Chapter 10.3, Problem 8IP

To determine

To calculate:The mean absolute deviation of the given set of values.

Answer to Problem 8IP

The Mean Absolute Deviation for the given sets are

  $\begin{array}{l}\left\{1,2\right\},M.A.D=0.5\\ \left\{1,2,3,4\right\},M.A.D=1\\ \left\{1,2,3,4,5,6\right\},M.A.D=1.5\\ \left\{1,2,3,4,5,6,7,8,9,10\right\},M.A.D=2.5\end{array}$

Explanation of Solution

Given information:The given set of values are

  $\left\{1,2\right\},\left\{1,2,3,4\right\},\left\{1,2,3,4,5,6\right\}\&\left\{1,2,3,4,5,6,7,8,9,10\right\}$

Formula used:

The Mean of a given set of values is the average of the given set of values.

  $Mean=\frac{\text{sum}\text{of}\text{values}}{\text{number}\text{of}\text{values}}=\frac{{\displaystyle \sum x}}{n}$

The Mean Absolute Deviation of a given set of values is the average of deviation of each value from the mean value.

  $M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}$

Calculation:

The value of $\text{mean}\left(\overline{x}\right)\&\text{mean}\text{absolute}\text{deviation}\left(M.A.D\right)$ for the set $\left\{1,2\right\}$ are shown below

  $\begin{array}{l}\text{set=}\left\{1,2\right\},n=2\left(even\right)\\ Mean=\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{\left(1+2\right)}{2}=1.5\\ \\ \text{New}\text{set}\text{of}\text{values}\text{of}\left(|x-\overline{x}|\right)\text{=}\left\{0.5,0.5\right\}\\ M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}=\frac{0.5+0.5}{2}=0.5\\ M.A.D=0.5\mathrm{............}\left(1\right)\end{array}$

The Mean Absolute Deviation for the set $\left\{1,2\right\}$ is 0.5

The value of $\text{mean}\left(\overline{x}\right)\&\text{mean}\text{absolute}\text{deviation}\left(M.A.D\right)$ for the set $\left\{1,2,3,4\right\}$ are shown below

  $\begin{array}{l}\text{set=}\left\{1,2,3,4\right\},n=4\left(even\right)\\ Mean=\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{\left(1+2+3+4\right)}{4}=2.5\\ \\ \text{New}\text{set}\text{of}\text{values}\text{of}\left(|x-\overline{x}|\right)\text{=}\left\{1.5,0.5,0.5,1.5\right\}\\ M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}=\frac{1.5+0.5+0.5+1.5}{4}=1\\ M.A.D=1\mathrm{........}\left(2\right)\end{array}$

The Mean Absolute Deviation for the set $\left\{1,2,3,4\right\}$ is 1.

The value of $\text{mean}\left(\overline{x}\right)\&\text{mean}\text{absolute}\text{deviation}\left(M.A.D\right)$ for the set $\left\{1,2,3,4,5,6\right\}$ are shown below

  $\begin{array}{l}\text{set=}\left\{1,2,3,4,5,6\right\},n=6\left(even\right)\\ Mean=\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{\left(1+2+3+4+5+6\right)}{6}=3.5\\ \\ \text{New}\text{set}\text{of}\text{values}\text{of}\left(|x-\overline{x}|\right)\text{=}\left\{2.5,1.5,0.5,0.5,1.5,2.5\right\}\\ M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}=\frac{2\times (0.5+1.5+2.5)}{6}=1.5\\ M.A.D=1.5\mathrm{.........}\left(3\right)\end{array}$

The Mean Absolute Deviation for the set $\left\{1,2,3,4,5,6\right\}$ is 1.5.

By observing the pattern from equation (1),(2) & (3) it can be concluded that for every 2 new consecutive values added in the data-set the mean absolute deviation increases by 0.5.

Therefore the Mean Absolute Deviation for the set $\left\{1,2,3,4,5,6,7,8,9,10\right\}$ is 2.5.

Hence The Mean Absolute Deviation for the given sets are

  $\begin{array}{l}\left\{1,2\right\},M.A.D=0.5\\ \left\{1,2,3,4\right\},M.A.D=1\\ \left\{1,2,3,4,5,6\right\},M.A.D=1.5\\ \left\{1,2,3,4,5,6,7,8,9,10\right\},M.A.D=2.5\end{array}$