Question

If the median of the following frequency distribution is $32.5$. Find the values of ${f_1}$ and ${f_2}$

Class	Frequency
$0 - 10$	${f_1}$
$10 - 20$	$5$
$20 - 30$	$9$
$30 - 40$	$12$
$40 - 50$	${f_2}$
$50 - 60$	$3$
$60 - 70$	$2$
$70 - 80$	$40$

Answer 1

In this question, we are given a table with class interval and frequency. The median of the data is also given. Using the median, find the median class. Draw the table calculating cumulative frequency. Use the median class and formula, Median $= l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h$ to find ${f_1}$ . After ${f_1}$ has been found, use total frequency from the table to find ${f_2}$.

Formula used: Median = $l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h$ where, $l$ = lower limit of median class, $n = \sum {{f_i}}$, $cf$= cumulative frequency of the class before median class, $h$ = class interval, $f$ = frequency of median class.

Complete step-by-step solution:
We are given a table with class intervals and its frequency and we are asked to find the value of ${f_1}$ and ${f_2}$. First, we will make a table with a cumulative frequency (cf).

Class	Frequency	Cumulative frequency
$0 - 10$	${f_1}$	${f_1}$
$10 - 20$	$5$	$5 + {f_1}$
$20 - 30$	$9$	$9 + 5 + {f_1} = 14 + {f_1}$
$30 - 40$	$12$	$14 + 12 + {f_1} = 26 + {f_1}$
$40 - 50$	${f_2}$	$26 + {f_1} + {f_2}$
$50 - 60$	$3$	$26 + 3 + {f_1} + {f_2} = 29 + {f_1} + {f_2}$
$60 - 70$	$2$	$29 + 2 + {f_1} + {f_2} = 31 + {f_1} + {f_2}$
Total	$40$

Now, we know that median = $32.5$. Since median lies in the median class, its median class is $30 - 40$. So, we will apply the formula, Median $= l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h$ in this median class.
In this question, $l = 30$, $cf = 14 + {f_1}$, $f = 12$, $h = 10 - 0 = 10$ and we know that $n = \sum {{f_1} = 40}$.
Therefore, $\dfrac{n}{2} = \dfrac{{40}}{2} = 20$.
Putting all the values in the formula,
$\Rightarrow 32.5 = 30 + \dfrac{{20 - (14 + {f_1})}}{{12}} \times 10$
Solving for x,
$\Rightarrow 32.5 - 30 = \left( {\dfrac{{20 - 14 - {f_1}}}{{12}}} \right) \times 10$
Shifting and solving,
$\Rightarrow \dfrac{{2.5 \times 12}}{{10}} = 6 - {f_1}$
$\Rightarrow 3 = 6 - {f_1}$
$\Rightarrow {f_1} = 6 - 3 = 3$
Now, we know from the table that $31 + {f_1} + {f_2} = 40$. Putting ${f_1} = 3$ to find the value of ${f_2}$.
$\Rightarrow 31 + 3 + {f_2} = 40$
Shifting to find the value of ${f_2}$,
$\Rightarrow {f_2} = 40 - 34 = 6$

$\therefore$ The value of ${f_1}$ is 3 and ${f_2}$ is 6.

Note: The median is the middle number in a sorted, ascending or descending list of numbers and can be more descriptive of that data set than the average. The median is sometimes used as opposed to the mean when there are outliers in the sequence that might skew the average of the values. If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above. If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.