Mean and Median: Definition, Properties and Solved Examples

The mean or average of a number of observations is the sum of the values of all observations divided by the total number of observations. It is denoted by \(\overline x \) and read as \(x\) bar. The median is the value of the given number of observations, which divides it into exactly two parts. So, when the data is arranged in ascending or descending order, the median of ungrouped data is calculated based upon whether a number of terms are odd or even.

An average is called a measure of central tendency. The commonly used measures of central tendency are mean, median, and mode.

Mean and Median Definition

Mean

The mean of a set of observations is equal to their sum divided by the total number of observations.
In other words, if \({x_1},\,{x_2},\,{x_3},\,….,{x_n}\) are \(n\) values of a variable \(X,\) then the mean of these values is denoted by \(\overline X \) and is defined as

\(\overline X = \frac{{{x_1} + {x_2} + {x_3} + …. + {x_n}}}{n} = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)\)
Here the symbol \({\sum\limits_{i = 1}^n {{x_i}} }\) denotes the sum \({{x_1} + {x_2} + {x_3} + …. + {x_n}}\)

Median

Median of distribution is the value of the variable which divides the distribution into two equal parts, i.e. it is the value of the variable such that the number of observations above it is equal to the number of observations below it. So, when the data is arranged in ascending or descending order, the median of ungrouped data is calculated as below:
When the number of observations \(\left( n \right)\) is odd, then the median is the value of the observation.
\({\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th}}}}\)

When the number of observations \(\left( n \right)\) is even, the median is the mean of the \({\left( {\frac{n}{2}} \right)^{{\rm{th}}}}\) and the \({\left( {\frac{n}{2} + 1} \right)^{{\rm{th}}}}\) observation.

Median \( = \frac{{{\rm{Value}}\,{\rm{of}}\,{{\left( {\frac{n}{2}} \right)}^{th}}\,{\rm{observation}} + {\rm{value}}\,{\rm{of}}{{\left( {\frac{n}{2} + 1} \right)}^{{\rm{th}}}}\,{\rm{observation}}}}{2}\)

Properties of Mean

1. If \(\overline X \) is the mean of observations \({x_1},\,{x_2},\,{x_3},\,…,\,{x_{n,}}\) then the algebraic sum of the deviations from the mean is zero. i.e \(\sum\limits_{i = 1}^n {\left( {{x_i} – \overline X } \right)} = 0\)

2. If \(\overline X \) is the mean of observations \({x_1},\,{x_2},\,{x_3},\,…,\,{x_{n,}}\) then the mean of the observations \({x_1} + a,\,{x_2} + a,\,{x_3} + a,\,…,\,{x_n} + a\) is\(\overline X + a.\) i.e. if each observation is increased by \(a,\) then the mean is also increased by \(a.\)

3. If \(\overline X \) is the mean of observations \({x_1},\,{x_2},\,{x_3},\,…,\,{x_{n,}}\) then the mean of the observations of \(a{x_1},\,a{x_2},\,a{x_3},\,…,\,a{x_n}\) is \(a\overline X ,\) where \(a\) is any number different from zero, i.e., if each observation is multiplied by a non-zero number \(a,\) then the mean is also multiplied by \(a.\)

4. If \(\overline X \) is the mean of observations \({x_1},\,{x_2},\,{x_3},\,…,\,{x_{n,}}\) then the mean of \(\frac{{{x_1}}}{a},\,\frac{{{x_2}}}{a},\,\frac{{{x_3}}}{a},\,…,\frac{{{x_n}}}{a}\) is \(\frac{{\overline X }}{a}\) Where \(a\) is any non-zero number.

5. If \(\overline X \) is the mean of observations \({x_1},\,{x_2},\,{x_3},\,…,\,{x_{n,}}\) then the mean of the observations \({x_1} – a,\,{x_2} – a,\,{x_3} – a,….,\,{x_n} – a\) is \(\overline X – a,\) where \(a\) is any non-zero number.

