Measures of Central Tendency Class 11 OP Malhotra Exe-20A ISC Maths Ch-20 Solutions. In this article you would learn about Arithmetic Mean and its Properties. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Measures of Central Tendency Class 11 OP Malhotra Exe-20A ISC Maths Solutions Ch-20

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-20	Measures of Central Tendency
Writer	O.P. Malhotra
Exe-20(A)	Arithmetic Mean and its Properties.

Properties of Arithmetic Mean Example with Solutions

Measures of Central Tendency Class 11 OP Malhotra Exe-20A ISC Maths Ch-20 Solutions.

Que-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.

Sol: (i) First six natural numbers are ; 1, 2, 3, 4, 5, 6
∴ required mean = (1+2+3+4+5+6)/6 = 21/6 = 7/2

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

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

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

(v) Squares of first n natural numbers are ; 1², 2², 3²,…, n²
∴ required mean = (1², 2², 3²,…+ n²)/n = ∑n²/n = {n(n+1)(2n+1)}/6n
= {(n+1)(2n+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 = (2³+4³+6³+…+(2n)³}/n = 2³[1³+2³+3³+….+n³]/n
= (8Σn³)/n = {8n²(n+1)²}/4n = 2n(n + 1)²

Que-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.

Sol: (i) (a) Required mean = (2.5+2.4+3.5+2.8+2.9+3.3+3.6)/7
= 21/7 = 3
(b) Required Mean = {(−6)+(−2)+(−1)+0+1+2+5+9}/8
= 8/8 = 1

(ii) Q. 4 on P-744 (understanding)

Que-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 x¯
(ii) Check ∑10i = 1(xi−x¯) = 0.

Sol: 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) ∴ x¯ = Σx/n = {5.1+4.7+6.1+1.7+5.4+5.0+11.7+11.6+11.9+3.2}/10
= 332/50 = 6.64
(ii) ∑10 i=1 (xi−x¯) = ∑10i=1 xi−∑10i=1 x¯ = (x1 + x2 + ….. + x10)
= 10x¯ = 66.4 – 10 × 6.64 = 66.4 – 66.4 = 0

Que-4: M being the mean of x₁, x₂, x₃, x₄, x₅ and x₆. Find the value of ∑6i=1(xi−M).

Sol: Given observations are x₁, x₂, x₃, x₄, x₅ and x₆
∴ Mean = M = (x₁+ x₂+ x₃+ x₄+ x₅+x₆)/6
⇒ x₁ + x₂ + x₃ + x₄ + x₅ + x₆ 6M …(1)
∴ ∑6i=1 (xi−M)= (x₁ – M) + (x₂ – M) + (x₃ – M) + (x₄ – M) + (x₅ – M) + (x₆ – M)
= (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) – 6M = 6M – 6M = 0

Que-5: Find the mean of the following frequency distributions :
(i)

(ii)

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

Then by short-cut method, we have required mean = x¯ = A + (Σfd/Σf) = 25 + (0/106) = 25

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

Then by short-cut method, we have required Mean = x¯ = Σfd/Σf = 12.5 + (140/40)
= 12.5 + 3.5 = 16

Que-6: Calculate the mean of the following data by short cut method :

Sol: The table of values is given as under :

Then by short-cut method, we have
required Mean = x¯ = A + (Σfd/Σf)
= 12.5 + (140/40) = 12.5 + 3.5 = 16

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

Sol: The table of values is given as under :

Then by step deviation method, we have Mean x¯ = A + (Σfu/Σf) × i
= 25 – (214/384) × 5 = 22.21

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

Sol: The table of values is given as under :

Then by step deviation method,
Mean x¯ = A + (Σfiui/Σfi) × i
= 20 + (17/40) × 8
= 20 + 3.4 = 23.4

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

Sol: The table of values is given as under :

(i) By direct method x¯ = Σfixi / Σfi = 2/2
(ii) By short cut method, x¯ = A + (Σfidi/Σfi)
= 27.5 – (275/50) = 27.5 – 5.5
⇒ x¯ = 22

Que-10: In a city, the following weekly observations were made in a study of cost of living index for year 1970-71.
(i) Calculate the average weekly cost of living index.
(ii) Verify Σfi (xi – x¯) = 0, where x is the mid-value.

Sol: The table of values is given as under :

(i) ∴ Mean x¯ = Σfixi / Σfi = 8710/52 = 167.5
(ii) ∴Σfi (xi – x¯) = 0

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

Sol: The table of values is given as under :

Then by step deviation method, we have
x¯ = A + (Σfiui/Σfi) × i
= 152.5 + (19/100) × 5
= 152.5 + 0.95 = 153.45

Que-12: Given:

Find the mean variable by taking 11 as the assumed mean, and verify by the direct method.

Sol: The table of values is given as under:

Then by direct method, x¯ = Σfix / iΣfi = 777/50 = 15.54
By short at method, mean x¯ = A + (Σfidi/Σfi) = 15 + (27/50) = 15.54

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

Estimate (i) the total turnover, (ii) the mean value of a single order.

Sol: The table of values is given as under :

∴ required turn over = Σfixi = 19140
Thus, required mean value = Σfixi / Σfi = 19140/1000 = 19.14

Que-14: The mean of the following data is 20.5. Find the missing frequency.

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

∴ by direct method, Mean = Σfx/Σf ⇒ Mean = (635+20f)/(130+f1)
Also given mean be 20.5.
∴ 20.5 = (635+20f1)/(30+f1)
⇒ 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.

Que-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.

Sol: Given no. of observation = 40

Que-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.

Sol: 

Given Σf_i = 120 ⇒ 120 = 70 + f₁ + f₂ ⇒ f₁ + f₂ = 50 …(1)
By direct method, we have
Mean = Σfixi / Σfi
⇒ 50 = (3500+30f1+70f2)/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

Que-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.

Sol: Given no. of observations = 30

Que-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.

Sol: 

–: End of Measures of Central Tendency Class 11 OP Malhotra Exe-20A ISC Maths Ch-20 Solutions. :–

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