OP Malhotra Class 11 Maths Solutions Chapter 20 Measures of Central Tendency Ex 20(a)

Effective S Chand ISC Maths Class 11 Solutions Chapter 20 Measures of Central Tendency Ex 20(a) can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 20 Measures of Central Tendency Ex 20(a)

Question 1.
Find the mean of :
(i) the first 6 natural numbers.
(ii) the first ten odd natural numbers.
(iii) the first eight even natural numbers.
(iv) x, x + 2, x + 3, x + 6 and x + 9.
(v) Squares of the first n natural numbers.
(vi) Squares of the first 10 natural numbers.
(vii) Cubes of first n even natural numbers.
Solution:
(i) First six natural numbers are ; 1, 2, 3, 4, 5, 6
∴ required mean = \(\frac{1+2+3+4+5+6}{6}\) = \(\frac{21}{6}\) = \(\frac{7}{2}\)

(ii) First ten add natural numbers are = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
∴ required mean = \(\frac{Sum of all observations}{No. of observations}\) = \(\frac{1+3+5+7+9+11+13+15+17+19}{10}\) = \(\frac{100}{10}\) = 10

(iii) First eight even natural numbers are ; 2, 4, 6, 8, 10, 12, 14, 16
∴ required mean = \(\frac{2+4+6+8+10+12+14+16}{8}\) = \(\frac{72}{8}\) = 9

(iv) Required mean = \(\frac{Sum of all observations}{No. of observations}\) = \(\frac{x+x+2+x+3+x+6+x+9}{5}\) = \(\frac{5 x+20}{5}\) = \(\frac{5(x+4)}{5}\) = x + 4

(v) Squares of first n natural numbers are ; 1², 2², 3²,…, n²
∴ required mean = \(\frac{1^2+2^2+\ldots+n^2}{n}\) = \(\frac{\sum n^2}{n}\) = \(\frac{n(n+1)(2 n+1)}{6 n}\) = \(\frac{(n+1)(2 n+1)}{6}\)

(vi) Squares of first 10 natural numbers are ; 1², 2², 3²,…, 10²

(vii) Class of first n even natural numbers are 2³, 4³, 6³,…, (2n)³
∴ required mean = \(\frac{2^3+4^3+6^3+\ldots+(2 n)^3}{n}\) = \(\frac{2^3\left[1^3+2^3+3^3+\ldots .+n^3\right]}{n}\) = \(\frac{8 \Sigma n^3}{n}\) = \(\frac{8 n^2(n+1)^2}{4 n}\) = 2n(n + 1)²

Question 2.
(i) Find the mean of the following sets of numbers :
(a) 2.5, 2.4, 3.5, 2.8, 2.9, 3.3, 3.6
(b) -6, -2, -1, 0, 1, 2, 5, 9.
(ii) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.
Solution:
(i) (a) Required mean = \(\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}\) = \(\frac{21}{7}\) = 3
(b) Required Mean = \(\frac{(-6)+(-2)+(-1)+0+1+2+5+9}{8}\) = \(\frac{8}{8}\) = 1
(ii) Q. 4 on P-744 (understanding)

Question 3.
Duration of sunshine (in hours) in Delhi for first ten days of month as reported by Meteorological department are as under:
5.1, 4.7, 6.1, 1.7, 5.4, 5.0, 11.7, 11.6, 11.9, 3.2
(i) Calculate \(\bar{x}\);
(ii) Check \(\sum_{i=1}^{10}\left(x_i-\bar{x}\right)\) = 0.
Solution:
Given observations are ; 5.1, 4.7, 6.1, 1.7, 5.4, 5.0, 11.7, 11.6, 11.9, 3.2
Here n = 10
(i) ∴ \(\bar{x}\) = \(\frac{\Sigma x}{n}\) = \(\frac{5.1+4.7+6.1+1.7+5.4+5.0+11.7+11.6+11.9+3.2}{10}\) = \(\frac{332}{50}\) = 6.64
(ii) \(\sum_{i=1}^{10}\left(x_i-\bar{x}\right)\) = \(\sum_{i=1}^{10} x_i-\sum_{i=1}^{10} \bar{x}\) = (x₁ + x₂ + ….. + x₁₀) = 10\(\bar{x}\) = 66.4 – 10 × 6.64 = 66.4 – 66.4 = 0

