Imagine you and your friends are comparing your weekly mobile data usage. One friend uses 2 GB daily, another uses 3 GB, while someone else barely uses 1 GB. Now, if you want to find the “middle” usage value that best represents the group without getting affected by extremely high or low values, you use the concept of median. In Mathematics and Statistics, the median is one of the most important measures of central tendency used to analyze both grouped and ungrouped data in real life, from student marks and salary distribution to business reports and survey analysis. Understanding the Median of Grouped and Ungrouped Data helps students solve statistical problems quickly using formulas, median calculators, and step-by-step methods. In this article, you will learn the median formula for grouped and ungrouped data, solved examples, shortcut tricks, median calculation methods, and important practice questions. This topic is highly important for CBSE Maths, CUET, SSC, Bank Exams, CAT, IPMAT, JIPMAT, NDA, and other competitive quantitative aptitude and statistics-based examinations.
The median in statistics is the middle value of a data set after arranging all observations in ascending or descending order. It is one of the most important measures of central tendency used in mathematics and statistics to represent the central or typical value of data. Unlike the mean, the median is not heavily affected by extremely large or small values, which makes it highly useful in practical data analysis.
For example, if the ages of five students are 12, 14, 15, 18, and 20 years, then the median age is 15 because it lies exactly in the middle of the arranged data.
The concept of median is widely used in school-level statistics, competitive exam mathematics, economics, business reports, and data interpretation questions. It helps students understand how data is distributed and how to identify the central position of observations quickly and accurately.
The median is defined as the value that divides a data set into two equal halves after arranging the observations in proper order.
Half of the observations lie below the median.
Half of the observations lie above the median.
If the total number of observations is odd, the median is the middle observation.
$\text{Median} = \left(\frac{n+1}{2}\right)^{th} \text{ observation}$
If the total number of observations is even, the median is the average of the two middle observations.
$\text{Median} = \frac{\left(\frac{n}{2}\right)^{th} \text{ observation} + \left(\frac{n}{2}+1\right)^{th} \text{ observation}}{2}$
The median provides a clear understanding of the central value of the data and is especially helpful when the observations are unevenly distributed.
The median is extremely important in real-life situations because many data sets contain unusually high or low values that can affect the average. In such cases, the median gives a more realistic representation of the data compared to the arithmetic mean.
For example, consider the monthly salaries of employees in a company:
₹18,000, ₹22,000, ₹25,000, ₹27,000, and ₹4,50,000
Here, the mean salary becomes very high because of one extremely large salary, but the median salary remains ₹25,000, which better represents the income of most employees.
The median is commonly used in:
Salary and income analysis
Student marks evaluation
Business and market research
Population studies
Real estate price analysis
Sports performance statistics
Survey and data interpretation reports
Because of its accuracy in representing the middle value, the median is widely preferred in statistical analysis and competitive aptitude questions.
The median is one of the three main measures of central tendency in statistics:
Mean
Median
Mode
A measure of central tendency is used to identify the central or representative value of a data set. Among these measures, the median is particularly useful for skewed data or data sets containing outliers.
Important features of median include:
It is easy to calculate and understand.
It is not affected significantly by extreme values.
It can be used for both grouped and ungrouped data.
It represents the positional average of observations.
It is suitable for statistical and data interpretation problems.
The concept of median is an important topic in:
CBSE Class 9 and Class 10 Statistics
NCERT Mathematics
CUET Quantitative Aptitude
SSC and Banking Exams
CAT, IPMAT, and JIPMAT
Data Interpretation and Logical Reasoning
A strong understanding of median helps students solve statistical questions faster and improves their overall problem-solving skills in mathematics and competitive exams.
Finding the median of ungrouped data is one of the most important concepts in statistics and measures of central tendency. In ungrouped data, all observations are given individually rather than in the form of class intervals. The median helps identify the middle value of the arranged data set and is widely used in school mathematics, competitive exams, and data interpretation questions.
Follow these simple steps to calculate the median of ungrouped data accurately:
Write all the observations from the smallest value to the largest value.
Count the total number of values in the data set. Let the total number of observations be represented by $n$.
