Question

Find the sum of the lower limit of the median class and modal class of the following distribution table:

Distribution	\[0 - 5\]	\[5 - 10\]	\[10 - 15\]	\[15 - 20\]	\[20 - 25\]
Frequency	\[10\]	\[15\]	\[12\]	\[20\]	\[9\]

A. \[25\]
B. \[20\]
C. \[30\]
D. \[50\]

Answer 1

Hint: The given problem revolves around the concepts of statistics calculating the median and mode. So, first of all draw the certain mathematical calculations to find the respective cumulative frequencies of each class interval as well as the total frequency ‘\[N\]’. As a result, to calculate the lower limit of the median class, consider the half the value of ‘\[N\]’ and then equating or approximating it with the respective cumulative frequency (observed from the table). Similarly, to find the (lower) limits in case of modal class which in term equals the maximum frequency among the entire given distribution table. Hence, the desired sum of lower limits is possible.

Complete step-by-step answer:
Since, we have given the ‘frequency table’ with respect to the ‘distribution’ that is class interval given or represented as below;

Distribution (\[C\])	Frequency (\[{f_i}\])
\[0 - 5\]	\[10\]
\[5 - 10\]	\[15\]
\[10 - 15\]	\[12\]
\[15 - 20\]	\[20\]
\[20 - 25\]	\[9\]

As a result, to find the values of median class, we should know the cumulative frequency,
Hence, taking cumulative additions of the frequencies (since, by adding the present term that is frequency with the previous term or, frequency)

Distribution (\[C\])	Frequency (\[{f_i}\])	\[c{f_i}\]
\[0 - 5\]	\[10\]	\[10\]
\[5 - 10\]	\[15\]	\[10 + 15 = 25\]
\[10 - 15\]	\[12\]	\[25 + 12 = 37\]
\[15 - 20\]	\[20\]	\[37 + 20 = 57\]
\[20 - 25\]	\[9\]	\[57 + 9 = 66\]
Total	\[\sum N = 66\]

Where, ‘\[c{f_i}\]’ is the cumulative frequency calculated by the formula ‘\[c{f_i} = \]present frequency\[ + \]previous frequencies’.
Hence, from the above calculations it is observed that
\[\sum {\left( {\dfrac{N}{2}} \right)} \]
Median is defined by the half of its total frequency that is,
\[ \Rightarrow \dfrac{{66}}{2} = 33\]
From the above table,
It is seems that the ‘\[33\]’ is near to cumulative frequency i.e. ‘\[37\]’,
Hence, its class interval is ‘\[10 - 15\]’
Where, the lower limit is ‘\[10\]’ and that of higher class is ‘\[15\]’ respectively.
\[ \Rightarrow \]Lower limit of the required median class is \[ \equiv 10\]. … (i)
Similarly,
To determine the modal class,
 It is equal to the ‘distribution’ with the ‘maximum frequency’.
Hence, from the above table,
Maximum frequency is ‘\[20\]’ for the class ‘\[15 - 20\]’ respectively.
 \[ \Rightarrow \]Lower limit of the required modal class is\[ \equiv 15\]. … (ii)
Now,
From (i) and (ii),
Required sum of the lower limit of median class as well as modal class seems to be,
\[ = 10 + 15\]
\[ = 25\]
So, the correct answer is “Option A”.

Note: One must be able to correlate the certain values for the respective distribution table particularly such as to equate (or, approximate) the respective frequencies or cumulative frequencies with respect to the class intervals (as we did here). Also, being able to distinguish the certain formulae for mean, mode, median, etc. properly, this might be confusing, so as to be sure of the final answer.