OP Malhotra Class 11 Maths Solutions Chapter 20 Measures of Central Tendency Chapter Test

Continuous practice using S Chand ISC Maths Class 11 Solutions Chapter 20 Measures of Central Tendency Chapter Test can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 20 Measures of Central Tendency Chapter Test

Question 1.
The weights of 50 apples were recorded as given below. Calculate the mean weight to the nearest gram, by step deviation method.

Weight in grams	80-85	85-90	90-95	95-100	100-105	105-110	110-115
Number of apples	5	8	10	12	8	4	3

Solution:

Weight (in gms)	No. of apples (frequency fi)	Mid-Marks xi	di = xi – 97.5 A = 97.5	ui = \(\frac{d_i}{i}\) i = 5	fiui
80-85	5	82.5	– 15	-3	– 15
85-90	8	87.5	– 10	-2	– 16
90-95	10	92.5	-5	– 1	– 10
95-100	12	97.5	0	0	0
101 – 105	8	102.5	5	1	8
105-110	4	107.5	10	2	8
110-115	3	112.5	15	3	9
	Σfi = 50				Σfiui = – 16

Then by step deviation method \(\bar{x}\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × i
⇒ \(\bar{x}\) = 97.5 + \(\frac{(-16)}{50}\) × 5 = 97.5 – 1.6 = 95.9
Thus the required mean weight be 95.9 grams.

Question 2.
Find the value of p if the mean of the following distribution is 7.5 :

X	2-4	4-6	6-8	8-10	10-12	12-14
f	6	8	15	P	8	4

Solution:
The table of values is given as under:

Class	frequency (fi)	Mid-Marks xi	fixi
2-4	6	3	18
4-6	8	5	40
6-8	15	7	105
8-10	P	9	9P
10-12	8	11	88
12-14	4	13	52
	Σfi =41 +P		Σfixi = 303 + 9p

Then by direct method, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) ⇒ 7.5 = \(\frac{303+9 p}{41+p}\)
⇒ 7.5 (41 + p) = 303 + 9p ⇒ 1.5p = 4.5 ⇒ p = 3

Question 3.
The mean of the following frequency distribution is 57.6 and the number of observations is 50. Find the missing frequencies f₁ and f₂.

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	7	A	12	A	8	5

Solution:
We construct the table of values is given as under :

=

=

Classes	Class Mark xi	Frequency fi	di = xi – A A = 50	ui = \(\frac{d_i}{C}\) C = 20	fiui
0-20	10	7	-40	-2	– 14
20-40	30	f1	-20	– 1	-h
40-60	50	12	0	0	0
60-80	70	f2	20	1	h
80-100	90	8	40	2	16
100-120	110	5	60	3	15
		Σfi = 32 + f1+f2			Σfiui = 17 – f1+f2

since sum of all observations = 50
∴ Σf_i = 50
⇒ 50 = 32 + f₁ + f₂
⇒ f₁ + f₂ = 18
By step deviation method, we have
Mean = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × C
⇒ 57.6 = 50 + \(\frac{17-f_1+f_2}{32+f_1+f_2}\) × 20
⇒ 57.6 – 50 = \(\frac{17-f_1+f_2}{50}\) × 20
⇒ 5 × 7.6 = 2 (17 – f₁ + f₂)
⇒ 38 = 34 – 2f₁ + 2f₂
⇒ 4 = -2f₁ + 2f₂
⇒ -f₁ + f₂ = 2 …(2)
On adding eqn. (1) and eqn. (2); we have
2f₂ = 20 ⇒ f₂ = 10
∴ from(1); f₁ = 18 – 10 = 8