Chapter 10.3, Problem 4IP

To determine

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

Answer to Problem 4IP

The mean absolute deviation is 9.6 and 11.1

Explanation of Solution

Given information:Score of bowling league in two different months.

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 $x_i\text{and}{y}_i$ represented the observations for the month of September and April.

Calculation of Mean

  $\begin{array}{l}\overline{x}=\frac{98+95+101+103+90+121+116+118+94}{9}\\ \overline{x}=\frac{936}{9}\\ \therefore \overline{x}=104\end{array}$

Similarly,

  $\begin{array}{l}\overline{y}=\frac{112+130+118+118+125+122+145+105+150}{9}\\ \overline{y}=\frac{1125}{9}\\ \therefore \overline{y}=125\end{array}$

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

Now, calculation of mean absolute deviation

$x_i$	$x_i-\overline{x}$	$\left|x_i-\overline{x}\right|$	$y_i$	$y_i-\overline{y}$	$\left|y_i-\overline{y}\right|$
98	$98-104=-6$	6	112	$112-125=-13$	13
95	$95-104=-9$	9	130	$130-125=5$	5
101	$101-104=-3$	3	118	$118-125=-7$	7
103	$103-104=-1$	1	118	$118-125=-7$	7
90	$90-104=-14$	14	125	$125-125=0$	0
121	$121-104=17$	17	122	$122-125=-3$	3
116	$116-104=12$	12	145	$145-125=20$	20
118	$118-104=14$	14	105	$105-125=-20$	20
94	$94-104=-10$	10	150	$150-125=25$	25

  $\begin{array}{l}MD=\frac{6+9+3+1+14+17+12+14+10}{9}\\ MD=\frac{86}{9}\\ MD=9.6\end{array}$

And

  $\begin{array}{l}MD=\frac{13+5+7+7+0+3+20+20+25}{9}\\ MD=\frac{100}{9}\\ MD=11.1\end{array}$

Thus, mean absolute deviation is 9.6 and 11.1

Hence, the required mean absolute deviation is 9.6 and 11.1

The observations of x are more variable than the observation of y

So, y is more consistent and dependable.