6 Most FAQ’s of Measures of Dispersion Chapter in Inter 2nd Year Maths-2A (TS/AP)

Table of Contents

7 Marks

Marks obtained	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

We take the assumed mean A = 25 Here C = 10 Hence we form the following table

Here, N = 50 So Mean $$x = A + C(\sum f_i d_i / N) = 25 + 10(10/50) = 25 + 2 = 27$$

Mean deviation about the mean $$M.D = 1/N \sum_{i=1}^5 f_i |x_i – x| = 1/50(472) = 9.44$$

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70
No. of students	6	5	8	15	7	6	3

We take assumed mean A = 35 Here C = 10 Hence we form the following table

Here N = 50 So, mean $$x = A + C(\sum f_i d_i / N) = 35 + 10(-8/50) = 35 – 1.6 = 33.4$$

Mean deviation about the mean $$M.D = 1/N \sum f_i |x_i – x| = 1/50(659.2) = 13.184$$

Marks	0-10	10-20	20-30	30-40	40-50	50-60
No. of boys	6	8	14	16	4	2

We form the following table from the given data

Median class = class containing (50/2)^th item = 25^th item

Thus Median class = (20-30) class

Here i = 20, C = 10, f = 14, m = 14, N = 50

Median $$M = l + \frac{C}{f}(N/2 – m) = 20 + \frac{10}{14}(25 – 14) = 20 + 7.86 = 27.86$$

Mean deviation about the median $$M.D = \frac{\sum f_i |x_i – M|}{N} = \frac{517.16}{50} = 10.34$$

x_i	6	9	3	12	15	13	21	22
f_i	4	5	3	2	5	4	4	3

We form the following table by writing the given data in ascending order

$$N/2 = (30/2) = 15$$

So Median $$M = 13$$

Median deviation about median $$M.D = \sum f_i |x_i – M|/N = 149/30 = 4.97$$

Class	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2

Here h = 10 we take the assumed mean A = 65

Here C = 10 then d_i = x_i– 65/10 we form the following table with the given data

Here $$N = 50$$ $$\sum f_i d_i = -15$$ $$\sum f_i d_i^2 = 105$$

Variance $$\sigma^2 = C^2[\sum (f_i d_i^2)/N – (\sum (f_i d_i)/N)^2] = 10^2[105/50 – (-15/50)^2]$$

$$= 100[105/50 – 225/2500]$$

$$= 100[(105)(50)-225/2500]$$

x_i	4	8	11	17	20	24	32
f_i	3	5	9	5	4	3	1

Here $$N = \sum f_i = 3 + 5 + 9 + 5 + 4 + 3 + 1 = 30$$

Also $$\sum f_i x_i = 4(3) + 8(5) + 11(9) + 17(5) + 20(4) + 24(3) + 32(1) = 420$$

$$x = \frac{\sum f_i x_i}{N} = \frac{420}{30} = 14$$

Variance $$\sigma^2 = \frac{1}{N}\sum f_i(x_i – x)^2 = \frac{1}{30}(1374) = 45.8$$

Standard Deviation $$\sigma = \sqrt{45.8} = 6.77$$