$\text{Median} = \left(\frac{n+1}{2}\right)^{th} \text{ observation}$
$\text{Median} = \frac{\left(\frac{n}{2}\right)^{th} \text{ observation} + \left(\frac{n}{2}+1\right)^{th} \text{ observation}}{2}$
Find the median of the following data set:
5, 2, 9, 1, 6
Arrange the observations in ascending order:
1, 2, 5, 6, 9
Number of observations:
$n = 5$
Since the number of observations is odd:
$\text{Median} = \left(\frac{5+1}{2}\right)^{th}$ observation
$\text{Median} = 3^{rd}$ observation
The 3rd observation is 5.
Therefore, the required median is:
$\boxed{5}$
This method is frequently used in CBSE Class 9 and 10 Statistics, CUET Quantitative Aptitude, SSC Maths, Banking Exams, and CAT preparation.
The median of grouped data is calculated when observations are presented in the form of class intervals along with their frequencies. In grouped frequency distribution, the exact values are not individually known, so we use the median formula for grouped data to determine the central value.
This topic is highly important for statistics chapters, data interpretation questions, and competitive aptitude exams.
Cumulative frequency is the running total of frequencies in a frequency distribution table. It helps in identifying the median class while calculating the median of grouped data.
For example, if the frequencies are 5, 7, and 9:
Cumulative frequency plays a key role in finding the median class accurately.
Calculating the median of grouped data is an important concept in statistics that helps determine the middle value from a continuous frequency distribution table. By using cumulative frequency and the median formula, students can easily find the central value of grouped observations. This method is widely used in CBSE Statistics, NCERT Maths, CUET, SSC, Banking, CAT, and other competitive exams.
Make a table containing:
Enter the frequencies corresponding to each class interval.
Find the cumulative frequency by adding frequencies successively.
Calculate:
$\sum f = n$
The last cumulative frequency is equal to the total number of observations.
Calculate:
$\frac{n}{2}$
Now identify the class interval whose cumulative frequency is just greater than $\frac{n}{2}$.
This class interval is known as the median class.
Use the median formula for grouped data:
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Where:
Consider the following grouped frequency distribution table:
| Class Interval | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
|---|---|---|
| 0–10 | 7 | 7 |
| 10–20 | 9 | 16 |
| 20–30 | 10 | 26 |
| 30–40 | 14 | 40 |
| 40–50 | 10 | 50 |
$\sum f = 50$
$\frac{n}{2} = \frac{50}{2} = 25$
The cumulative frequency nearest to and greater than 25 is 26.
Therefore, the median class is:
20–30
$\text{Median} = 20+\frac{25-16}{10}\times10$
$\text{Median} = 20+\frac{9}{10}\times10$
$\text{Median} = 20+9$
$\text{Median} = 29$
Hence, the required median is: $\boxed{29}$
The median signifies the middle value of a data set and divides the observations into two equal halves.
The median provides a better representation of central tendency when the data contains outliers or extreme values. This is why median is commonly used in economics, salary analysis, business statistics, and survey reports.
For example:
The median is one of the most reliable measures of central tendency in statistics because it represents the middle value of a data set accurately, even when extreme values are present. Understanding the important properties and features of median helps students solve statistics questions more effectively in CBSE exams, competitive aptitude tests, and data interpretation problems.
Unlike arithmetic mean, the median remains stable even if some observations are extremely high or low.
For unevenly distributed data, the median gives a more realistic central value.
The concept of median applies to both individual observations and frequency distribution tables.
In a perfectly symmetrical distribution: $\text{Mean} = \text{Median}$
Learning the best tips and tricks to solve median questions helps students improve speed and accuracy in statistics problems. Whether you are preparing for CBSE board exams, CUET, SSC, Banking, CAT, IPMAT, or other competitive aptitude exams, these shortcut techniques can help you solve median-based questions quickly with fewer calculation mistakes.
Competitive exams often include direct and concept-based questions on median, grouped data, and measures of central tendency. Using smart shortcut methods can save valuable time during the exam.
Before finding the median, arrange all observations in ascending or descending order. Many students directly apply formulas without sorting the data, which leads to incorrect answers.
