This exercise teaches the approach to compute median values within datasets divided into categories. The median functions as the central value in datasets after performing an ascending order arrangement. A deep understanding of median calculation from frequency distributions becomes vital since this method provides key information about data centralisation. This exercise shows why median measurements matter in practical applications, which include both income distribution analysis and examination scores and population age examinations.
This Story also Contains
Students will identify the median class and apply the right formula to compute the median through this assignment. The NCERT Solutions preserve their structure based on the newest books introduced for the educational year 2025–26. The given solutions show students how to apply step-by-step procedures for median computation, which improves their ability to understand data described in the NCERT Books. Knowing this concept enables learners to use their apply the learned skills in economics and healthcare sectors and social studies since knowledge of data central values remains vital.
First, we need to find the cumulative frequency and also let the assumed mean be a = 130 and h = 20 and will make the table as follows:
Class | Number of consumers $f_i$ | Cumulative Frequency | Classmark $x_i$ | $d_i = x_i -a$ | $u_i = \frac{d_i}{h}$ | $f_iu_i$ |
65-85 | 4 | 4 | 70 | -60 | -3 | -12 |
85-105 | 5 | 9 | 90 | -40 | -2 | -10 |
105-125 | 13 | 22 | 110 | -20 | -1 | -13 |
125-145 | 20 | 42 | 130 | 0 | 0 | 0 |
145-165 | 14 | 56 | 150 | 20 | 1 | 14 |
165-185 | 8 | 64 | 170 | 40 | 2 | 16 |
185-205 | 4 | 68 | 190 | 60 | 3 | 12 |
$\sum f_i = N$ = 68 | $\sum f_ix_i$ = 7 |
MEDIAN:
As, $N= 68 \implies \frac{N}{2} = 34$
Therfore, Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125; c.f. = 22; f = 20; h = 20
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting the values, we get:
$\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12$
$= 137$
Thus, the median of the data is 137
MODE:
The class having the maximum frequency is the modal class.
The maximum frequency is 20, and hence the modal class = 125 - 145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( $f_1$ ) of the modal class = 20
Frequency ( $f_0$ ) of class preceding the modal class = 13
Frequency ( $f_2$ ) of class succeeding the modal class = 14.
Therefore, $Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$
After putting the values, we get:
$\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20$
$= 135.76$
Thus, the Mode of the data is 135.76
MEAN:
$\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h$
After putting the values, we get:
$= 130 + \frac{7}{68}\times20 = 137.05$
Thus, the Mean of the data is 137.05
Q2 If the median of the distribution given below is 28.5, find the values of x and y.
Class | Number of consumers $f_i$ | Cumulative Frequency |
0-10 | 5 | 5 |
10-20 | x | 5+x |
20-30 | 20 | 25+x |
30-40 | 15 | 40+x |
40-50 | y | 40+x+y |
50-60 | 5 | 45+x+y |
$\sum f_i = N$ = 60 |
As, $N= 60 \implies \frac{N}{2} = 30$
Given median = 28.5, which lies in the class 20-30
Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10
Thus, $Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting the values, we get:
$\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8$
Also,
$\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7$
Therefore, the required values are: x=8 and y=7
Class | Frequency $f_i$ | Cumulative Frequency |
15-20 | 2 | 2 |
20-25 | 4 | 6 |
25-30 | 18 | 24 |
30-35 | 21 | 45 |
35-40 | 33 | 78 |
40-45 | 11 | 89 |
45-50 | 3 | 92 |
50-55 | 6 | 98 |
55-60 | 2 | 100 |
Given, $N= 100 \implies \frac{N}{2} = 50$
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10
Therefore, $Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting in the values, we get:
$\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\$
$= 35.75$
Thus, the median age is 35.75 years.
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes.
Class | Frequency $f_i$ | Cumulative Frequency |
117.5-126.5 | 3 | 3 |
126.5-135.5 | 5 | 8 |
135.5-144.5 | 9 | 17 |
144.5-153.5 | 12 | 29 |
153.5-162.5 | 5 | 34 |
162.5-171.5 | 4 | 38 |
171.5-180.5 | 2 | 40 |
As,$ N= 40 \implies \frac{N}{2} = 20$
Therefore, Median class = 144.5-153.5
Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting in the values, we get:
$\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\$
$= 146.75$
Thus, the median length of the leaves is 146.75 mm
Q5 The following table gives the distribution of the lifetime of 400 neon lamps:
Find the median lifetime of a lamp.
