Chapter 3.2, Problem 63E

Mean absolute deviation: A measure of spread that is an alternative to the standard deviation (SD) is the mean absolute deviation (MAD). For a data set containing values $x_1,\mathrm{....}{x}_n$, the mean absolute deviation is given by $\text{mean absolute deviation = }\frac{\sum |x-\bar{x}|}{n}$

1. Compute the mean $\bar{x}$ for the data set 1, 3, 4, 7, 9.
2. Construct a table like Table 3.5 that contains an additional column for the values $|x-\bar{x}|$.
3. Use the table to compute the SD and the MAD.
4. Now consider the data set 1, 3, 4, 7, 9, 30. Compute the SD and the MAD for this data set.
5. Which measure of spread is more resistant, the SD or the MAD? Explain.

To determine

To find mean of the data

Answer to Problem 63E

Mean of the data is

Explanation of Solution

Given:

$x$

1

3

4

7

9

Formula:

  $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Calculation:

Here n = 5.

Values of Xi are data values.

Putting all values in formula,

  $\overline{x}=\frac{24}{5}=4.8$

Therefore, mean is 4.8

To determine

To construct table for finding standard deviation and mean absolute deviation

Explanation of Solution

Creating table for finding standard deviation and mean absolute deviation:

$x$	$x-\overline{x}$	${(x-\overline{x})}^2$	$|x-\overline{x}|$
1	-3.8	14.44	3.8
3	-1.8	3.24	1.8
4	-0.8	0.64	0.8
7	2.2	4.84	2.2
9	4.2	17.64	4.2

To determine

To find standard deviation and mean absolute deviation

Answer to Problem 63E

Standard deviation is 3.19 and mean absolute deviation is 2.56

Explanation of Solution

Formula:

Sample standard deviation:

  $s=\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{x})}^2}}{n-1}}$

And

Mean absolute deviation:

  $MAD=\frac{{\displaystyle \sum |x-\overline{x}|}}{n}$

Calculation:

Sample standard deviation:

From table,

  ${\displaystyle \sum {(x_i-\overline{x})}^2}=40.8$

Put in a formula,

  $s=\sqrt{\frac{40.8}{4}}=3.19$

Mean absolute deviation:

  $MAD=\frac{12.8}{5}=2.56$

To determine

To find standard deviation and mean absolute deviation for new data set

Answer to Problem 63E

Standard deviation is 10.68 and mean absolute deviation is 8.4

Explanation of Solution

Formula:

Mean:

  $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Sample standard deviation:

  $s=\sqrt{\frac{{\displaystyle \sum {(x_i-\overline{x})}^2}}{n-1}}$

And

Mean absolute deviation:

  $MAD=\frac{{\displaystyle \sum |x-\overline{x}|}}{n}$

Calculation:

Creating table for new data set:

$x$	$x-\overline{x}$	${(x-\overline{x})}^2$	$|x-\overline{x}|$
1	-8	64	8
3	-6	36	6
4	-5	25	5
7	-2	4	2
9	0	0	0
30	21	441	21

Mean:

Here n =6

Values of Xi are data values.

Putting all values in formula,

  $\overline{x}=\frac{54}{6}=9$

Therefore, mean is 9

Sample standard deviation:

From table,

  ${\displaystyle \sum {(x_i-\overline{x})}^2}=570$

Put in a formula,

  $s=\sqrt{\frac{570}{5}}=10.68$

Mean absolute deviation:

  $MAD=\frac{42}{6}=8.4$

To determine

To determine which measure of spread is more resistant

Answer to Problem 63E

MAD seems to be good measure of spread for given data

Explanation of Solution

Standard deviation and mean absolute deviation for original data:

SD = 3.19 and MAD = 2.56

Standard deviation and mean absolute deviation for modified data:

SD = 10.68 and MAD = 8.4

Standard deviation for modified data increases by 7.49 as compare to original data. Also, MAD for modified data increases by 5.84 as compared to original data.

In this data set, MAD seems to be more capable as it seems to be good measure of spread.