For example:
Given data: 8, 2, 11, 5, 7
Correct arrangement:
2, 5, 7, 8, 11
Now the median can be identified easily.
If the number of observations is odd, the median is the middle term.
If the number of observations is even, take the average of the two middle terms.
This basic observation helps solve many objective-type statistics questions instantly.
In grouped frequency distribution questions:
First calculate $\frac{n}{2}$
Then identify the class whose cumulative frequency is just greater than $\frac{n}{2}$
That class interval is always the median class.
This trick is highly useful in SSC, Banking, and Data Interpretation questions.
The grouped data median formula should be remembered clearly for fast calculations.
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Students who memorize the formula properly can solve statistics questions much faster during exams.
Using efficient calculation methods can significantly reduce the time required to solve median questions.
While preparing the cumulative frequency table:
Add frequencies step-by-step
Double-check the final total frequency
Avoid skipping any values
A small mistake in cumulative frequency changes the entire median calculation.
In MCQ-based exams, estimation techniques often help eliminate wrong options quickly.
For example:
If the median class is 20–30, the answer will usually lie within this interval.
Options outside the interval can often be rejected immediately.
In symmetrical distributions:
$\text{Mean} = \text{Median} = \text{Mode}$
This shortcut is useful in conceptual and theory-based aptitude questions.
Repeated practice improves calculation speed and helps students recognize common exam patterns in median and statistics problems.
Median questions are frequently asked in:
CBSE Class 9 and 10 Statistics
CUET Quantitative Aptitude
SSC CGL and CHSL
Banking and Insurance Exams
CAT and MBA Entrance Tests
IPMAT and JIPMAT
Students should focus on the following important concepts while preparing median for competitive exams and board exams:
| Important Concept | Explanation |
|---|---|
| Median divides data into two equal halves | Half the observations lie above and below the median |
| Median is a positional average | It depends on the position of observations |
| Median is not affected by outliers | Extreme values do not significantly affect the median |
| Median class concept | Important in grouped frequency distribution |
| Cumulative frequency | Essential for grouped data calculations |
| Ordered data requirement | Ungrouped data must always be arranged properly |
| Skewed distributions | Median works better than mean in skewed data |
| Application-based questions | Frequently asked in data interpretation and statistics |
Mean, median, and mode are the three major measures of central tendency in statistics. These statistical measures help represent the central or typical value of a data set. Understanding the difference between mean, median, and mode is extremely important for solving statistics questions in school mathematics and competitive aptitude exams.
The following table explains the major differences between mean, median, and mode in statistics:
| Basis of Comparison | Mean | Median | Mode |
|---|---|---|---|
| Definition | Arithmetic average of observations | Middle value of arranged data | Most frequently occurring value |
| Formula | $\frac{\sum x}{n}$ | Middle observation | Highest frequency observation |
| Based On | Numerical values | Position of values | Frequency of occurrence |
| Effect of Extreme Values | Highly affected | Least affected | Not affected significantly |
| Best Used For | Symmetrical data | Skewed data | Repetitive data |
| Data Requirement | Numerical data | Ordered data | Repeated observations |
| Usage in Statistics | Average calculation | Central position analysis | Frequency analysis |
| Real-Life Example | Average marks | Median salary | Most common shoe size |
The median is preferred over the mean when the data contains outliers or extreme values.
For example:
Suppose the salaries of employees are:
₹20,000, ₹25,000, ₹28,000, ₹30,000, ₹10,00,000
Here:
Mean becomes extremely large because of one very high salary.
Median gives a better idea of the salary earned by most employees.
Therefore, median is widely used in:
Income and salary analysis
Population studies
Business statistics
Economic surveys
Real estate market analysis
Data interpretation questions
Median is also more reliable when the data distribution is uneven or skewed.
The median is considered one of the most practical and reliable measures of central tendency in statistics.
Extreme values do not change the median significantly, making it highly useful in real-life statistical analysis.
For unevenly distributed data sets, the median gives a more accurate representation of the central value than the mean.
The concept of median is simple and can be calculated easily for both grouped and ungrouped data.
Median is commonly used in survey analysis, business reports, and economics because it works effectively even for large data sets.
