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 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:

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

Read  More:

• Classmark and Discrete Frequency Distribution
• Cumulative Frequency in statistics

Question 2:
We prepare the frequency table, given below

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.

More Resources

• RS Aggarwal Solutions Class 10
• RD Sharma Class 10 Solutions
• NCERT Solutions

Question 3:
We prepare the frequency table given below:

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:

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:

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:

\(N=100\\ \Longrightarrow \frac { N }{ 2 } =50  \)
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.

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 f₁ and f₂ 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 +f₁ = (30 + f₁)


Question 9:
Let f₁ and f₂ 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 = f₁ +5+9= (14 + f₁)

Question 10:
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

\(N=340\\ \Longrightarrow \frac { N }{ 2 } =170  \)
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

\(N=98\\ \Longrightarrow \frac { N }{ 2 } =49  \)
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

\(N=50\\ \Longrightarrow \frac { N }{ 2 } =25  \)
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

\(N=123\\ \Longrightarrow \frac { N }{ 2 } =61.5  \)
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:

\(N=200\\ \Longrightarrow \frac { N }{ 2 } =100  \)
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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