Mean of Grouped Data

In a discrete frequency distribution, the mean may be computed by any one of the following methods:
(i) Direct Method
(ii) Shortcut Method
(iii) Step-Deviation Method

Direct Method

If a variable \(X\) takes the values \({x_1},\,{x_2},\,{x_3},\,…,\,{x_n}\) with corresponding frequencies \({f_1},\,{f_2},\,{f_3},\,…,\,{f_n}\) respectively, then mean of these values is
\(\bar X = \frac{{{f_1}{x_1} + {f_2}{x_2} + ……… + {f_n}{x_n}}}{{{f_1} + {f_2} + …….. + {f_n}}}\)

Shortcut Method

If the values of \(x\) or (and) \(f\) are large, the calculation of mean by the direct method is quite tricky and time-consuming because the calculations involved are lengthy. To minimize the time involved in the computation, we take deviations from an arbitrary point using the below formula.

Let \({x_1},\,{x_2},\,{x_3},…,\,{x_n}\) be the values of variable \(X\) with corresponding frequencies \({f_1},\,{f_2},\,{f_3},\,…..,\,{f_n}\) respectively. Taking deviations about an arbitrary point \(A,\) we have
\({d_1} = {x_i} – A,\) where \(i = 1,\,2,\,3,\,….,\,n\)
\( \Rightarrow {f_i}{d_i} = {f_i}\left( {{x_i} – A} \right),\) where \(i = 1,\,2,\,3,\,…..,\,n\)
\( \Rightarrow \sum\limits_{i = 1}^n {{f_i}{d_i}} = \sum\limits_{i = 1}^n {{f_i}\left( {{x_i} – A} \right)} \)
\( \Rightarrow \sum\limits_{i = 1}^n {{f_i}{d_i}} = \sum\limits_{i = 1}^n {{f_i}{x_i} – A} \sum\limits_{i = 1}^n {{f_i}} \)
\( \Rightarrow \sum\limits_{i = 1}^n {{f_i}{d_i}} = \sum\limits_{i = 1}^n {{f_i}{x_i} – A} N\left[ {N = \sum\limits_{i = 1}^n {{f_i}} } \right]\)
\( \Rightarrow \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{d_i}} = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{x_i} – \frac{{AN}}{N}} \)
\( \Rightarrow \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{d_i}} = \overline X – A\)
\( \Rightarrow \overline X = A + \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{d_i}} \)

Here, \(A\) is generally known as the assumed mean and is usually chosen so that the deviations are minor.

Step-Deviation Method

The formula to find the mean using the step-deviation method is
\(\overline X = A + h\left[ {\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{u_i}} } \right]\)
Where, \({u_i} = \frac{{{x_i} – A}}{h},h\) is a common factor.

Mean of a Continuous Frequency Distribution

Till now, we have been discussing various methods for computing the mean of a discrete frequency distribution. For a continuous frequency distribution or a frequency distribution with class intervals, the mean may be calculated by applying any of the methods discussed so far. The values of \({x_1},\,{x_2},\,{x_3},\,…..,\,{x_n}\) are taken as the mid-points or class-marks of the various classes. It should be noted that the mid-value or class-marks of a class interval is equal to \(\frac{1}{2}\left( {{\rm{lower}}\,{\rm{limit}} + {\rm{upper}}\,{\rm{limit}}} \right)\)

Median

The median is the middle value of a distribution, i.e., the median of a distribution is the variable’s value, which divides it into two equal parts. Thus, it is the variable’s value such that the number of observations above it is similar to the number of observations below it.

Median of Discrete Frequency Distribution

In the case of a discrete frequency distribution, we calculate the median using the below steps:
1. Find the cumulative frequency \(cf.\)
2. Find \(\frac{N}{2},\) where \(N = \sum\limits_{i = 1}^n {{f_i}} .\)
3. See the cumulative frequency \(\left( {cf} \right)\) just greater than \(\frac{N}{2}\) and determine the corresponding value of the variable.
4. The value obtained in the above step is a median.

Median of a Grouped or Continuous Distribution

To calculate the median of a grouped or continuous distribution, we will follow the below steps.
1. Obtain the frequency distribution.
2. Prepare the cumulative frequency column and obtain \(N = \sum\limits_{i = 1}^n {{f_i}} \)
3. Find \(\frac{N}{2}\)
4. See the cumulative frequency greater than \(\frac{N}{2}\) and determine the corresponding class. This class is known as the median class.
5. Apply the formula Median \( = l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
Where \(l = \)lower frequency of the median class
\(f= \)frequency of the median class
\(h = \)size of the median class
\(F = \)Cumulative frequency of the class preceding the median class \(N = \sum\limits_{i = 1}^n {{f_i}} \)
Let us understand the concept using Mean and Median Practice Problems.

Solved Examples – Mean and Median

Q.1. If the heights of \(5\) persons are \(144\,{\rm{cm}},\,152\,{\rm{cm}},\,151\,{\rm{cm}},\,158\,{\rm{cm}}\) and \(155\,{\rm{cm}}\) respectively. Find the mean height.
Ans: We know that if \({x_1},\,{x_2},\,{x_3},\,…..,\,{x_n}\) are \(n\) values of a variable \(X.\) Then the mean \(\overline X = \frac{{{x_1} + {x_2} + {x_3} + …. + {x_n}}}{n}.\)
So, mean height \( = \frac{{144 + 152 + 151 + 158 + 155}}{5} = \frac{{760}}{5} = 152\,{\rm{cm}}\)

Q.2. The mean of the \(10\) numbers is \(20.\) If \(5\) is subtracted from every number, what will be the new mean?
Ans: Let \({x_1},\,{x_2},\,{x_3},\,…..,\,{x_{10}}\) be \(10\) numbers with their mean equal to \(20.\) Then,
\(\overline X = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)\)
\( \Rightarrow 20 = \frac{{{x_1} + {x_2} + {x_3} + …. + {x_{10}}}}{{10}}\)
\( \Rightarrow {x_1} + {x_2} + {x_3} + …. + {x_{10}} = 200\)
New numbers are \({x_1} – 5,\,{x_2} – 5,\,{x_3} – 5,\,….,\,{x_{10}} – 5.\) Let \(\overline {X’} \) be the mean of new numbers. Then
\(\overline {X’} = \frac{{\left( {{x_1} – 5} \right) + \left( {{x_2} – 5} \right) + \left( {{x_3} – 5} \right) + …. + \left( {{x_{10}} – 5} \right)}}{{10}}\)
\(\overline {X’} = \frac{{\left( {{x_1} + {x_2} + {x_3} + …. + {x_{10}}} \right) – 5 \times 10}}{{10}} = \frac{{200 – 50}}{{10}}\)
\(\overline {X’} = 15\)
Hence the new mean is \(15.\)

Q.3. Find the mean of the following distribution:

Ans:

Mean\( = \overline X = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} = \frac{{360}}{{40}} = 9\)

Q.4. Find the median of the following data:
\(25,\,34,\,31,\,23,\,22,\,26,\,25,\,35,\,28,\,20,\,32\)
Ans: Arranging the data in ascending order, we get
\(20,\,22,\,23,\,25,\,26,\,28,\,31,\,32,\,34,\,35\)
Here, the number of observations \(n = 10\) is even.

So, median \( = \frac{{{\rm{Value}}\,{\rm{of}}\,{{\left( {\frac{{10}}{2}} \right)}^{{\rm{th}}}}{\rm{observation}} + {\rm{value}}\,{\rm{of}}\,{{\left( {\frac{{10}}{2} + 1} \right)}^{{\rm{th}}}}{\rm{observation}}}}{2}\)
Median \( = \frac{{{\rm{Value}}\,{\rm{of}}\,{{\left( 5 \right)}^{{\rm{th}}}}{\rm{observation}} + {\rm{value}}\,{\rm{of}}\,{{\left( 6 \right)}^{{\rm{th}}}}{\rm{observation}}}}{2}\)
Median\( = \frac{{26 + 28}}{2} = 27\)
Therefore, median \( = 27\)

Q.5. Calculate the median from the following data

Ans: Here, the class intervals are of unequal width. If the class intervals are unequal, the frequencies need not be adjusted to equal the class intervals.

