OP Malhotra Class 10 Maths Solutions Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(c)

Students can cross-reference their work with ICSE Class 10 Maths Solutions S Chand Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(c) to ensure accuracy.

S Chand Class 10 ICSE Maths Solutions Chapter 18 Arithmetic Mean, Median, Mode and Quartiles Ex 18(c)

Question 1.
Find Q₁, Q₃ for the following sets data :
(i) 38, 7, 43, 25, 20, 15, 12, 18, 11
(ii) 5, 12, 17, 23, 28, 31, 37, 41, 42, 49, 54, 58, 65, 68, 73, 77
(iii) 22, 21, 12, 15, 17, 18, 18, 20, 19, 1, 6, 25
Solution:
(i) 38, 7, 43, 25, 20, 15, 12, 18, 11
Arranging in ascending order,
7, 11, 12. 15, 18, 20, 25, 38, 43
Here n = 9 which is odd

∴ Size of \(\frac { n + 1 }{ 2 }\)th term = \(\frac { 9 + 1 }{ 2 }\) = 2.5th term
Which lies in 12
Q₁ = Size of 2nd term + 0.5 of (Size of 3rd term – 2nd term)
= 11 + 0.5 (12 – 11) = 11 + 0.5 × 1 = 11 + 0.5 = 11.5
Q₃ = Size of \(\frac{3(n+1)}{4}\)th term = \(\frac{(9+1)}{4} \times 3\) = \(\frac { 30 }{ 4 }\) = 7.5th term
= 7th term + 0.5 (8th term – 7th term)
= 25 + 0.5 (38 – 25) = 25 + \(\frac { 1 }{ 2 }\) × 13 = 25 + 6.5 = 31.5

(ii) 5, 12. 17,23,28,31,37,41,42,49, 54, 58, 65, 68, 73, 77
There are 16 term which is even
\(\frac{n+1}{4}\) = \(\frac{16+1}{4}\) = \(\frac { 17 }{ 4 }\) = 4.25th term
∴ Q₁ = size of 4.25th term = 0
= Size of 4th term + 0.25 (28 – 23) = 23 + 0.25 × 5 = 23 + 1.25 = 24.25
Q₃ = size of \(\frac{16+1}{4} \times 3=\frac{51}{4} \text { th }\) term = 12.75th term
= size of 12th term + 0.75 (13th term – 12th term) = 58 + 0.75 (65 – 58) = 58 + 0.75 × 7 = 58 + 5.25 = 63.25

(iii) 22, 21, 12, 15, 17. 18, 18. 20, 19. 1, 6. 25
Arranging in ascending order
1, 6, 12, 15, 17, 18, 18, 19, 20, 21, 22. 25
Here n = 12
∴ \(\frac{n+1}{4} \text { th }\) term = \(\frac { 12 + 1 }{ 4 }\) = \(\frac { 13 }{ 4 }\) = 3.25th term
Q₁ = size of 3rd term + 0.25 (4th term – 3rd term)
= 12 + 0.25 (15 – 12) = 12 + 0.25 × 3 = 12 + 0.75 = 12.75
Now Q₃ = \(\frac { n + 1 }{ 4 }\) × 3 = \(\frac { 12 + 1 }{ 4 }\) × 3 = \(\frac { 39 }{ 4 }\) = 9.75th term
Q₃ = size of 9th term + 0.75 (10th term – 9th term)
= 20 + 0.75 (21 – 20) = 20 + 0.75 × 1 = 20 + 0.75 = 20.75

Question 2.
In a class of ten students, the marks obtained by each student are shown against their Roll No’s given below:

Roll No.	1	2	3	4	5	6	7	8	9	10
Marks obtained	12	30	20	15	25	10	2	40	4	8

Calculate the lower and the upper quartiles.
Solution:
Arranging the marks obtain by the students in ascending order
2,4,8,10,12.15,20,25,30,40
Here n = 10
Lower Quartile (Q₁) = \(\frac { n + 1 }{ 4 }\) = \(\frac { 10 + 1 }{ 4 }\) = \(\frac { 11 }{ 4 }\) = 2.75th term
= 2nd term + 0.75 (3rd term – 2nd term) = 4 + 75 (8 – 4) = 4 + \(\frac { 75 }{ 4 }\) × 4 = 4 + 3 = 7 marks
Upper Quartile (Q₃) = \(\frac{3(n+1)}{4}\) = \(\frac{11 \times 3}{4}\) = \(\frac{11}{4}\) = 8.25th term
∴ Q₃ = 8th term + .25 (9th term – 8th term) = 25 + 0.25 (30 – 25) = 25 + .25 × 5 = 25 + 1.25 = 26.25 marks

Question 3.
From the data given below calculate the value of 3rd quartile.

