Question

The following table gives the distribution of monthly salary of 900 employees. However, the frequencies of the classes \[40 - 50\] and \[60 - 70\] are missing. If the median of the distribution is Rs \[59.25\] . Find the missing frequencies.

Salaries Rs in \['000\]	\[30 - 40\]	\[40 - 50\]	\[50 - 60\]	\[60 - 70\]	\[70 - 80\]
No. of employees	\[120\]	?	\[200\]	?	\[185\]

Answer 1

Hint: In order to solve the question given above we have to use the concepts of median and cumulative frequency. Cumulative frequency is the total frequency of all the values less than the upper-class boundary of a given class interval and median is the middle number in a sorted list of numbers.

Complete step by step solution:
We have to calculate the frequencies for \[40 - 50\] and \[60 - 70\].
We know that, Cumulative frequency is calculated by adding the frequencies of each interval.
So, the frequency of \[30 - 40\] is \[120\]
The frequency of \[40 - 50\] becomes \[120 + {f_1}\]
The frequency of \[50 - 60\] becomes \[320 + {f_1}\]
And the frequency of \[60 - 70\] will be \[320 + {f_1} + {f_2}\]
Lastly, the frequency of \[70 - 80\] will be \[900\]

Salaries	No. of employees	Cumulative Frequency
\[30 - 40\]	\[120\]	\[120\]
\[40 - 50\]	\[{f_1}\]	\[120 + {f_1}\]
\[50 - 60\]	\[200\]	\[320 + {f_1}\]
\[60 - 70\]	\[{f_2}\]	\[320 + {f_1} + {f_2}\]
\[70 - 80\]	\[185\]	\[900\]

Now, we are given the question that the median is \[59.25\]. We can clearly see that the median class is \[50 - 60\].
Now, we will solve to calculate the value of \[{f_1}\].
We know,
\[median = L + \dfrac{{\dfrac{n}{2} - CF}}{f} \times c\],
Now we have:
Median \[ = 59.25\]
L \[ = \]Lower boundary of median class \[ = 50\]
n \[ = \]Number of employees \[ = 900\]
CF\[ = \]Cumulative frequency of median class \[ = 120 + {f_1}\]
c\[ = \]Class interval \[ = 10\]
f\[ = \]Frequency of median class \[ = 200\]
Putting these value we get:
\[
  59.25 = 40 + \dfrac{{\dfrac{{900}}{2} - (120 + {f_1})}}{{200}} \times 10 \\
  9.25 = \dfrac{{450 - 120 - {f_1}}}{{200}} \times 10 \\
  1850 = (450 - 120 - {f_1}) \times 10 \\
  10{f_1} = 3300 - 1850 \\
  {f_1} = \dfrac{{1450}}{{10}} \;
 \]
We get that, \[{f_1} = 145\].
Now, we have to calculate the value of \[{f_2}\].
\[{f_2} = 900 - \left( {120 + 145 + 200 + 185} \right)\]
 \[ = 200\].
So, the missing frequencies are \[{f_1} = 145\] and \[{f_2} = 200\].
So, the correct answer is “ \[{f_1} = 145\] and \[{f_2} = 200\] ”.

Note: While solving questions similar to the one given above, you have to remember the concepts of frequency, cumulative frequency and median. Draw a cumulative frequency table to simplify your answer. This will help you in solving your question in a very less amount of time. Assume the values of the missing frequencies.