Chapter 10.3, Problem 16STP

To determine

To find:Which data set has a mean absolute deviation of $4$ .

Answer to Problem 16STP

The data set of option H has the mean absolute deviation of $4$ .

Explanation of Solution

Given information:

Data set F is $1,6,11$ .

Data set G is $0,12,24$ .

Data set H is $3,10,14$ .

Data set J is $4,5,9$ .

Calculation:

Finding the mean of F,

  $\begin{array}{l}\text{mean}=\frac{1+6+11}{3}\\ \text{mean}=6\end{array}$

Finding the mean absolute deviation of F,

  $\begin{array}{c}\left|1-6\right|=5\\ \left|6-6\right|=0\\ \left|11-6\right|=5\\ \text{mean}\text{absolute}\text{deviation}=\frac{5+0+5}{3}\\ \text{mean}\text{absolute}\text{deviation}=3.\overline{3}\end{array}$

Therefore option F is does not have the mean absolute deviation of $4$ .

Finding the mean of G,

  $\begin{array}{l}\text{mean}=\frac{0+12+24}{3}\\ \text{mean}=12\end{array}$

Finding the mean absolute deviation of G,

  $\begin{array}{c}\left|0-12\right|=12\\ \left|12-12\right|=0\\ \left|24-12\right|=12\\ \text{mean}\text{absolute}\text{deviation}=\frac{12+0+12}{3}\\ \text{mean}\text{absolute}\text{deviation}=8\end{array}$

Therefore option G is does not have the mean absolute deviation of $4$ .

Finding the mean of H,

  $\begin{array}{l}\text{mean}=\frac{3+10+14}{3}\\ \text{mean}=9\end{array}$

Finding the mean absolute deviation of H,

  $\begin{array}{c}\left|3-9\right|=6\\ \left|10-9\right|=1\\ \left|14-9\right|=5\\ \text{mean}\text{absolute}\text{deviation}=\frac{6+1+5}{3}\\ \text{mean}\text{absolute}\text{deviation}=4\end{array}$

Therefore option H is does have the mean absolute deviation of $4$ .

Finding the mean of J,

  $\begin{array}{l}\text{mean}=\frac{4+5+9}{3}\\ \text{mean}=6\end{array}$

Finding the mean absolute deviation of J,

  $\begin{array}{c}\left|4-6\right|=1\\ \left|5-6\right|=1\\ \left|9-6\right|=3\\ \text{mean}\text{absolute}\text{deviation}=\frac{1+1+3}{3}\\ \text{mean}\text{absolute}\text{deviation}=2\end{array}$

Therefore option J is does not have the mean absolute deviation of $4$ .

Thus we can conclude that option H has the mean absolute deviation of $4$ .