Class | Frequency $f_i$ | Cumulative Frequency |
1500-2000 | 14 | 14 |
2000-2500 | 56 | 70 |
2500-3000 | 60 | 130 |
3000-3500 | 86 | 216 |
3500-4000 | 74 | 290 |
4000-4500 | 62 | 352 |
4500-5000 | 48 | 400 |
$ N= 400 \implies \frac{N}{2} = 200$
Therefore, Median class = 3000-3500
Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting in the values, we get:
$\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97$
$= 3406.97$
Thus, the median lifetime of a lamp is 3406.97 hours
$= 146.75$
Thus, the median length of the leaves is 146.75 mm
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Class | Number of surnames $f_i$ | Cumulative Frequency | Classmark $x_i$ | $f_ix_i$ |
1-4 | 6 | 6 | 2.5 | 15 |
4-7 | 30 | 36 | 5.5 | 165 |
7-10 | 40 | 76 | 8.5 | 340 |
10-13 | 16 | 92 | 11.5 | 184 |
13-16 | 4 | 96 | 14.5 | 51 |
16-19 | 4 | 100 | 17.5 | 70 |
$\sum f_i = N$ = 100 | $\sum f_ix_i$ = 825 |
MEDIAN:
$N= 100 \implies \frac{N}{2} = 50$
$\therefore$ Median class = 7-10; Lower limit, l = 7;
Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting in the values, we get:
$\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\$
$= 8.05$
Thus, the median of the data is 8.05
MODE:
The class having the maximum frequency is the modal class.
The maximum frequency is 40, and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( $f_1$ ) of the modal class = 40
Frequency ( $f_0$ ) of class preceding the modal class = 30
Frequency ( $f_2$ ) of class succeeding the modal class = 16
$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$
After putting in the values, we get:
$\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3$
$= 7.88$
Thus, the Mode of the data is 7.88
MEAN:
$\overline x =\frac{\sum f_ix_i}{\sum f_i}$
$= \frac{825}{100} = 8.25$
Thus, the Mean of the data is 8.25
Class | Number of students $f_i$ | Cumulative Frequency |
40-45 | 2 | 2 |
45-50 | 3 | 5 |
50-55 | 8 | 13 |
55-60 | 6 | 19 |
60-65 | 6 | 25 |
65-70 | 3 | 28 |
70-75 | 2 | 30 |
MEDIAN:
$N= 30 \implies \frac{N}{2} = 15$
Therefore, Median class = 55-60; Lower limit, l = 55;
Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$
After putting in the values, we get:
$\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5$
$= 56.67$
Thus, the median weight of the student is 56.67 kg
Also Read-
1. Understanding Median for Grouped Data: The concept of Median for grouped data demonstrates how central values separate statistical sets into two balanced sections for data that includes class intervals.
2. Identifying the Median Class: The process of identifying the Median Class enables you to find which interval contains the median because it serves as the fundamental step for calculating the median in grouped data.
3. Applying the Median Formula: When calculating the median, apply statistical formulas that include the lower end of the median class, its width and cumulative frequency figures.
4. Solving Real-Life Based Word Problems: The resolution of real-life word problems requires frequency tables that stem from actual survey results or performance scores to find the central value.
5. Interpreting Data Trends: Students should learn to recognise meaningful conclusions based on trends and patterns within frequency groups of raw data.
Check Out:
Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
The concepts related to median is discussed in class 10 maths ex 14.3. Formula of median of even observation is (n/2+ 1 )th observation. Also practice class 10 ex 14.3 which is discussed in this article to get deeper understanding of the concepts.
3 Median = Mode + 2 Mean is the relation between mean, medium and mode as mentioned in Class 10 maths chapter 14 exercise 14.3.
These concepts are discussed in 10th class maths exercise 14.3 answers. practice them to command the concepts. Cumulative Frequency Table is the cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors.
Median gives the value of the observation which is at the center so it is dependent on either the observation is odd number or even number. Practice ex 14.3 class 10 to command these concepts.
The concept related to discusses in class 10 ex 14.3. Formula of median of odd observation is n/2th observation. Students can practice ex 14.3 class 10 to get deeper understanding of concepts.
On Question asked by student community
Hello,
For your information, the class 7 2025 result of the CM Shri School is expected to be published on September 20, 2025. So, if you want to see and download the results, then you can visit the official website of the CM Shri School: edudel.nic.in.
I hope it will clear your query!!
Hello
Visit the official website of the Rajasthan Education Department or the Shala Darpan portal.
Look for the “Latest News” or “Examination” section. Check school notice boards or ask your class teacher for updates.
Some district education office websites may also upload the key. Avoid unofficial websites as they may provide incorrect or fake keys.
Hrllo,
If you want to view the CM Shri School 2025-26 admission test result , visit the official website, click on the "CM Shri Schools Admission Test - 2025." then select the "CM Shri Admission Test Result 2025 - Merit List" link. Here you need to log in with your credentials and view or download the merit list pdf.
I hope it will clear your query!!
Hello aspirant,
Fairness and equal opportunity are guaranteed by the rigorous merit-based admissions process at CM Shri Schools. The merit list will be made public on edudel.nic.in, the official DOE website. Based on how well they did on the admission exam, students will be shortlisted.
To know the result, you can visit our site through following link:
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
Thank you
Hello
The CM Shri School Admission Test was conducted for students seeking quality education.
The result and selection list are expected to be released in the last week of September 2025.
Students who qualified will get admission into CM Shri Model Schools across the state.
The official list of selected students will be available on the edudel.nic.in website.
Candidates can check their names in the merit list PDF once uploaded.
Selection is based on the student’s performance in the entrance exam.
Keep checking the official site or the school notice board for timely updates.
Check - https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters