# On the annual day of school age-wise participation of students is given in the following frequency distribution table. Find the median of the students.Age (in years)Number of studentLess than $6$ $2$Less than $8$$6Less than 10$$12$Less than $12$$22Less than 14$$42$Less than $16$$67Less than 18$$76$

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Hint: First we have to find the class interval and the frequency with the given information. Then we need to find the median class by using the below mentioned formula. We can find the required answer by using the median formula mentioned below.

Formula used: To find median class: ${\left( {\dfrac{{\text{n}}}{2}} \right)^{th}}$ observation.
${\text{Median = L + }}\dfrac{{\dfrac{{\text{n}}}{{\text{2}}}{\text{ - cf}}}}{{\text{f}}}{{ \times c}}$

Complete step-by-step solution:
The distribution is given as less than format and contains cumulative frequency.
It means that there are $2$ students whose age is less than $6$.
There are $12$ students whose age is less than $8$(this includes students whose age is also less than $6$ i.e., $2$ students)
Also, there are $76$ students whose age is less than $18$ which means all the students whose age is less than $18$ (it contains all the students) this is called cumulative frequency.
First we have to find the class interval and frequency table.
Clearly, we can see that the age is categorized in the difference of $2$ i.e., Less than $6$, Less than $8$, Less than $10$, etc. So the class length is $2$.
Now, the given data as changed in the form as follows:
 Age (in years) ${\text{cf}}$ ${\text{f}}$ ${\text{f}}$ $4 - 6$ $2$ $2$ $2$ $6 - 8$ $6$ $6 - 2 = 4$ $4$ $8 - 10$ $12$ $12 - 6 = 6$ $6$ $10 - 12$ $22$ $22 - 12 = 10$ $10$ $12 - 14$ $42$ $42 - 22 = 20$ $20$ $14 - 16$ $67$ $67 - 42 = 25$ $25$ $16 - 18$ $76$ $67 - 76 = 9$ $9$ $\sum {\text{f}} = 76$

First we need to find the median class, which is the value of ${\left( {\dfrac{{\text{n}}}{2}} \right)^{th}}$ observation.
On putting the values and we get,
$\Rightarrow {\left( {\dfrac{{{\text{76}}}}{{\text{2}}}} \right)^{th}}$ Observation
$\Rightarrow {\text{3}}{{\text{8}}^{th}}$ 4Observation
From the above table, in the column of ${\text{cf}}$, ${\text{3}}{{\text{8}}^{th}}$ observation lies in the class $12 - 14$
$\therefore$ The median class = $12 - 14$
Now, we need to apply the formula,
${\text{M = L + }}\dfrac{{\dfrac{{\text{n}}}{{\text{2}}}{\text{ - cf}}}}{{\text{f}}}{{ \times c}}$
Here,
${\text{L }}$ = lower limit of the median class
${\text{n}}$ = total frequency i.e., $\sum {\text{f}}$
${\text{cf}}$= cumulative frequency of the class preceding the median class.
${\text{f}}$ = frequency of median class
${\text{c}}$= class length of median class
Now, we write the data as follows:
${\text{L = 12}}$
${\text{n = 76}}$
${\text{cf = 22}}$
${\text{f = 20}}$
${\text{c = 2}}$
On putting the values in the formula and we get,
${\text{M = L + }}\dfrac{{\dfrac{{\text{n}}}{{\text{2}}}{\text{ - cf}}}}{{\text{f}}}{{ \times c}}$
$\Rightarrow$${\text{M = 12 + }}\dfrac{{\dfrac{{76}}{{\text{2}}}{\text{ - 22}}}}{{20}}{{ \times 2}}$
On divide the term in the numerator term and we get,
$\Rightarrow$${\text{M = 12 + }}\dfrac{{{\text{38 - 22}}}}{{20}}{{ \times 2}}$
Let us subtracting the numerator term and we get,
$\Rightarrow$${\text{M = 12 + }}\dfrac{{{\text{16}}}}{{20}}{{ \times 2}}$
On multiply the term and we get,
$\Rightarrow$${\text{M = 12 + }}\dfrac{{32}}{{20}}$
Let us divide the term and we get
$\Rightarrow$${\text{M = 12 + 1}}{\text{.6}}$
On add the term and we get,
$\Rightarrow$${\text{M = 1}}3.6$

Therefore the median of the given series is ${\text{1}}3.6$

Note: It is simple to remember, if it is less than frequency then the given number is upper limit and if it is more than frequency the given number is lower limit.
Less than = upper limit
More than = lower limit
Example:
Less than $3$ the class interval is $0 - 3$
Less than $6$ the class interval is $3 - 6$
Less than $9$ the class interval is $6 - 9$
More than $3$ the class interval is $3 - 6$
More than $6$ the class interval is $6 - 9$
More than $9$ the class interval is $9 - 12$
Always remember that while doing the less than and more than frequency, it is mandatory to find the class interval.
Students may go wrong or get confusion in finding the class interval and its difference for both less than and more than frequency.