Questions based on mean, median, and mode frequently appear in:
CBSE Statistics Chapters
NCERT Mathematics
CUET and SSC Exams
Banking Quantitative Aptitude
CAT and MBA Entrance Exams
Data Interpretation and Logical Reasoning Tests
A strong understanding of the difference between mean, median, and mode helps students solve statistical questions more accurately and improve overall quantitative aptitude skills.
Q.1. The median of the following data will be _________.
32, 25, 33, 27, 35, 29, and 30
a) 32
b) 27
c) 30
d) 29
Hint: The median of this data is the middlemost number because the total number of observations is odd.
Solution:
Given numbers:
32, 25, 33, 27, 35, 29, 30
Arrange the data in ascending order:
25, 27, 29, 30, 32, 33, 35
Total number of observations:
$n = 7$
Since the number of observations is odd:
$\text{Median} = \left(\frac{7+1}{2}\right)^{th}$ observation
$= \left(\frac{8}{2}\right)^{th}$ observation
$= 4^{th}$ observation
The 4th observation is 30.
Correct Answer: c) 30
Q.2. The median of a set of 11 distinct observations is 73.2. If each of the largest five observations of the set is increased by 3, then the median of the new set __________.
a) is 3 times that of the original set
b) is increased by 3
c) remains the same as that of the original set
d) is decreased by 3
Hint: Use the formula:
$\text{Median} = \left(\frac{n+1}{2}\right)^{th}$ observation
Solution:
Total number of observations:
$n = 11$
Median position:
$\text{Median} = \left(\frac{11+1}{2}\right)^{th}$ observation
$= \left(\frac{12}{2}\right)^{th}$ observation
$= 6^{th}$ observation
The median depends on the 6th observation.
Only the largest 5 observations are increased by 3.
Therefore, the 6th observation remains unchanged.
Hence, the median also remains unchanged.
Correct Answer: c) remains the same as that of the original set
Q.3. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set is:
a) is decreased by 2
b) is two times the original median
c) remains the same as that of the original set
d) is increased by 2
Hint: Use the formula:
$\text{Median} = \left(\frac{n+1}{2}\right)^{th}$ observation
Solution:
Total number of observations:
$n = 9$
Median position:
$\text{Median} = \left(\frac{9+1}{2}\right)^{th}$ observation
$= \left(\frac{10}{2}\right)^{th}$ observation
$= 5^{th}$ observation
The median depends on the 5th observation.
Only the largest 4 observations are increased by 2.
Therefore, the 5th observation remains unchanged.
Hence, the median also remains unchanged.
Correct Answer: c) remains the same as that of the original set
Q.4. Calculate the value of the median for the following data distribution:
| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 7 |
| 20–30 | 12 |
| 30–40 | 10 |
| 40–50 | 6 |
a) 26.67
b) 24.47
c) 22
d) 19.67
Hint: Use the formula for the median:
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Where:
$l$ = lower boundary of median class
$n$ = total number of observations
$cf$ = cumulative frequency preceding the median class
$f$ = frequency of median class
$h$ = class interval size
Solution:
| Class Interval | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 7 | 5 + 7 = 12 |
| 20–30 | 12 | 5 + 7 + 12 = 24 |
| 30–40 | 10 | 5 + 7 + 12 + 10 = 34 |
| 40–50 | 6 | 5 + 7 + 12 + 10 + 6 = 40 |
Total frequency:
$\sum f = 40$
Now,
$\frac{n}{2} = \frac{40}{2} = 20$
The cumulative frequency just greater than 20 is 24.