Marks obtained	10	11	12	13	14	15	16	17	18	19	20
No. of students	2	4	6	8	10	16	10	9	6	4	1

Solution:

Marks obtained	No. of Students (f)	c.f.
10	2	2
11	4	6
12	6	12
13	8	20
14	10	30
15	16	46
16	10	56
17	9	65
18	6	71
19	4	75
20	1	76
Total	76	Which is even

∴ Third Quartile (Q₃) = \(\frac{3n}{4}\) = \(\frac{3 \times 76}{4} \text { th }\) = 57th term
Which is 17
∵ 57 lies between 57 – 65 in c.f.
∴ Q₃ = 17 marks

Question 4.
Find both quartiles for the following distribution.
(i)

Marks	30	40	50	60	70	80	90
No. of students	4	6	11	19	20	26	14

(ii)

Size	4	5	6	7	8
Frequency	2	5	8	9	6

Solution:
(i)

Marks	No. of Students (f)	c.f.
30	4	4
40	6	10
50	11	21
60	19	40
70	20	60
80	26	86
90	14	100
Total	100	which is odd

Here n = 100
∴ Q₁ = \(\frac { n + 1 }{ 4 }\) = \(\frac { 100 + 1 }{ 4 }\)th term = \(\frac { 101 }{ 4 }\) = 25.25th term
which lies in 40 c.f.
∴ Q₁ = 60
Q₃ = \(\frac { n + 1 }{ 4 }\) × 3 = \(\frac { 100 + 1 }{ 4 }\) × 3 = \(\frac{101 \times 3}{4} \text { th }\) term = \(\frac{303}{4}\)th = 75.75th term which lies in 86 c.f.
∴ Q₃ = 80

(ii)

Size	Frequency (f)	c.f.
4	2	2
5	5	7
6	8	15
7	9	24
8	6	30
Total	30

Here n = 30
∴ Q₁th term = \(\frac { n + 1 }{ 4 }\) = \(\frac { 30 + 1 }{ 4 }\)th = \(\frac { 31 }{ 4 }\)th = 7.75th term
Which in 15 c.f.
∴ Q₁ = 6
Q₃th term = \(\frac { n + 1 }{ 4 }\) × 3 = \(\frac { 30 + 1 }{ 4 }\) × 3 = \(\frac{31 \times 3}{4}\) = \(\frac { 93 }{ 4 }\) = 23.25th term which lies in 24th c.f.
∴ Q₃ = 7

Question 5.
Find the median size of the shoes from figures given below :

Size of shoes	4.5	5	5.5	6	6.5	7	7.5	8	8.5	9	9.5	10	10.5	11
Frequency	1	2	4	5	15	30	60	95	82	75	44	25	15	4

Also calculate the Quartiles.
Solution:

Size of shoes	Frequency	c.f.
4.5	1	1
5	2	3
5.5	4	7
6	5	12
6.5	15	27
7	30	57
7.5	60	117
8	95	212
8.5	82	294
9	75	309
9.5	44	413
10	25	438
10.5	15	453
11	4	457

Total 457 which is odd
∴ Median =\(\frac { n + 1 }{ 2 }\)th = \(\frac { 457 + 1 }{ 2 }\)th = \(\frac { 45 8 }{ 2 }\) term = 229th term
Which lies between 213 – 294 in c.f.
∴ Median = 8.5
Q₁ = \(\frac { n + 1 }{ 2 }\)th = \(\frac { 458 }{ 4 }\)th = 114.5th term
Which lies between 58-117 in c.f.
∴ Q₁ = 7.5
Q₃ = \(\frac{3(n+1)}{4} \text { th }\)
\(=\frac{3(457+1)}{4} \text { th }\)
\(=\frac{3 \times 458}{4} \text { th }\)
\(=\frac{1374}{4} \text { th }\) = 343.5th term
Which lies between 294 and 369
∴ Q₃ = 9