Question

Find the value of y and if the median of the following data is 31 \[x\]

CI	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Frequency	$5$	$x$	$6$	$y$	$6$	$5$	$60$

Answer 1

Hint: As we know that the formula of median of grouped data is Median $ = L + \dfrac{{\dfrac{n}{2} - B}}{G} \times w$ where L is the lower class boundary of the group containing the median , that is $30$ n is the total number of values , that is $40$ B is the cumulative frequency of the groups before the median group, that is $11 + x$ G is the frequency of the median group , that is $y$ w is the group width that is $10$ and it is given that total frequency is $40$ therefore , $22 + x + y$ = $40$ hence from here we can solve it .
Complete step-by-step answer:
As we have to find the median of the given data For Median of the group data we use formula Median = $ = L + \dfrac{{\dfrac{n}{2} - B}}{G} \times w$ where:
L is the lower class boundary of the group containing the median ,
n is the total number of values ,
B is the cumulative frequency of the groups before the median group,
G is the frequency of the median group ,
w is the group width
 So for the median we have to find the cumulative frequency of the data ,

CI	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Frequency	$5$	$x$	$6$	$y$	$6$	$5$	$40$
Cumulative frequency	$5$	$5 + x$	$11 + x$	$11 + x + y$	$17 + x + y$	$22 + x + y$

Hence it is given that the total frequency is $40$
Therefore , $22 + x + y$ = $40$
On solving
$x + y = 18$ .......(i)
 here the median class = $30 - 40$, the median is in the class where the cumulative frequency reaches half the sum of the absolute frequencies .
So the
L is the lower class boundary of the group containing the median , that is $30$
n is the total number of values , that is $40$
B is the cumulative frequency of the groups before the median group, that is $11 + x$
G is the frequency of the median group , that is $y$
w is the group width that is $10$
So put it in the median equation we get ,
Median = $30 + \left[ {\dfrac{{20 - (11 + x)}}{y}} \right] \times 10$
It is given that median = $31$
$\Rightarrow$ $31$ = $30 + \left[ {\dfrac{{20 - (11 + x)}}{y}} \right] \times 10$
$\Rightarrow$ $1 = \dfrac{{9 - x}}{y} \times 10$
On cross multiplication we get ,
$\Rightarrow$ $y = (9 - x) \times 10$
$\Rightarrow$ $10x + y = 90$ ......(ii)
$\Rightarrow$ $x + y = 18$ .......(i)
Now subtract both equation ,
$\Rightarrow$ $9x = 72$
and $x = 8$
Now put it in equation (i) we get $y = 10$
Note: Mode of the grouped data is Estimated mode = $L + \dfrac{{{f_m} - {f_{m - 1}}}}{{({f_m} - {f_{m - 1}}) + ({f_m} - {f_{m + 1}})}} \times w$ Modal group is the group which have maximum frequency where L is the lower class boundary of the modal group ${f_{m - 1}}$ is the frequency of the group before the modal group ${f_m}$ is the frequency of the modal group ${f_{m + 1}}$ is the frequency of the group after the modal group w is the group width .