Therefore, the median class is:
20–30
Now,
$l = 20$
$cf = 12$
$f = 12$
$h = 10$
Substituting values in the formula:
$\text{Median} = 20+\frac{20-12}{12}\times10$
$= 20+\frac{8}{12}\times10$
$= 20+6.67$
$= 26.67$
Correct Answer: a) 26.67
Q.5. Find the median age of employees working at XYZ organization based on the following data:
| Ages (in years) | No. of Employees |
|---|---|
| 25–30 | 8 |
| 30–35 | 12 |
| 35–40 | 10 |
| 40–45 | 5 |
| 45–50 | 3 |
a) 44.58
b) 35.58
c) 40.58
d) 34.58
Hint: Use the formula for median:
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Where:
$l$ = lower boundary of median class
$n$ = total number of observations
$cf$ = cumulative frequency preceding the median class
$f$ = frequency of median class
$h$ = class interval size
Solution:
| Ages (in years) | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
|---|---|---|
| 25–30 | 8 | 8 |
| 30–35 | 12 | 8 + 12 = 20 |
| 35–40 | 10 | 8 + 12 + 10 = 30 |
| 40–45 | 5 | 8 + 12 + 10 + 5 = 35 |
| 45–50 | 3 | 8 + 12 + 10 + 5 + 3 = 38 |
Total frequency:
$\sum f = 38$
Now,
$\frac{n}{2} = \frac{38}{2} = 19$
The cumulative frequency just greater than 19 is 20.
Therefore, the median class is:
30–35
Now,
$l = 30$
$cf = 8$
$f = 12$
$h = 5$
Substituting values in the formula:
$\text{Median} = 30+\frac{19-8}{12}\times5$
$= 30+\frac{11}{12}\times5$
$= 30+4.58$
$= 34.58$
Correct Answer: d) 34.58
Q.6. Find the median score of a cricket team in the past 20 matches based on the following data:
| Scores | No. of Matches |
|---|---|
| 80–100 | 3 |
| 100–120 | 6 |
| 120–140 | 4 |
| 140–160 | 4 |
| 160–180 | 3 |
a) 123
b) 125
c) 124
d) 126
Hint: Use the formula for median:
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Solution:
| Scores | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
|---|---|---|
| 80–100 | 3 | 3 |
| 100–120 | 6 | 3 + 6 = 9 |
| 120–140 | 4 | 3 + 6 + 4 = 13 |
| 140–160 | 4 | 3 + 6 + 4 + 4 = 17 |
| 160–180 | 3 | 3 + 6 + 4 + 4 + 3 = 20 |
Total frequency:
$\sum f = 20$
Now,
$\frac{n}{2} = \frac{20}{2} = 10$
The cumulative frequency just greater than 10 is 13.
Therefore, the median class is:
120–140
Now,
$l = 120$
$cf = 9$
$f = 4$
$h = 20$
Substituting values in the formula:
$\text{Median} = 120+\frac{10-9}{4}\times20$
$= 120+\frac{1}{4}\times20$
$= 120+5$
$= 125$
Correct Answer: b) 125
Q.7. Find the median height of students in a class based on the following data:
| Heights (in cms) | No. of Students |
|---|---|
| 152–156 | 8 |
| 156–160 | 7 |
| 160–164 | 12 |
| 164–168 | 2 |
| 168–172 | 1 |
a) 162
b) 158
c) 160
d) 155
Hint: Use the formula for median:
$\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$
Where:
$l$ = lower boundary of median class
$n$ = total number of observations
$cf$ = cumulative frequency preceding the median class
$f$ = frequency of median class
$h$ = class interval size
Solution:
| Heights (in cms) | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
|---|---|---|
| 152–156 | 8 | 8 |
| 156–160 | 7 | 8 + 7 = 15 |
| 160–164 | 12 | 8 + 7 + 12 = 27 |
| 164–168 | 2 | 8 + 7 + 12 + 2 = 29 |
| 168–172 | 1 | 8 + 7 + 12 + 2 + 1 = 30 |
Total frequency:
$\sum f = 30$
Now,
$\frac{n}{2} = \frac{30}{2} = 15$
The cumulative frequency just greater than 15 is 27.
Therefore, the median class is:
160–164
Now,
$l = 160$
$cf = 15$
$f = 12$
$h = 4$
Substituting values in the formula:
$\text{Median} = 160+\frac{15-15}{12}\times4$
$= 160+\frac{0}{12}\times4$
$= 160+0$
$= 160$
Correct Answer: c) 160
Understanding the important median formulas and shortcut tricks helps students solve statistics questions faster and more accurately in board exams and competitive aptitude tests. These formulas are extremely useful for CBSE Maths, CUET, SSC, Banking, CAT, IPMAT, and Data Interpretation questions.
