Answer to Question #281822 in Statistics and Probability for Neha

Question

Find mean, median, mean deviation about mean, mean deviation about median for the

following data

Age in Years 20 – 22 22 – 24 24 – 26 26 – 28 28 – 30 30 – 32 32 – 34

No. of Employees 70 90 110 140 130 80 80

15. Find mean deviation from mode and its coefficient for the following data

Class – Interval 14 – 18 18 – 22 22 – 26 26 – 30 30 – 34

Frequencies 5 12 14 12 7

Answer 1

14.

$n=700$

$$mean=\bar{x}=\dfrac{\sum _if_ix_i}{n}=\dfrac{18960}{700}=27.0857$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Class & f_i & Mid\ value & f_ix_i & |x_i-\bar{x}| & |x_i-\bar{x}|f_i\\ & & x_i & & & & \\ \hline 20-22 & 70 & 21 & 1470 & 6.0857 & 426 \\ \hdashline 22-24 & 90 & 23 & 2070 & 4.0857 & 367.7143 \\ \hdashline 24-26 & 110 & 25 & 2750 & 2.0857 & 229.4286 \\ \hdashline 26-28 & 140 & 27 & 3780 & 0.0857 & 12 \\ \hdashline 28-30 & 130 & 29 & 4030 & 1.9143 & 248.8571 \\ \hdashline 30-32 & 80 & 31 & 2640 & 3.9143 & 313.1429 \\ \hdashline 32-34 & 80 & 33 & 2800 & 5.9143 & 473.1429 \\ \hdashline & 700 & & 18960 & & 2070.2857 \\ \hdashline \end{array}$$

Mean deviation of Mean

$$\delta \bar{x}=\dfrac{\sum _if_i|x_i-\bar{x}|}{n}=\dfrac{2070.2857}{700}=2.9576$$

Coefficient of Mean deviation

$$\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{2.9576}{27.0857}=0.1092$$

$$median=M$$

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

$$=\text{value of } (350)^{th}\text{ observation}$$

The median class is $26-28.$

$L=$ lower boundary point of median class $=26$

$n=$ Total frequency $=700$

$cf=$ Cumulative frequency of the class preceding the median class $=270$

$f=$ Frequency of the median class $=140$

$c=$ class length of median class $=2$

Median 

$$M=L+\dfrac{\dfrac{n}{2}-cf}{f}\cdot c=26+\dfrac{\dfrac{700}{2}-270}{140}\cdot 2$$

$$=27.1429$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} Class & f_i & Mid\ value & cf & |x_i-M| & |x_i-M|f_i\\ & & x_i & & & & \\ \hline 20-22 & 70 & 21 & 70 & 6.1429 & 430 \\ \hdashline 22-24 & 90 & 23 & 160 & 4.1429& 372.8571 \\ \hdashline 24-26 & 110 & 25 & 270 & 2.1429 & 235.7143 \\ \hdashline 26-28 & 140 & 27 & 410 & 0.1429 & 20 \\ \hdashline 28-30 & 130 & 29 & 540 & 1.8571 & 241.4286 \\ \hdashline 30-32 & 80 & 31 & 620 & 3.8571 & 308.5714 \\ \hdashline 32-34 & 80 & 33 & 700 & 5.8571& 468.5714 \\ \hdashline & 700 & & & & 2077.1429 \\ \hdashline \end{array}$$

Mean deviation of Median

$$\delta \bar{x}=\dfrac{\sum _if_i|x_i-M|}{n}=\dfrac{2077.1429}{700}=2.9673$$

Coefficient of Mean deviation

$$\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{2.9673}{27.1429}=0.1093$$

15.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Class & f_i & Mid\ value & |x_i-Z| & |x_i-Z|f_i\\ & & x_i & & & \\ \hline 14-18 & 5 & 16 & 8 & 40 \\ \hdashline 18-22 & 12 & 20 & 4 & 48 \\ \hdashline 22-26 & 14 & 24 & 0 & 0 \\ \hdashline 26-30 & 12 & 28 & 4 & 48 \\ \hdashline 30-34 & 7 & 32 & 8 & 56 \\ \hdashline & 50 & & & 192 \\ \hdashline \end{array}$$

To find Mode Class

Here, maximum frequency is $14.$

The mode class is $22-26.$

$L=$ lower boundary point of mode class $=22$

$f_1=$  frequency of the mode class $=14$

$f_0=$  frequency of the preceding class $=12$

$f_2=$  frequency of the succedding class $=12$

$c=$  class length of mode class $=4$

$$Z=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c$$

$$=22+(\dfrac{14-12}{2(14)-12-12})\cdot 4=24$$

Mean deviation of Mode

$$\delta \bar{x}=\dfrac{\sum _if_i|x_i-Z|}{n}=\dfrac{192}{50}=3.84$$

Coefficient of Mean deviation

$$\dfrac{\delta \bar{x}}{\bar{x}}=\dfrac{3.84}{24}=0.16$$