Question 4.
M being the mean of x₁, x₂, x₃, x₄, x₅ and x₆. Find the value of \(\sum_{i=1}^6\left(x_i-M\right)\).
Solution:
Given observations are x₁, x₂, x₃, x₄, x₅ and x₆
∴ Mean = M = \(\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\) ⇒ x₁ + x₂ + x₃ + x₄ + x₅ + x₆ 6M …(1)
∴ \(\sum_{i=1}^6\left(x_i-\mathrm{M}\right)\) = (x₁ – M) + (x₂ – M) + (x₃ – M) + (x₄ – M) + (x₅ – M) + (x₆ – M)
= (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) – 6M = 6M – 6M = 0

Question 5.
Find the mean of the following frequency distributions :
(i)

X	19	21	23	25	27	29	31
f	13	15	16	18	16	15	13

(ii)

X	2.5	7.5	12.5	17.5	22.5
f	4	5	7	12	7

Solution:
(i) The table of values is given as under :

x	f	d = x – 25 A = 25	fd
19	13	-6	-68
21	15	-4	-60
23	16	-2	-32
25	18	0	0
27	16	2	32
29	15	4	50
31	13	6	78
	N = Σf = 106		Σfd = 0

Then by short-cut method, we have required mean = \(\bar{x}\) = A + \(\frac{\Sigma f d}{\Sigma f}\) = 25 + \(\frac{0}{106}\) = 25

(ii) The table of values is given as under :

x	f	d = x – A A = 12.5	fd
2.5	4	-10	-40
7.5	5	-5	-25
12.5	7	0	0
17.5	12	5	60
22.5	7	10	70
27.5	5	15	75
	Σf = 40		Σfd = 140

Then by short-cut method, we have required Mean = \(\bar{x}\) = \(\frac{\Sigma f d}{\Sigma f}\) = 12.5 + \(\frac { 140 }{ 40 }\) = 12.5 + 3.5 = 16

Question 6.
Calculate the mean of the following data by short cut method :

x	2.5	7.5	12.5	17.5	22.5
f	4	5	7	12	7

Solution:
The table of values is given as under :

x	f	d = x – A A = 12.5	fd
2.5	4	-10	-40
7.5	5	-5	-25
12.5	7	0	0
17.5	12	5	60
22.5	7	10	70
27.5	5	15	75
Σf = 40			Σfd = 140

Then by short-cut method, we have
required Mean = \(\bar{x}\) = A + \(\frac{\Sigma f d}{\Sigma f}\) = 12.5 + \(\frac{140}{40}\) = 12.5 + 3.5 = 16

Question 7.
Calculate by step-deviation method, the arithmetic mean of the following marks obtained by students in English.

Marks	5	10	15	20	25	30	35	40	45	50
No. of students	20	43	75	67	72	45	39	9	8	6

Solution:
The table of values is given as under :

X	f	d = x – A A = 25	Ui = d/i i = 5	fu
5	20	-20	-4	-80
10	43	-15	-3	– 129
15	75	-10	-2	– 150
20	67	-5	– 1	-67
25	72	0	0	0
30	45	5	1	45
35	39	10	2	78
40	9	15	3	27
45	8	20	4	32
50	6	25	5	30
	Σf = 384			Σfu = – 214

Then by step deviation method, we have Mean \(\bar{x}\) = A + \(\frac{\Sigma f u}{\Sigma f}\) × i = 25 – \(\frac{214}{384}\) × 5 = 22.21

Question 8.
The frequency distribution of marks obtained by 40 students of a class is as under. Calculate the Arithmetic Mean.