Here, \(N = 60.\) So\(\frac{N}{2} = 30.\)
The cumulative frequency just greater than \(\frac{N}{2} = 30\) is \(50,\) and the corresponding class is\(30 – 60.\)
Hence, \(30 – 60.\) is the median class.

Therefore, \(l = 30,\,f = 30,\,F = 20,\,h = 30\)
Now, \({\rm{Median}} = l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
\( \Rightarrow {\rm{Median}} = 30 + \frac{{30 – 20}}{{30}} \times 30\)
\( \Rightarrow {\rm{Median}} = 40\)
The above examples help in understanding the Mean and Median Facts.

Summary

In this article, we studied the definition of mean and median. We also highlighted the properties of the mean, formulas to calculate the mean, and different methods to find the mean. We learnt how to find the mean and median of discrete frequency distribution and how to calculate the mean and median of grouped data. The discussed mean and median notes help solve the questions quickly.

Frequently Asked Questions (FAQ) – Mean and Median

Q.1. How do I calculate the median?
Ans: Median of distribution is the value of the variable which divides the distribution into two equal parts, i.e., it is the value of the variable such that the number of observations above it is equal to the number of observations below it. So, when the data is arranged in ascending or descending order, the median of ungrouped data is calculated as below:
When the number of observations \(\left( n \right)\) is odd. The median is the value of the \({\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th}}}}\)
Observation.
When the number of observations \(\left( n \right)\) is even. The median is the mean of the \({\left( {\frac{n}{2}} \right)^{{\rm{th}}}}\) and the \({\left( {\frac{n}{2} + 1} \right)^{{\rm{th}}}}\) observation.

Q.2. What is the mean vs median?
Ans: The mean is the average of given data, whereas the median is the middle value of the given data.

Q.3. Explain mean and median with example.
Ans: The mean of a set of observations is equal to their sum divided by the total number of observations.
Example: The marks of \(5\) students are \(20,\,30,\,32,\,37,\,40.\)
Therefore, the mean marks of \(5\) students \( = \frac{{20 + 30 + 32 + 37 + 40}}{5} = \frac{{159}}{5}\)
So, mean\( = 31.8\)
Median: Median of distribution is the value of the variable which divides the distribution into two equal parts, i.e. it is the value of the variable such that the number of observations above it is equal to the number of observations below it.
Example: Find the median of the following values:
\(37,\,31,\,42,\,43,\,46,\,25,\,39,\,45,\,32\)
Solution: Arranging the given data in ascending order, we get
\(25,\,31,\,32,\,37,\,39,\,42,\,43,\,45,\,46\)
Here the number of observations \(n = 9\) which is odd
So, Median\( = {\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th}}}} = {\left( {\frac{{9 + 1}}{2}} \right)^{{\rm{th}}}} = {5^{{\rm{th}}}}\) observation
Therefore, the median of the given data is \(39.\)

Q.4. What is the median of these numbers; \(2,\,4,\,6,\,8,\,10?\)
Ans: Since the number of observations of the given data is odd, the median is the value of \({\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th}}}}\) term.
So, \(n = 5,\) median \({\left( {\frac{{5 + 1}}{2}} \right)^{{\rm{th}}}}\) term i.e. \({3^{{\rm{rd}}}}\) term.
Therefore, the median of \(2,\,4,\,6,\,8,\,10\) is \(6.\)

Q.5. How do you find the mean and mode?
Ans:
Mean: If \({x_1},\,{x_2},\,{x_3},….,\,{x_n}\) are \(n\) values of a variable \(X,\) then the mean of these values is denoted by \(\overline X \) and is defined as
\(\overline X = \frac{{{x_1} + {x_2} + {x_3} + …. + {x_n}}}{n} = \frac{1}{n}\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)\)
Mode: Mode is the value that occurs most frequently in a set of observations and around which the other items of the set cluster densely.
Thus, the mode of a frequency distribution is the value of the variable which has a maximum frequency.

We hope this article on mean and median has provided significant value to your knowledge. If you have any queries or suggestions, feel to write them down in the comment section below. We will love to hear from you. Embibe wishes you all the best of luck!