Chapter 10.3, Problem 3IP

To determine

To Find: To find the mean absolute deviation of the given data.

Answer to Problem 3IP

The mean absolute deviation is 3.5

Explanation of Solution

Given information:Weights of guinea pigs in a pet store.

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:

Calculation of Mean

  $\begin{array}{l}\overline{x}=\frac{33+39+42+33+39+31+32+37}{8}\\ \overline{x}=\frac{286}{8}\\ \therefore \overline{x}=35.75\end{array}$

Thus, mean is 35.75

Now, calculation of mean absolute deviation

$x_i$	$x_i-\overline{x}$	$\left|x_i-\overline{x}\right|$
33	$33-35.75=-2.75$	2.75
39	$39-35.75=3.25$	3.25
42	$42-35.75=6.25$	6.25
33	$33-35.75=-2.75$	2.75
39	$39-35.75=3.25$	3.25
31	$31-35.75=-4.75$	4.75
32	$32-35.75=-3.75$	3.75
37	$37-35.75=1.25$	1.25
Total	0	28

  $\begin{array}{l}MD=\frac{28}{8}\\ MD=3.5\end{array}$

Thus, mean absolute deviation is 3.5

Hence, the required mean absolute deviation is 3.5

The mean absolute deviation represents the data scattered from mean is on average of 3.5.