Measures Of Dispersion (LAQs)

Maths-2A | 8. Measures Of Dispersion – LAQs:
Welcome to LAQs in Chapter 8: Measures Of Dispersion. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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LAQ-1 : Find the mean deviation about the mean for the given data using ‘step deviation method’

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

Class interval	frequency fi	Midpoint xi	di = xi-25/10	fidi	|xi – x|	fi |xi – x|
0-10	5	5	-2	-10	22	110
10-20	8	15	-1	-8	12	96
20-30	15	25 A	0	0	2	30
30-40	16	35	1	16	8	128
40-50	6	45	2	12	18	108
	∑fi = 50 = N			∑fidi = 10		∑fi |xi – x| = 472

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

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LAQ-2 : Find the mean deviation about the mean for the following continuous distribution:

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

Class Interval	Midpoint (xi)	Number of students (fi)	di = xi – 25/10	fidi	|xi – x|	fi |xi – x|
0-10	5	6	-3	-18	28.4	170.4
10-20	15	5	-2	-10	18.4	92
20-30	25	8	-1	-8	8.4	67.2
30-40	35	15	0	0	1.6	24.0
40-50	45	7	1	7	11.6	81.2
50-60	55	6	2	12	21.6	129.6
60-70	65	3	3	9	31.6	94.8
		∑fi = 50 = N		∑fidi = -8		659.2

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

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LAQ-3 : Find the mean deviation about median for the following data:

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

Marks	Number of Boys (fi)	Cumulative frequency (c.f)	Midpoints xi	|xi – M|	fi |xi – M|
0-10	6	6	5	22.86	137.16
10-20	8	14	15	12.86	102.88
20-30	14	28	25	2.86	40.04
30-40	16	44	35	7.14	114.24
40-50	4	48	45	17.14	68.56
50-60	2	50	55	27.14	54.28
	∑fi = 50 = N				517.16

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 $$M.D = \frac{\sum f_i |x_i – M|}{N} = \frac{517.16}{50} = 10.34$$

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LAQ-4 : Find the mean deviation about median for the following data:

xi	6	9	3	12	15	13	21	22
fi	4	5	3	2	5	4	4	3

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

xi	fi	c.f	|xi – M|	fi |xi – M|
3	3	3	10	30
6	4	7	7	28
9	5	12	4	20
12	2	14	1	2
13	4	18	0	0
15	5	23	2	10
21	4	27	8	32
22	3	30	9	27
	N=30			149

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

So Median $$M = 13$$

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

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LAQ-5 : Calculate the variance and standard deviation for the following distribution:

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

Class Interval (C.I)	frequency (fi)	Midpoint of C.I. (xi)	di = xi – 65/10	di2	fidi	fidi2
30-40	3	35	-3	9	-9	27
40-50	7	45	-2	4	-14	28
50-60	12	55	-1	1	-12	15
60-70	15	65	0	0	0	0
70-80	8	75	1	1	8	8
80-90	3	85	2	4	6	12
90-100	2	95	3	9	6	18
	∑fi = 50 = N				∑fidi = -15	∑fidi2 = 105

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

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

$$= 100[105/50 – 225/2500]$$

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

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LAQ-6 : Find the variance and standard deviation of the following frequency distribution

xi	4	8	11	17	20	24	32
fi	3	5	9	5	4	3	1

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

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

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

xi	fi	fixi	xi – x	(xi – x)2	fi(xi – x)2
4	3	12	-10	100	300
8	5	40	-6	36	180
11	9	99	-3	9	81
17	5	85	3	9	45
20	4	80	6	36	144
24	3	72	10	100	300
32	1	32	18	324	324
		∑fixi = 420			∑fi(xi – x)2 = 1374

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

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