Question

If the median of the following data is \[16\]. Then, find the missing frequencies ‘\[a\]’ and ‘\[b\]’ respectively such that the total frequency is \[70\].

Class	\[0 - 5\]	\[5 - 10\]	\[10 - 15\]	\[15 - 20\]	\[20 - 25\]	\[25 - 30\]	\[30 - 35\]	\[35 - 40\]
Frequency	\[12\]	\[a\]	\[12\]	\[15\]	\[b\]	\[6\]	\[6\]	\[4\]

A. \[7\] and \[8\]
B. \[8\] and \[7\]
C. Doesn’t exist
D. None of the above

Answer 1

Hint: The given problem revolves around the concepts of statistics involving certain terms such as mean, mode, median, etc. As a result of using the respective formula for the median i.e. \[Median = I + \left[ {h \times \dfrac{{\left( {\dfrac{N}{2} - cf} \right)}}{f}} \right]\] (since, the value of median is given). Substituting the respective values by considering the approximate or the nearby value of the frequency. Then, adding up all the given frequencies that is “cumulative frequency” which is equal to \[70\], so as to obtain the remaining value/s.

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

Class intervals (\[C\])	Frequency (\[{f_i}\])
\[0 - 5\]	\[12\]
\[5 - 10\]	\[a\]
\[10 - 15\]	\[12\]
\[15 - 20\]	\[15\]
\[20 - 25\]	\[b\]
\[25 - 30\]	\[6\]
\[30 - 35\]	\[6\]
\[35 - 40\]	\[4\]

As a result, to find the values of ‘\[a\]’ and ‘\[b\]’ respectively
Therefore, taking cumulative additions of the frequencies (since, by adding the present term that is frequency with the previous term or, frequency)

Class intervals (\[C\])	Frequency (\[{f_i}\])	\[c{f_i}\]
\[0 - 5\]	\[12\]	\[12\]
\[5 - 10\]	\[a\]	\[a + 12\]
\[10 - 15\]	\[12\]	\[24 + a\]
\[15 - 20\]	\[15\]	\[39 + a\]
\[20 - 25\]	\[b\]	\[39 + a + b\]
\[25 - 30\]	\[6\]	\[45 + a + b\]
\[30 - 35\]	\[6\]	\[51 + a + b\]
\[35 - 40\]	\[4\]	\[55 + a + b\]

Where, ‘\[c{f_i}\]’ is the cumulative frequency calculated by the formula ‘\[c{f_i} = \] present frequency \[ + \] previous frequencies’.
Hence, taking the summations of the given frequencies and cumulative frequencies, we get

Class intervals (\[C\])	Frequency (\[{f_i}\])	\[c{f_i}\]
\[0 - 5\]	\[12\]	\[12\]
\[5 - 10\]	\[a\]	\[a + 12\]
\[10 - 15\]	\[12\]	\[24 + a\]
‘\[15 - 20\]’	‘\[15\]’	‘\[39 + a\]’
\[20 - 25\]	\[b\]	\[39 + a + b\]
\[25 - 30\]	\[6\]	\[45 + a + b\]
\[30 - 35\]	\[6\]	\[51 + a + b\]
\[35 - 40\]	\[4\]	\[55 + a + b\]
		\[\sum {c{f_i}} = 70\]

Where, it is given that total or, the cumulative frequency is \[70\] that is mathematically represented as \[\sum {c{f_i}} = 55 + a + b = 70\] … (i)
Also, we have given that the value for the respective median is \[16\].
\[Median = 16\]
As a result, from the above table of frequency median it seems that – the nearby value/s for the given respective median\[ = 16\] is at,
Class interval, \[ \equiv 15 - 20\] and,
Frequency, \[f \equiv 15\]
Also, at preceding \[cf = 24 + a\] respectively.
Hence, using the formula for median that is,
\[Median = I + \left[ {h \times \dfrac{{\left( {\dfrac{{\sum {c{f_i}} }}{2} - cf} \right)}}{{{f_i}}}} \right]\]
Where, \[I = \] initial class interval (considered class) that is equal to \[15\],
And, \[h = \] difference between the class intervals (considered class) that is equal to \[5\].
As a result, substituting all the required values considered above, in the above formula, we get
 \[ \Rightarrow 16 = 15 + \left[ {5 \times \dfrac{{\left( {\dfrac{{70}}{2} - \left( {24 + a} \right)} \right)}}{{15}}} \right]\]
Hence, solving the equation algebraically that is as per BODMAS method, we get
 \[ \Rightarrow 16 = 15 + \dfrac{{\left( {35 - 24 - a} \right)}}{3}\]
Multiplying the equation by \[3\], we get
\[ \Rightarrow 48 = 45 + 11 - a\]
\[ \Rightarrow 48 = 56 - a\]
Hence, the equation becomes
\[ \Rightarrow a = 56 - 48\]
\[ \Rightarrow a = 8\] … (ii)
Now, Hence, substituting the value of \[a = 8\] in from (i), we get
 \[ \Rightarrow \sum {c{f_i}} = 55 + 8 + b = 70\]
\[ \Rightarrow 63 + b = 70\]
Hence, the required value for ‘\[b\]’ is
\[ \Rightarrow b = 70 - 63\]
\[ \Rightarrow b = 7\] … (iii)
\[\therefore \]From (ii) and (iii), we can say that option (B) is correct.

Note:
One must able to know the parameters or the formulae used to measure the statistical problems (involving graphical analysis) such as ‘\[mean = \dfrac{{number{\text{ }}of{\text{ }}terms}}{{total{\text{ }}number{\text{ }}of{\text{ }}the{\text{ }}terms{\text{ }}in{\text{ }}a{\text{ }}set}}\]’ or the average, mode, median, etc. Remember that, the cumulative frequency i.e. ‘\[c{f_i}\]’ used here need to be understand to calculate the term by the formula ‘\[c{f_i} = present{\text{ }}term + previous{\text{ }}terms\]’, so as to be sure of the correct answer.