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.


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

Marks obtained0-1010-2020-3030-4040-50
No. of students5815166

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

Class intervalfrequency fiMidpoint xidi = xi-25/10fidi|xi – x|fi |xi – x|
0-1055-2-1022110
10-20815-1-81296
20-301525 A00230
30-4016351168128
40-5064521218108
fi = 50 = Nfidi = 10fi |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 $$M.D = 1/N \sum_{i=1}^5 f_i |x_i – x| = 1/50(472) = 9.44$$


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

Marks0-1010-2020-3030-4040-5050-6060-70
No. of students65815763

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

Class IntervalMidpoint (xi)Number of students (fi)di = xi – 25/10fidi|xi – x|fi |xi – x|
0-1056-3-1828.4170.4
10-20155-2-1018.492
20-30258-1-88.467.2
30-403515001.624.0
40-504571711.681.2
50-6055621221.6129.6
60-706533931.694.8
fi = 50 = Nfidi = -8659.2

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


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

Marks0-1010-2020-3030-4040-5050-60
No. of boys68141642

We form the following table from the given data

MarksNumber of Boys (fi)Cumulative frequency (c.f)Midpoints xi|xi – M|fi |xi – M|
0-1066522.86137.16
10-208141512.86102.88
20-301428252.8640.04
30-401644357.14114.24
40-504484517.1468.56
50-602505527.1454.28
fi = 50 = N517.16

Median class = class containing (50/2)th item = 25th 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 $$M.D = \frac{\sum f_i |x_i – M|}{N} = \frac{517.16}{50} = 10.34$$


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

xi6931215132122
fi45325443

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

xific.f|xi – M|fi |xi – M|
3331030
647728
9512420
1221412
1341800
15523210
21427832
22330927
N=30149

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

So Median $$M = 13$$

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


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

Class30-4040-5050-6060-7070-8080-9090-100
Frequency371215832

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

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

Class Interval (C.I)frequency (fi)Midpoint of C.I. (xi)di = xi – 65/10di2fidifidi2
30-40335-39-927
40-50745-24-1428
50-601255-11-1215
60-7015650000
70-808751188
80-9038524612
90-10029539618
fi = 50 = Nfidi = -15fidi2 = 105

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

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

$$= 100[105/50 – 225/2500]$$

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


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

xi481117202432
fi3595431

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

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

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

xififixixi – x(xi – x)2fi(xi – x)2
4312-10100300
8540-636180
11999-3981
175853945
20480636144
2437210100300
3213218324324
fixi = 420fi(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 $$\sigma = \sqrt{45.8} = 6.77$$