Marks	0-8	8-16	16-24	24-32	32-40	40-48
Students	6	3	10	16	4	2

Solution:
The table of values is given as under :

Marks	Frequency fi	Class Marks xi	di= Xi – A A = 20	Ui = di/8 i = 8	fiui
0-8	5	4	– 16	-2	– 10
8-16	3	12	-8	– 1	-3
16-24	10	20	0	0	0
24-32	16	28	8	1	16
32-40	4	36	16	2	8
40-48	2	44	24	3	6
	Σfi = 40				Σfiui = 17

Then by step deviation method,
Mean \(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i = 20 + \(\frac{17}{40}\) × 8 = 20 + 3.4 = 23.4

Question 9.
Compute the mean of the following frequency table by :
(i) direct method and
(ii) short-cut method

Class	Frequency	Class	Frequency
5-10	10	30-35	4
10-15	6	35-40	2
15-20	4	45-45	1
20-25	12	45-50	3
25-30	8

Solution:
The table of values is given as under :

Class	frequency fi	Mid Marks xi	fixi	di = Xi – A A = 27.5	ui = \(\frac{d_i}{i}\) i = 5	fidi
5-10	10	7.5	75	-20	-4	-200
10-15	6	12.5	75	– 15	-3	-90
15-20	4	17.5	70	– 10	-2	-40
20-25	12	22.5	270	-5	– 1	-60
25-30	8	27.5	220	0	0	0
30-35	4	32.5	130	5	1	20
35-40	2	37.5	75	10	2	20
40-45	1	42.5	42.5	15	3	15
45-50	3	47.5	142.5	20	4	60
	Σfi = 50		Σfixi = 1100			Σfidi = – 275

(i) By direct method \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = 22
(ii) By short cut method, \(\bar{x}\) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\) = 27.5 – \(\frac { 275 }{ 50 }\) = 27.5 – 5.5 ⇒ \(\bar{x}\) = 22

Question 10.
In a city, the following weekly observations were made in a study of cost of living index for year 1970-71.

Cost of living index	140-150	150-160	160-170	170-180	180-190	190 – 200
Number of weeks	5	10	18	9	6	4

(i) Calculate the average weekly cost of living index.
(ii) Verify Σf_i (x_i – \(\bar{x}\)) = 0, where x is the mid-value.
Solution:
The table of values is given as under :

Cost of living Index	No. of weeks fi	Mid-Marks xi	fixi	xi – \(\bar{x}\)	fi (xi – \(\bar{x}\))
140-150	5	145	725	-22.5	– 112.5
150-160	10	155	1550	-12.5	– 125
160- 170	18	165	2970	-2.5	-45
170-180	9	175	1575	7.5	67.5
180-190	6	185	1110	17.5	105
190-200	4	195	780	27.5	110
	Σfi = 52		Σfixi = 8710		Σfi (xi – \(\bar{x}\)) = 0

(i) ∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac { 8710 }{ 52 }\) = 167.5
(ii) ∴Σf_i (x_i – \(\bar{x}\)) = 0

Question 11.
Calculate the mean by step-deviation method for the following data :

Height (in cm)	No. of boys	Height (in cm)	No. of boys
135-140	4	155-160	24
140 -145	9	160-165	10
145-150	18	165-170	5
150-155	28	170-175	2

Solution:
The table of values is given as under :

Height (in cm)	No. of boys fi	Mid-Marks xi	di = xi – A A = 152.5	ui = \(\frac{d_i}{i}\) i = 5	fiui
135-140	4	137.5	– 15	-3	– 12
140-145	9	142.5	– 10	-2	– 18
145-150	18	147.5	-5	– 1	– 18
150-155	28	152.5	0	0	0
155-160	24	157.5	5	1	24
160-165	10	162.5	10	2	20
165-170	5	167.5	15	3	15
170-175	2	172.5	20	4	8
	Σfi=100				Σfiui = 19

Then by step deviation method, we have
\(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i = 152.5 + \(\frac { 19 }{ 100 }\) × 5 = 152.5 + 0.95 = 153.45

Question 12.
Given:

Variable	20	19	18	17	16	15	14	13	12	11
Frequency	1	2	4	8	11	10	7	4	2	1

Find the mean variable by taking 11 as the assumed mean, and verify by the direct method.
Solution:
The table of values is given as under:

Variable xi	Frequency fi	fixi	di = xi – A A = 15	fidi
20	1	20	5	5
19	2	38	4	8
18	4	72	3	12
17	8	136	2	16
16	11	176	1	11
15	10	150	0	0
14	7	98	– 1	-7
13	4	52	-2	-8
12	2	24	-3	-6
11	1	11	-4	-4
	Σfi = 50	Σfixi = 777		Σfidi = 27

Then by direct method, \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{777}{50}\) = 15.54
By short at method, mean \(\bar{x}\) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\) = 15 + \(\frac{27}{50}\) = 15.54

Question 13.
The following table shows the distribution of orders of a firm, according to their value :

Value	Under ?10	? 10 and Under ?20	? 20 and Under ?30	? 30 and Under ?40	? 40 and Under ? 50	? 50 and Under ? 60	? 60 and Under ? 70
No. of buyers	245	383	205	89	47	21	10

Estimate (i) the total turnover, (ii) the mean value of a single order.
Solution:
The table of values is given as under :

Classes	No. of buyers fi	Mid-value xi	fixi
0-10	245	5	1225
10-20	383	15	5745
20-30	205	25	5125
30-40	89	35	3115
40-50	47	45	2115
50-60	21	55	1155
60-70	10	65	650
	Σfi = 1000		Σfixi = 19140

∴ required turn over = Σf_ix_i = 19140
Thus, required mean value = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{19140}{1000}\) = 19.14

Question 14.
The mean of the following data is 20.5. Find the missing frequency.

x	10	15	20	25	30
f	5	7	•••	12	6

Solution:
Let the missing frequency be f₁
The table of values is given as under :

X	f	fx
10	5	50
15	7	105
20	f1	20f1
25	12	300
30	6	180
	Σf = 30 + f1	Σfx = 635 + 20f1

∴ by direct method, Mean = \(\frac{\Sigma f x}{\Sigma f}\) ⇒ Mean = \(\frac{635+20 f_1}{30+f_1}\)
Also given mean be 20.5.
∴ 20.5 = \(\frac{635+20 f_1}{30+f_1}\)
⇒ 20.5 = (30 + f₁) = 635 + 20f₁
⇒ 615 + 20.5 f₁ = 625 + 20f₁
⇒ 0.5 f₁ = 20 ⇒ f₁ = 40
Hence the required missing frequency be 40.

Question 15.
The mean of 40 observations was 160. It was detected on re-checking that the value 125 was wrongly copied as 165 for the computation of the mean. Find the
correct mean.
Solution:
Given no. of observation = 40

Question 16.
The mean of the following frequency table is 50. But the frequencies f₁ and f₂ in classes 20 – 40 and 60 – 80 are missing. Find the missing frequencies.

Class	0-20	20-40	40-60	60-80	80 -100
Frequency	19	f1	32	f2	19	(Total 120)

Solution:

Class	frequency fi	Mid-class xi	fixi
0-20	19	10	190
20-40	f1	30	30f1
40-60	32	50	1600
60-80	f2	70	70f2
80-100	19	90	1710
Total	70 + f1 + f2 = Σf		Σfixi = 3500 + 30f1 + 70f2

Given Σf_i = 120 ⇒ 120 = 70 + f₁ + f₂ ⇒ f₁ + f₂ = 50 …(1)
By direct method, we have
Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) ⇒ 50 = \(\frac{3500+30 f_1+70 f_2}{120}\) ⇒ 3500 + 30f₁ + 70f₂ = 6000
⇒ 30f₁ + 70f₂ = 2500 ⇒ 3f₁ + 7f₂ = 250 …(2)
On solving eqn. (1) by eqn. (2); we have
4f₂ = 100 ⇒ f₂ = 25 and f₁ = 25

Question 17.
The mean of 30 values was 150. It was detected on re-checking that the value 165 was wrongly copied as 135 for the computation of the mean. Find the correct mean.
Solution:
Given no. of observations = 30

Question 18.
The sum of deviation of set of values x₁, x₂,…, x_n measured from 50 is – 10 and the sum of deviation of values from 46 is 70. Find the value of n and the mean.
Solution: