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Median of Grouped Data PYQs

Class 10 Statistics Board Questions (2014 – 2026)

Topic Overview

Master the Median. Identify the Median Class using $N/2$. Use the formula $L + \left(\frac{\frac{N}{2}-cf}{f}\right)h$. Pay special attention to Missing Frequency problems (finding $x, y$).

Q1 2024
00:00
The median class of the following distribution is: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |-------|------|-------|-------|-------|-------| | Freq | 8 | 10 | 12 | 22 | 18 |
(a)(A) 10-20
(b)(B) 20-30
(c)(C) 30-40
(d)(D) 40-50
CF: 8, 18, 30, 52, 70. $N/2 = 35$. Median class is 30-40.
Ans: (C) 30-40
Q2 2026
00:00
In the formula $\text{Median} = L + \left(\frac{N/2 - cf}{f}\right)h$, the symbol '$cf 00:00
The median class of the following distribution is: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |-------|------|-------|-------|-------|-------| | Freq | 8 | 10 | 12 | 22 | 18 |
(a)(A) 10-20
(b)(B) 20-30
(c)(C) 30-40
(d)(D) 40-50
CF: 8, 18, 30, 52, 70. $N/2 = 35$. Median class is 30-40.
Ans: (C) 30-40
Q2 2026
00:00
In the formula $\text{Median} = L + \left(\frac{N/2 - cf}{f}\right)h$, the symbol '$cf
stands for:
(a)(A) CF of the median class
(b)(B) CF of the class preceding the median class
(c)(C) CF of the class succeeding the median class
(d)(D) Sum of frequencies
By definition, $cf$ is the cumulative frequency of the class preceding the median class.
Ans: (B)
Q3 2026
00:00
Calculate the median heart rate for the following data obtained from a school health check-up: | Heart Rate (bpm) | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 | |------------------|-------|-------|-------|-------|-------| | No. of students | 12 | 30 | 25 | 18 | 15 |
$N = 100, N/2 = 50$. CF: 12, 42, 67, 85, 100.
Median class: 75-80.
$L=75, cf=42, f=25, h=5$.
Median $= 75 + [(50-42)/25] \times 5 = 75 + 1.6 = 76.6$ bpm.
Ans: 76.6 bpm
Q4 2025
00:00
If the median of the distribution given below is 28.5, find the values of $x$ and $y$, if the total frequency is 60. | Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | |----------------|------|-------|-------|-------|-------|-------| | Frequency | 5 | $x$ | 20 | 15 | $y$ | 5 |
$x+y=15$. $28.5 = 20 + [(30-(5+x))/20] \times 10 \Rightarrow x=8, y=7$.
Ans: $x=8, y=7$
Q5 2018
00:00
The following distribution gives the marks of 100 students. Find the median marks. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | |-------|------|-------|-------|-------|-------|-------|-------|-------| | No. | 5 | 9 | 17 | 28 | 24 | 10 | 4 | 3 |
$N=100, N/2=50$. CF: 5, 14, 31, 59.
Median class: 30-40.
Median $= 30 + [(50-31)/28] \times 10 = 36.79$.
Ans: 36.79
stands for:
(a)(A) CF of the median class
(b)(B) CF of the class preceding the median class
(c)(C) CF of the class succeeding the median class
(d)(D) Sum of frequencies
By definition, $cf$ is the cumulative frequency of the class preceding the median class.
Ans: (B)
Q3 2026
00:00
Calculate the median heart rate for the following data obtained from a school health check-up: | Heart Rate (bpm) | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 | |------------------|-------|-------|-------|-------|-------| | No. of students | 12 | 30 | 25 | 18 | 15 |
$N = 100, N/2 = 50$. CF: 12, 42, 67, 85, 100.
Median class: 75-80.
$L=75, cf=42, f=25, h=5$.
Median $= 75 + [(50-42)/25] \times 5 = 75 + 1.6 = 76.6$ bpm.
Ans: 76.6 bpm
Q4 2025
00:00
If the median of the distribution given below is 28.5, find the values of $x$ and $y$, if the total frequency is 60. | Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | |----------------|------|-------|-------|-------|-------|-------| | Frequency | 5 | $x$ | 20 | 15 | $y$ | 5 |
$x+y=15$. $28.5 = 20 + [(30-(5+x))/20] \times 10 \Rightarrow x=8, y=7$.
Ans: $x=8, y=7$
Q5 2018
00:00
The following distribution gives the marks of 100 students. Find the median marks. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | |-------|------|-------|-------|-------|-------|-------|-------|-------| | No. | 5 | 9 | 17 | 28 | 24 | 10 | 4 | 3 |
$N=100, N/2=50$. CF: 5, 14, 31, 59.
Median class: 30-40.
Median $= 30 + [(50-31)/28] \times 10 = 36.79$.
Ans: 36.79
00:00
The median class of the following distribution is: | Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | |-------|------|-------|-------|-------|-------| | Freq | 8 | 10 | 12 | 22 | 18 |
(a)(A) 10-20
(b)(B) 20-30
(c)(C) 30-40
(d)(D) 40-50
CF: 8, 18, 30, 52, 70. $N/2 = 35$. Median class is 30-40.
Ans: (C) 30-40
Q2 2026
00:00
In the formula $\text{Median} = L + \left(\frac{N/2 - cf}{f}\right)h$, the symbol '$cf stands for:
(a)(A) CF of the median class
(b)(B) CF of the class preceding the median class
(c)(C) CF of the class succeeding the median class
(d)(D) Sum of frequencies
By definition, $cf$ is the cumulative frequency of the class preceding the median class.
Ans: (B)
Q3 2026
00:00
Calculate the median heart rate for the following data obtained from a school health check-up: | Heart Rate (bpm) | 65-70 | 70-75 | 75-80 | 80-85 | 85-90 | |------------------|-------|-------|-------|-------|-------| | No. of students | 12 | 30 | 25 | 18 | 15 |
$N = 100, N/2 = 50$. CF: 12, 42, 67, 85, 100.
Median class: 75-80.
$L=75, cf=42, f=25, h=5$.
Median $= 75 + [(50-42)/25] \times 5 = 75 + 1.6 = 76.6$ bpm.
Ans: 76.6 bpm
Q4 2025
00:00
If the median of the distribution given below is 28.5, find the values of $x$ and $y$, if the total frequency is 60. | Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | |----------------|------|-------|-------|-------|-------|-------| | Frequency | 5 | $x$ | 20 | 15 | $y$ | 5 |
$x+y=15$. $28.5 = 20 + [(30-(5+x))/20] \times 10 \Rightarrow x=8, y=7$.
Ans: $x=8, y=7$
Q5 2018
00:00
The following distribution gives the marks of 100 students. Find the median marks. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | |-------|------|-------|-------|-------|-------|-------|-------|-------| | No. | 5 | 9 | 17 | 28 | 24 | 10 | 4 | 3 |
$N=100, N/2=50$. CF: 5, 14, 31, 59.
Median class: 30-40.
Median $= 30 + [(50-31)/28] \times 10 = 36.79$.
Ans: 36.79