RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B
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 9B
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 9B
Question 1:
Class | Frequency f_{i} | C.F. |
0 – 10 | 3 | 3 |
10 – 20 | 6 | 9 |
20 – 30 | 8 | 17 |
30 – 40 | 15 | 32 |
40 – 50 | 10 | 42 |
50 – 60 | 8 | 50 |
N=∑ f_{i}=50 |
Now, \(N=50\\ \Longrightarrow \frac { N }{ 2 } =25 \)
The cumulative frequency just greater than 25 is 32 and corresponding class is 30 – 40.
Thus, the median class is 30 – 40
l = 30, h = 10, f = 15, c = C.F. preceding class 30 – 40 is 17 and \(\frac { N }{ 2 } \)= 25
Hence the median is 35.33
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Question 2:
We prepare the frequency table, given below
Class | Frequency f_{i} | C.F. |
0 – 7 | 3 | 3 |
7 – 14 | 4 | 7 |
14 – 21 | 7 | 14 |
21 – 28 | 11 | 25 |
28 – 35 | 0 | 25 |
35 – 42 | 16 | 41 |
42 – 49 | 9 | 50 |
N=∑ f_{i}=50 |
Now,\(N=50\\ \Longrightarrow \frac { N }{ 2 } =25 \)
The cumulative frequency is 25 and corresponding class is 21 – 28.
Thus, the median class is 21 – 28
l = 21, h = 7, f = 11, c = C.F. preceding class 21 – 28 is 14 and \(\frac { N }{ 2 } \)= 25
Hence the median is 28.
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Question 3:
We prepare the frequency table given below:
Daily wages | Frequency f_{i} | C.F. |
0 – 100 | 40 | 40 |
100 – 200 | 32 | 72 |
200 – 300 | 48 | 120 |
300 – 400 | 22 | 142 |
400 – 500 | 8 | 150 |
N=∑ f_{i}=150 |
Now, \(N=150\\ \Longrightarrow \frac { N }{ 2 } =75 \)
The cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.
Thus, the median class is 200 – 300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and \(\frac { N }{ 2 } \)= 75
Hence the median of daily wages is Rs. 206.25.
Question 4:
We prepare the frequency table, given below:
Class | Frequency f_{i} | C.F. |
5 – 10 | 5 | 5 |
10 – 15 | 6 | 11 |
15 – 20 | 15 | 26 |
20 – 25 | 10 | 36 |
25 – 30 | 5 | 41 |
30 – 35 | 4 | 45 |
35 – 40 | 2 | 47 |
40 – 45 | 2 | 49 |
N=∑ f_{i}=49 |
Now, \(N=49\\ \Longrightarrow \frac { N }{ 2 } =24.5 \)
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.
Thus, the median class is 15 – 20
l = 15, h = 5, f = 15
c = CF preceding median class = 11 and \(\frac { N }{ 2 } \)=24.5
Median of frequency distribution is 19.5
Question 5:
We prepare the cumulative frequency table as given below:
Consumption | Frequency f_{i} | C.F. |
65 – 85 | 4 | 4 |
85 – 105 | 5 | 9 |
105 – 125 | 13 | 22 |
125 – 145 | 20 | 42 |
145 – 165 | 14 | 56 |
165 – 185 | 7 | 63 |
185 – 205 | 4 | 67 |
N=∑ f_{i}=67 |
Now, \(N=67\\ \Longrightarrow \frac { N }{ 2 } =33.5 \)
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.
Thus, the median class is 125 – 145
l = 125, h = 20, and c = CF preceding the median class = 22, \(\frac { N }{ 2 } \)= 33.5
Hence median of electricity consumed is 136.5
Question 6:
Frequency table is given below:
Hieght | Frequency f_{i} | C.F. |
135 – 140 | 6 | 6 |
140 – 145 | 10 | 16 |
145 – 150 | 18 | 34 |
150 – 155 | 22 | 56 |
155 – 160 | 20 | 76 |
160 – 165 | 15 | 91 |
165 – 170 | 6 | 97 |
170 – 175 | 3 | 100 |
N=∑ f_{i}=100 |
\(N=100\\ \Longrightarrow \frac { N }{ 2 } =50 \)
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155
Thus, the median class is 150 – 155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
Hence, Median = 153.64
Question 7:
The frequency table is given below. Let the missing frequency be x.
