Chapter 10.3, Problem 7IP

To determine

To calculate:The mean absolute deviation of the given information.

Answer to Problem 7IP

The months of January has the greater mean absolute deviation ( $M.A.D=8^{\circ }F$ ) than the month of July which has the mean absolute deviation = ${2.5}^{\circ }F$ .It is because the data set of the month of January is widely spread while that of the month of July is very closely packed.

Explanation of Solution

Given information:The table below shows the high temperatures of the first eight days for January and July.

High Temperatures (in F)
January	July
4	83
13	83
16	85
19	86
21	87
23	88
27	90
45	92

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 $x,\overline{x},\left(|x-\overline{x}|\right),\text{mean}\text{absolute}\text{deviation}\left(M.A.D\right)$ for the month of January are shown below

$x$	$\left(|x-\overline{x}|\right)$
4	17
13	8
16	5
19	2
21	0
23	2
27	6
45	24
${\displaystyle \sum x}=168$	${\displaystyle \sum \left(|x-\overline{x}|\right)}=64$
$\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{168}{8}=21$	$M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}=\frac{64}{8}=8$

The Mean Absolute Deviation for the month of January is $M.A.D=8$ .

It shows for the month of January the average distance between the actual temperature on each day and the mean temperature on each day is $8^{\circ }F$ .

The value of $x,\overline{x},\left(|x-\overline{x}|\right),\text{mean}\text{absolute}\text{deviation}\left(M.A.D\right)$ for the month of July are shown below

$x$	$\left(|x-\overline{x}|\right)$
83	3.75
83	3.75
85	1.75
86	0.75
87	0.25
88	1.25
90	3.25
92	5.25
${\displaystyle \sum x}=694$	${\displaystyle \sum \left(|x-\overline{x}|\right)}=20$
$\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{694}{8}=86.75$	$M.A.D=\frac{{\displaystyle \sum \left(|x-\overline{x}|\right)}}{n}=\frac{20}{8}=2.5$

The Mean Absolute Deviation for the month of January is $M.A.D=2.5$ .

It shows for the month of July the average distance between the actual temperature on each day and the mean temperature on each day is ${2.5}^{\circ }F$ .

Hence the months of January has the greater mean absolute deviation ( $M.A.D=8^{\circ }F$ )than the month of Julywhich hasthe mean absolute deviation, ${2.5}^{\circ }F$ .It is because the data set of the month of January is widely spread while that of the month of July is very closely packed.