Chapter 10.3, Problem 11HP

To determine

To Find: To find the data whose mean absolute deviation is 5.

Answer to Problem 11HP

Conducting a test of 5 students in two different days

Explanation of Solution

Given information:Given mean absolute deviation is 5.

Formula used: For Mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{N}$ and for mean absolute deviation $MD=\frac{{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}}{N}$

Calculation:

Let $x_i\text{and}{y}_i$ are the marks represented by 5 students in two different tests out of 20 marks.

$x_i$	2.5	5	10	15	17.5
$y_i$	0	7.5	12.5	13	17

Calculation of Mean

  $\begin{array}{l}\overline{x}=\frac{2.5+5+10+15+17.5}{5}\\ \overline{x}=\frac{50}{5}\\ \therefore \overline{x}=10\end{array}$

Similarly,

  $\begin{array}{l}\overline{y}=\frac{0+7.5+12.5+13+17}{5}\\ \overline{y}=\frac{50}{5}\\ \therefore \overline{y}=10\end{array}$

Thus, the average score in both the months are 104 and 125 respectively.

Now, calculation of mean absolute deviation

$x_i$	$x_i-\overline{x}$	$\left|x_i-\overline{x}\right|$	$y_i$	$y_i-\overline{y}$	$\left|y_i-\overline{y}\right|$
2.5	$2.5-10=-7.5$	7.5	0	$0-10=-10$	10
5	$5-10=-5$	5	7.5	$7.5-10=-2.5$	2.5
10	$10-10=0$	0	12.5	$12.5-10=2.5$	2.5
15	$15-10=5$	5	13	$13-10=3$	3
17.5	$17.5-10=7.5$	7.5	17	$17-10=7$	7

  $\begin{array}{l}\text{MD}\text{of}x=\frac{7.5+5+0+5+7.5}{5}\\ \text{MD}\text{of}x=\frac{25}{5}\\ \text{MD}\text{of}x=5\end{array}$

And

  $\begin{array}{l}\text{MD}\text{of}y=\frac{10+2.5+2.5+3+7}{5}\\ \text{MD}\text{of}y=\frac{25}{5}\\ \text{MD}\text{of}y=5\end{array}$

Thus, mean absolute deviation is 5 in both situations

Hence, the required data satisfies the given conditions.