Class | Frequency f_{i} | C.F. |
0 – 10 | 5 | 5 |
10 – 20 | 25 | 30 |
20 – 30 | x | 30+x |
30 – 40 | 18 | 48+x |
40 – 50 | 7 | 55+x |
Median = 24
Median class is 20 – 30
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
Hence, the missing frequency is 25.
Question 8:
Let f_{1} and f_{2} be the frequencies of classes 20 – 30 and 40 – 50 respectively, then
Median is 35, which lies in 30 – 40, so the median class is 30 – 40.
l = 30, h = 10, f = 40, N = 170 and c = 10 + 20 +f_{1} = (30 + f_{1})
Question 9:
Let f_{1} and f_{2 }be the frequencies of class intervals 0 – 10 and 40 – 50
Median is 32.5 which lies in 30 – 40, so the median class is 30 – 40
l = 30, h = 10, f = 12, N = 40 and c = f_{1} +5+9= (14 + f_{1})
Question 10:
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Class | Frequency f_{i} | C.F. |
18.5 – 25.5 | 35 | 35 |
25.5 – 32.5 | 96 | 131 |
32.5 – 39.5 | 68 | 199 |
39.5 – 46.5 | 102 | 301 |
46.5 – 53.5 | 35 | 336 |
53.5 – 60.5 | 4 | 340 |
N=∑ f_{i}=340 |
\(N=340\\ \Longrightarrow \frac { N }{ 2 } =170 \)
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
Median class is 32.5 – 39.5
l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
Hence median is 36.5 years
Question 11:
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
Wages per day
(in Rs.) |
Frequency f_{i} | C.F. |
60.5 – 70.5 | 5 | 5 |
70.5 – 80.5 | 15 | 20 |
80.5 – 90.5 | 20 | 40 |
90.5 – 100.5 | 30 | 70 |
100.5 – 110.5 | 20 | 90 |
110.5 – 120.5 | 8 | 98 |
N=∑ f_{i}=98 |
\(N=98\\ \Longrightarrow \frac { N }{ 2 } =49 \)
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.
median class is 90.5 – 100.5
l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
Hence, Median = Rs 93.50
Question 12:
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks | Frequency f_{i} | C.F. |
10.5 – 15.5 | 2 | 2 |
15.5 – 20.5 | 3 | 5 |
20.5 – 25.5 | 6 | 11 |
25.5 – 30.5 | 7 | 18 |
30.5 – 35.5 | 14 | 32 |
35.5 -40.5 | 12 | 44 |
40.5 -45.5 | 4 | 48 |
45.5 -50.5 | 2 | 50 |
N=∑ f_{i}=50 |
\(N=50\\ \Longrightarrow \frac { N }{ 2 } =25 \)
The cumulative frequency just greater than 25 is 32.
The corresponding class is 30.5 – 35.5.
Thus, the median class is 30.5 – 35.5
l = 30.5, h = 5, f = 14, c = C.F preceding median class = 18
Hence, Median = 33
Question 13:
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks | Frequency f_{i} | C.F. |
0.5 – 5.5 | 7 | 7 |
5.5 – 10.5 | 10 | 17 |
10.5 – 15.5 | 16 | 33 |
15.5 – 20.5 | 32 | 65 |
20.5 – 25.5 | 24 | 89 |
25.5 – 30.5 | 16 | 105 |
30.5 – 35.5 | 11 | 116 |
35.5 – 40.5 | 5 | 121 |
40.5 – 45.5 | 2 | 123 |
N=∑ f_{i}=123 |
\(N=123\\ \Longrightarrow \frac { N }{ 2 } =61.5 \)
The cumulative frequency just greater than 61.5 is 65.
The corresponding median class is 15.5 – 20.5.
Then the median class is 15.5 – 20.5
l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
Hence, Median = 19.95
Question 14:
Marks | Frequency f_{i} | C.F. |
0 – 10 | 12 | 12 |
10 – 20 | 20 | 32 |
20 – 30 | 25 | 57 |
30 – 40 | 23 | 80 |
40 – 50 | 12 | 92 |
50 – 60 | 24 | 116 |
60 – 70 | 24 | 116 |
70 – 80 | 36 | 200 |
N=∑ f_{i}=200 |
\(N=200\\ \Longrightarrow \frac { N }{ 2 } =100 \)
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus the median class is 50 – 60
l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, \(\frac { N }{ 2 } \)= 100
Hence, Median = 53.33
Hope given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9B are helpful to complete your math homework.
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