Answer to Question #281821 in Statistics and Probability for Neha

Question

Find the mean deviation from mean and median for the following data

13, 17, 9, 21, 15, 25, 30, 35, 31

12. Find the mean deviation from mean and its coefficient for the following data representing

age distribution of 50 boys.

Age (in years): 17 18 19 20 21 22 23

No. of Boys 8 12 15 8 4 2 1

13. Find the mean deviation from median and its coefficient for the following data:

Height (in cm) 160 164 168 172 176 180

Number of students 3 7 12 15 6 2

Answer 1

11.

$$9,13, 15,17, 21, 25, 30, 31, 35$$

$$mean=\bar{x}=\dfrac{\sum _ix_i}{n}=\dfrac{1}{9}(13+17+9+21$$

$$+15+25+ 30+35+31)=\dfrac{196}{9}\approx21.7778$$

$$median=M=21$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & x_i & |x_i-\bar{x}| & |x_i-M| \\ \hline & 9 & 12.7778 &12 \\ \hdashline & 13 & 8.7778 & 8 \\ \hdashline & 15 & 6.7778 & 6 \\ \hdashline & 17 & 4.7778 & 4 \\ \hdashline & 21 & 0.7778 & 0 \\ \hdashline & 25 & 3.2222 & 4 \\ \hdashline &30 & 8.2222 & 9 \\ \hdashline & 31 & 9.2222 & 10 \\ \hdashline & 35 & 13.2222 &14 \\ \hdashline sum= & 196 & 67.7778 & 67 \\ \hdashline \end{array}$$

Mean deviation of Mean

$$\delta \bar{x}=\dfrac{\sum _i|x_i-\bar{x}|}{n}=\dfrac{67.7778}{9}=7.5309$$

Mean deviation of Median

$$\delta \bar{x}=\dfrac{\sum _i|x_i-M|}{n}=\dfrac{67}{9}=7.4444$$

12.

$$mean=\bar{x}=\dfrac{\sum _if_ix_i}{n}=\dfrac{948}{50}=18.96$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x_i & f_i & f_ix_i & |x_i-\bar{x}| & |x_i-\bar{x}|\cdot f_i \\ \hline & 17 & 8 &136 & 1.96 & 15.68 \\ \hdashline & 18 & 12 & 216 & 0.96 & 11.52 \\ \hdashline & 19 & 15 & 275 & 0.04 & 0.6 \\ \hdashline & 20 & 8 & 160 & 1.04 & 8.32 \\ \hdashline & 21 & 4 & 84 & 2.04 & 8.16 \\ \hdashline & 22 & 2 & 44 & 3.04 & 6.08 \\ \hdashline & 23 & 1 & 23 & 4.04 & 4.04 \\ \hdashline & & 50 & 948 & & 54.4 \\ \hdashline \end{array}$$

Mean deviation of Mean

$$\delta \bar{x}=\dfrac{\sum _if_i|x_i-\bar{x}|}{n}=\dfrac{54.4}{50}=1.088$$

Coefficient of Mean deviation

$$\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{1.088}{18.96}=0.0574$$

13.

$$median=M$$

$$=\text{value of } (\dfrac{n+1}{2})^{th}\text{ observation}$$

$$=\text{value of } (23)^{th}\text{ observation}=172$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x_i & f_i & cf & |x_i-M| & |x_i-M|\cdot f_i \\ \hline & 160 & 3 & 3 & 12 & 36 \\ \hdashline & 164 & 7 & 10 & 8 & 56 \\ \hdashline & 168 & 12 & 22 & 4 & 48 \\ \hdashline & 172 & 15 & 37 & 0 & 0 \\ \hdashline & 176 & 6 & 43 & 4 & 24 \\ \hdashline & 180 & 2 & 45 & 8 & 16 \\ \hdashline & & 45 & & & 180 \\ \hdashline \end{array}$$

Mean deviation of Median

$$\delta \bar{x}=\dfrac{\sum _if_i|x_i-M|}{n}=\dfrac{180}{45}=4$$

Coefficient of Mean deviation

$$\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{4}{172}=0.0233$$