Chapter 3.2, Problem 63E

a.

To determine

Find the mean of the data set.

Answer to Problem 63E

The mean of the data set is 4.8.

Explanation of Solution

Calculation:

If $x_1,x_2\mathrm{...}{x}_n$ be n values, the mean absolute deviation is given by,

$\text{Mean absolute deviation}=\frac{{\displaystyle \sum \left|x-\overline{x}\right|}}{n}$. One data set is given.

Mean:

If $x_1,x_2\mathrm{...}{x}_n$ be n values.

The mean value is calculated as,

$\begin{array}{c}\text{Mean}=\frac{\text{sum of all values}}{\text{number of values}}\\ =\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\end{array}$

Substitute the values in the above formula,

$\begin{array}{c}\text{Mean}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{1+3+4+7+9}{5}\\ =\frac{24}{5}\\ =4.8\end{array}$

Hence, the mean of the data set is 4.8.

b.

To determine

Prepare one table like the Table 3.5 in the book with an additional column of  $\left|x-\overline{x}\right|$.

Answer to Problem 63E

The table is given below:

x	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$	$\left|x-\overline{x}\right|$
1	–3.8	${\left(-3.8\right)}^2=14.44$	3.8
3	–1.8	${\left(-1.8\right)}^2=3.24$	1.8
4	–0.8	${\left(-0.8\right)}^2=0.64$	0.8
7	2.2	${\left(2.2\right)}^2=4.84$	2.2
9	4.2	${\left(4.2\right)}^2=17.64$	4.2
Total	0	40.8	12.8

Explanation of Solution

Calculation:

The Table 3.5 consists of $x,\left(x-\overline{x}\right),{\left(x-\overline{x}\right)}^2$. One more column of $\left|x-\overline{x}\right|$ will be added here.

x	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$	$\left|x-\overline{x}\right|$
1	–3.8	${\left(-3.8\right)}^2=14.44$	3.8
3	–1.8	${\left(-1.8\right)}^2=3.24$	1.8
4	–0.8	${\left(-0.8\right)}^2=0.64$	0.8
7	2.2	${\left(2.2\right)}^2=4.84$	2.2
9	4.2	${\left(4.2\right)}^2=17.64$	4.2
Total	0	40.8	12.8

c.

To determine

Find SD and MAD from the table.

Answer to Problem 63E

The SD and the MD are 3.1937 and 2.56 respectively.

Explanation of Solution

Calculation:

Standard deviation:

Let $x_1,x_2\mathrm{...}{x}_n$ be the sample observations of size n. the sample mean is denoted by $\overline{x}$. Hence, the sample variance can be expressed as,

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

The sample standard deviation can be expressed as,

$\begin{array}{c}SD=\sqrt{s^2}\\ =s\end{array}$

The table from part (a), is,

x	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$	$\left|x-\overline{x}\right|$
1	–3.8	${\left(-3.8\right)}^2=14.44$	3.8
3	–1.8	${\left(-1.8\right)}^2=3.24$	1.8
4	–0.8	${\left(-0.8\right)}^2=0.64$	0.8
7	2.2	${\left(2.2\right)}^2=4.84$	2.2
9	4.2	${\left(4.2\right)}^2=17.64$	4.2
Total	0	40.8	12.8

Substitute the value in the above formula,

$\begin{array}{c}{s}^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}\\ =\frac{40.8}{4}\\ =10.2\end{array}$

$\begin{array}{c}SD=\sqrt{10.2}\\ =3.1937\end{array}$

Substitute the values in the MD formula,

$\begin{array}{c}\text{Mean absolute deviation}=\frac{{\displaystyle \sum \left|x-\overline{x}\right|}}{n}\\ =\frac{12.8}{5}\\ =2.56\end{array}$

Hence, the SD and the MD are 3.1937 and 2.56 respectively.

d.

To determine

Find SD and MAD for the new data set.

Answer to Problem 63E

The SD and the MD are 10.677 and 7 respectively.

Explanation of Solution

Calculation:

The data values are 1, 3, 4, 7, 9, and 30.

Substitute the values in the mean formula,

$\begin{array}{c}\text{Mean}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{1+3+4+7+9+30}{6}\\ =9\end{array}$

The table of $x,\left(x-\overline{x}\right),{\left(x-\overline{x}\right)}^2$ is,

x	$\left(x-\overline{x}\right)$	${\left(x-\overline{x}\right)}^2$	$\left|x-\overline{x}\right|$
1	–8	64	8
3	–6	36	6
4	–5	25	5
7	–2	4	2
9	0	0	0
30	21	441	21
Total	0	570	42

Substitute the value in the above formula,

$\begin{array}{c}{s}^2=\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}\\ =\frac{570}{5}\\ =114\end{array}$

$\begin{array}{c}SD=\sqrt{114}\\ =10.677\end{array}$

Substitute the values in the MD formula,

$\begin{array}{c}\text{Mean absolute deviation}=\frac{{\displaystyle \sum \left|x-\overline{x}\right|}}{n}\\ =\frac{42}{6}\\ =7\end{array}$

Hence, the SD and the MD are 10.677 and 7 respectively.

e.

To determine

Identify the resistant measure between standard deviation and mean deviation about mean.

Answer to Problem 63E

The mean deviation about mean is more resistant than standard deviation.

Explanation of Solution

From part (c), the SD and the MD are 3.1937 and 2.56 respectively.

From part (d), the SD and the MD for new data are 10.677 and 7 respectively.

Hence, after adding the outlier value, the mean deviation about mean has changed less comparative to standard deviation.

Hence, it can be said that the mean deviation about mean is more resistant than standard deviation.