

  $\begin{array}{l}\text{Player 1 batting averages: 0}\text{.265}0.3270.2940.3160.2810.318\\ \text{Player 2 batting averages: 0}\text{.304}0.2850.3120.2910.3030.314\end{array}$

For $\text{Player 1 batting averages: 0}\text{.265}0.3270.2940.3160.2810.318$ ,

Range of a data set is the difference between the greatest and least data value

  $\text{Range}=0.327-0.265=0.62$

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ \Rightarrow \text{Mean}=\frac{0.265+0.327+0.294+0.316+0.281+0.318}{6}\\ \Rightarrow \text{Mean}=0.300\end{array}$

For $\text{Player 2 batting averages: 0}\text{.304}0.2850.3120.2910.3030.314$

Range of a data set is the difference between the greatest and least data value

  $\text{Range}=0.314-0.285=0.29$

  $\begin{array}{l}\text{Mean}=\frac{\text{Sum of data values}}{\text{Total number of data values}}\\ \Rightarrow \text{Mean}=\frac{0.304+0.285+0.312+0.291+0.303+0.314}{6}\\ \Rightarrow \text{Mean}=0.301\end{array}$

Hence, for the given two data sets −

  $\begin{array}{l}\text{Player 1: Range}=0.62;\text{Mean}=0.300\\ \text{Player 2: Range}=0.29;\text{Mean}=0.301\end{array}$

Both players have approx. similar mean batting averages across six seasons but player $2$ is more consistent performer as ranges is less.