| Concept | Formula / Shortcut Trick | Usage |
|---|---|---|
| Median for Odd Number of Observations | $\text{Median} = \left(\frac{n+1}{2}\right)^{th}$ observation | Used when the total number of observations is odd |
| Median for Even Number of Observations | $\text{Median} = \frac{\left(\frac{n}{2}\right)^{th} + \left(\frac{n}{2}+1\right)^{th}}{2}$ | Used when the total number of observations is even |
| Median Formula for Grouped Data | $\text{Median}=l+\frac{\frac{n}{2}-cf}{f}\times h$ | Used for grouped frequency distribution |
| Shortcut to Find Median Class | Find the class whose cumulative frequency is just greater than $\frac{n}{2}$ | Helps identify the median class quickly |
| Total Frequency Formula | $n = \sum f$ | Used to calculate the total number of observations |
| Cumulative Frequency Trick | Add frequencies continuously from top to bottom | Helps save time in grouped data questions |
| Important Exam Tip | Always arrange ungrouped data in ascending order first | Prevents calculation mistakes |
| Fast Calculation Trick | In symmetrical distributions, Mean = Median | Useful in objective-type questions |
| Outlier Concept | Median is not affected by extreme values | Important theory-based property |
| Data Interpretation Shortcut | Median divides data into two equal halves | Helps solve conceptual questions quickly |
Choosing the right statistics and quantitative aptitude books helps students understand the concept of median more effectively through solved examples, shortcut methods, and practice questions. The following books are highly recommended for CBSE students and competitive exam aspirants.
| Book Name | Author | Best For | Key Features |
|---|---|---|---|
| NCERT Mathematics Class 9 | NCERT | CBSE Class 9 Students | Strong basics of statistics and median concepts |
| NCERT Mathematics Class 10 | NCERT | CBSE Board Preparation | Detailed grouped data and median formula questions |
| Quantitative Aptitude for Competitive Examinations | R.S. Aggarwal | SSC, Banking, CUET | Shortcut tricks and multiple practice questions |
| Fast Track Objective Arithmetic | Rajesh Verma | CAT, IPMAT, Banking | Advanced aptitude-based median questions |
| Magical Book on Quicker Maths | M. Tyra | Competitive Exams | Fast calculation methods and shortcut techniques |
| Quantitative Aptitude Quantum CAT | Sarvesh Kumar Verma | CAT and MBA Entrance Exams | High-level data interpretation and statistics concepts |
| Data Interpretation and Logical Reasoning | Arun Sharma | CAT, XAT, SNAP | DI-based median and statistical analysis questions |
| Objective Mathematics | R.D. Sharma | School and Competitive Exams | Step-by-step solved examples for statistics |
| Secondary School Mathematics | R.S. Aggarwal | CBSE and Foundation Level | Concept-building exercises and practice sets |
| SSC Mathematics Chapterwise Solved Papers | Kiran Publications | SSC Exams | Previous year median and statistics questions |
The concept of median is closely connected with several important quantitative aptitude and statistics topics that are frequently asked in CBSE board exams, CUET, SSC, Banking, CAT, IPMAT, and other competitive exams. Learning these related aptitude concepts helps students strengthen their problem-solving skills and improve overall performance in data interpretation and mathematics sections.
Quantitative Aptitude Topics | |||
Frequently Asked Questions (FAQs)
If $n$ is odd, the median equals the $(\frac{n+1}{2})$th observation.
If $n$ is even, the median is given by the mean of the $(\frac{n}{2})$th observation and the $(\frac{n}{2}+1)$th observation.
The mean is the average of a data set, calculated by adding all the values together and dividing by the number of values. The median is the middle value in a data set when the values are arranged in ascending or descending order.
The median class in a frequency distribution is the class interval that contains the median value of the data set. It is determined by locating the cumulative frequency that reaches or exceeds half the total number of observations.
Cumulative frequency is the running total of frequencies up to a certain class interval in a frequency distribution table. It helps in identifying the position of the median class in grouped data by showing how frequencies accumulate over the classes.
Yes, median is a central tendency.
Median should be used when the data contains outliers, extreme values, or skewed distributions. It provides a more accurate representation of the typical value in such cases.