Question

For the given frequency distribution match the column -I with Column-II.

Class	30-35	35-40	40-45	45-50	50-55	55-60	60-65
Frequency	14	16	18	23	18	8	3

h $=$ width of the class interval c $=$ cumulative frequency
f $=$ frequency of the class interval to which median belongs $l_1=$ lower limit of the median class interval

Answer 1

Class	35-35	35-40	40-45	45-50	50-55	55-60	60-65
Frequency	14	16	18	23	18	8	3
Cumulative Frequency	14	30	48	71	89	97	100 = N

Now, $\frac{N}{2}=\frac{100}{2}=50$
The median class is the class with C.F. just greater than $50$, which is $45-50$ and the median will lie in this class.
Hence, frequency of the class interval to which median belongs $f=23$
Cumulative frequency $=48$
Lower limit of the median class interval $=l_1=45$
Now, median $l_1+\left[\frac{\frac{N}{2}-cf}{f}\right]\times h$
where,

Here, $ { l }_{ 1 }= $ lower limit of the median class

$ f
= $ frequency of the median class

$ h
= $ width (size) of the median class

$ cf
= $ cumulative frequency of the class preceding the median class

So, Median $=45+\left[\frac{\frac{100}{2}-48}{23}\right]\times 5=45.4$

Hence, $A-4,B-6,C-2,D-1$