RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D
Other Exercises
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D
- RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9E
- Mean, Median, Mode of Grouped Dat
Exercise 9C
Question 1:
As the class 50 – 60 has maximum frequency, so it is the modal class:
Hence, mode = 53.33
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Question 2:
As the class 40 – 50 as maximum frequency, so it s modal class.
Hence, mode = 43.75 years
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Question 3:
As the class 26 – 30 has maximum frequency so it is modal class
Hence, mode = 28.5
Question 4:
As the class 1500 – 2000 has maximum frequency, so it os modal class
Hence the average expenditure done by maximum number of workers = Rs. 1820
Question 5:
As the class 5000 – 10000 has maximum frequency, so it is modal class
Hence, mode = Rs. 7727.27
Question 6:
As the class 15 – 20 has maximum frequency so it is modal class.
Hence mode = 17.3 years
Question 7:
As the class 85 – 95 has the maximum frequency it is modal class
Hence, mode = 85.71
Question 8:
The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get
Class | Frequency |
0.5 – 5.5 | 3 |
5.5 – 10.5 | 8 |
10.5 – 15. 5 | 13 |
15.5 – 20.5 | 18 |
20.5 – 25. 5 | 28 |
25.5 – 30.5 | 20 |
30.5 – 35.5 | 13 |
35.5 – 40.5 | 8 |
40.5 – 45.5 | 6 |
45.5 – 50.5 | 3 |
As the class 20.5 – 25.5 has maximum frequency, so it is modal class
Hence, mode = 23.28
Exercise 9D
Question 1:
Let assumed mean be 35, h = 10, now we have
Class | Frequency f_{i} | Mid value x_{i} | C.F | f_{i }u_{i} | |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 |
5
10 18 30 20 12 5 |
5
15 25 35 = A 45 55 65 |
-3
-2 -1 0 1 2 3 |
5
15 33 63 83 95 100 |
-15
-20 -18 0 20 24 15 |
N=100 | ∑(f_{i}_{ }u_{i})=6 |
(i) Mean
=
(ii) N = 100, = 50
Cumulative frequency just after 50 is 63
Median class is 30 – 40
l = 30, h = 10, N = 100, c = 33, f = 30
(iii) Mode = 3 × median – 2 × mean
= 3 × 35.67 – 2 × 35.6 = 107.01 – 71.2
= 35.81
Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81
Question 2:
Let assumed mean A be 8.5. Class interval h = 3
Class | Frequency f_{i} | Mid value x_{i} | f_{i }u_{i} | C.F | |
1-4 4-7 7-10 10-13 13-16 16-19 |
6
30 40 16 4 4 |
2.5
5.5 8.5 = A 11.5 14.5 17.5 |
-2
-1 0 1 2 3 |
-12
-30 0 16 8 12 |
6
36 76 92 96 100 |
N=100 | ∑(f_{i}_{ }u_{i})=-6 |
N = total frequency = 100
(i) Mean
=
(ii) = 50, Cumulative frequency just after 50 is 76
Median class is 7 – 10
l = 7, h = 3, N = 100, f = 40, c = 36
(iii) Mode = 3 × Median – 2 × Mean
= 3 × 8.05 – 2 × 8.32 = 24.15 – 16.64
= 7.51
Thus, mean = 8.32, Median = 8.05, Mode = 7.51
Question 3:
Let the assumed mean A be 145. Class interval h = 10.
Class | Frequency f_{i} | Mid value x_{i} | f_{i }u_{i} | C.F | |
120-130
130-140 140-150 150-160 160-170 |
2
8 12 20 8 |
125
135 145=A 155 165 |
-2
-1 0 1 2 |
-4
-8 0 20 16 |
2
10 22 42 50 |
N=50 | ∑(f_{i}_{ }u_{i})=24 |
(i) Mean
=
= 145 + 4.8 = 149.8
(ii) N = 50, = 25
Cumulative frequency just after 25 is 42
Corresponding median class is 150 – 160
Cumulative frequency before median class, c = 22
Median class frequency f = 20
(iii) Mode = 3 × median – 2 × mean
= 3 × 151.5 – 2 × 149.8 = 454.5 – 299.6
= 154.9
Thus, Mean = 149.8, Median = 151.5, Mode = 154.9
Question 4:
Let assumed mean A = 150 and h = 20
Class | Frequency f_{i} | Mid value x_{i} | f_{i }u_{i} | C.F | |
100-120
120-140 140-160 160-180 180-200 |
12
14 8 6 10 |
110
130 150= A 170 190 |
-2
-1 0 1 2 |
-24
-14 0 6 20 |
12
26 34 40 50 |
N=50 | ∑(f_{i}_{ }u_{i})=-12 |
(i) Mean
(ii) N = 50, = 25
Cumulative frequency just after 25 is 26
Corresponding frequency median class is 120 – 140
So, l = 120, f = 14, h = 20, c = 12
(iii) Mode = 3 × Median – 2 × Mode
= 3 × 138.6 – 2 × 145.2
= 415.8 – 190.4
= 125.4
Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4
Question 5:
Let assumed mean = 225 and h = 50
Class | Frequency f_{i} | Mid value x_{i} | f_{i }u_{i} | C.F | |
100-150
150-200 200-250 250-300 300-350 |
6
7 12 3 2 |
125
175 225 275 325 |
-2
-1 0 1 2 |
-12
-7 0 3 4 |
6
13 25 28 30 |
N=30 | ∑(f_{i}_{ }u_{i})=-12 |
(i) Mean
=
(ii) N = 30, = 15
Cumulative frequency just after 15 is 25
corresponding class interval is 200 – 250
Median class is 200 – 250
Cumulative frequency c just before this class = 13
So I=200, f=12, = 15, h=50, c=13
Hence, Mean = 205 and Median = 208.33
Hope given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9C & 9D are helpful to complete your math homework.
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