OP Malhotra Class 11 Maths Solutions Chapter 21 Measures of Dispersion Chapter Test

Students can track their progress and improvement through regular use of S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Chapter Test.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Chapter Test

Question 1.
Find the mean deviation from the mean for the following data :
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Mean \(\bar{x}\) = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\) = \(\frac{500}{10}\) = 50

xi	38	70	48	40	42	55	63	46	54	44
|xi – \(\bar{x}\) |	12	20	2	10	8	5	13	4	4	6	Σ |xi – \(\bar{x}\) | = 84

∴ M.D about mean = \(\frac{\Sigma\left|x_i-\bar{x}\right|}{n}\) = \(\frac{84}{10}\) = 8.4

Question 2.
Find the mean deviation from the mean for the following data:

xi	3	5	7	9	11	13
fi	6	8	15	25	8	4

Solution:

xi	fi	fixi	| xi – \(\bar{x}\) |	fi | xi – \(\bar{x}\) |
3	6	18	5	30
5	8	40	3	24
7	15	105	1	15
9	25	225	1	25
11	8	88	3	24
13	4	52	5	20
	Σxi = 66	Σfixi = 528		Σfi | xi – \(\bar{x}\) | = 138

By direct method, Mean \(\bar{x}\) = \( \frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{528}{66}\) = 8
Thus M.D from mean = \(\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}\) = \(\frac{138}{66}\) = 2.09

Question 3.
Find the mean deviation for the mean for the following data:

Classes	0-10	10-20	20-30	30-40	40-50	50-60
Frequencies	6	8	14	16	4	2

Solution:

Classes	Frequencies fi	Mid-Marks xi	fi xi	| xi – 27 |	fi | xi – 27 |
0-10	6	5	30	22	132
10-20	8	15	120	12	96
20-30	14	25	350	2	28
30-40	16	35	560	8	128
40-50	4	45	180	18	72
50-60	2	55	110	28	56
	Σfi = 50		Σfixi = 1350		Σfi | xi – 27| = 512

Thus by direct method, Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{1350}{50}\) = 27
∴ M.D about mean = \(\frac{\Sigma f_i\left|x_i-27\right|}{\Sigma f_i}\) = \(\frac{512}{50}\) = 10.24

Question 4.
Find the mean deviation about the median for the following data :
11, 3, 8, 7, 5, 14, 10, 2, 9
Solution:
Arranging the given data in ascending order ; we have
2, 3, 5, 7, 8, 9, 10, 11, 14
Here no. of observations = n = 9 (odd)
∴ M_d = \(\left(\frac{n+1}{2}\right)\)th observation = \(\left(\frac{9+1}{2}\right)\)th obs = 5th obs = 8

xi	| xi – Md |
2	6
3	5
5	3
7	1
8	0
9	1
10	2
11	3
14	6
	Σ | xi – Md | = 27

∴ M.D about Median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{27}{9}\) = 3

Question 5.
Find the variance and standard deviation of the following data :

xi	92	93	97	98	102	104	109
fi	3	2	3	2	6	3	3

Solution:
The table of values is given as under:

xi	fi	di = xi – A A = 98	\( d_i^2 \)	fidi	\( f_i d_i^2 \)
92	3	-6	36	-18	108
93	2	-5	25	-10	50
97	3	-1	1	-3	3
98	2	0	0	0	0
102	6	4	16	24	96
104	3	6	36	18	108
109	3	11	12	33	363
	Σfi = 22			Σ fidi = 44	Σ\( f_i d_i^2 \) = 728

∴ Variance = \(\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2\) = \(\frac{728}{22}\) – \(\left(\frac{44}{22}\right)^2\) = 33.090909 – 4 = 29.09
and S.D = \(\sqrt{\text { Variance }}\) = \(\sqrt{29.09}\) = 5.3935

Question 6.
Calculate the mean and variance after the following data :

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency (f)	2	3	5	10	3	5	2

Solution:
The table of values is given as under:

Classes	Frequency fi	xi	di = xi – A	Ui = \(\frac{d_i}{i}\)	fiui	\(f_i u_i^2\)
0-30	2	15	-90	-3	-6	18
30-60	3	45	-60	-2	-6	12
60-90	5	75	-30	– 1	-5	5
90-120	10	105	0	0	0	0
120-150	3	135	30	1	3	3
150-180	5	165	60	2	10	20
180-210	2	195	90	3	6	18
	Σfi = 30				Σfiui = 2	Σ\(f_i u_i^2\) = 76