Question

The following table shows the number of workers in a factory and their daily wages. Find the median of the daily wages.

Daily wages (rupees)	\[100 - 110\]	\[110 - 120\]	\[120 - 130\]	\[130 - 140\]	\[140 - 150\]	\[150 - 160\]
No. of workers	\[37\]	\[38\]	\[40\]	\[33\]	\[28\]	\[24\]

Answer 1

Hint:
We will first find the total frequency and cumulative frequencies by adding the previous frequency. Then, we will find the median class interval by dividing the total frequency by 2. Finally, we will use the formula for finding the median of grouped data and find the median of the given data.

Formula used:
Median \[ = l + \dfrac{{\left( {\dfrac{N}{2} - C} \right)}}{f} \times h\], where \[l\] is the lower limit of the median class interval, \[N\] is the total frequency, \[C\] is the cumulative frequency preceding the median class frequency, \[f\] is the frequency of the median class interval and \[h\] is the class width.

Complete step by step solution:
Let us find the total frequency and the cumulative frequencies and form the table below:

Daily wages (rupees)	No. of workers	Cumulative frequency
\[100 - 110\]	\[37\]	\[37\]
\[110 - 120\]	\[38\]	\[38 + 37 = 75\]
\[120 - 130\]	\[40\]	\[40 + 75 = 115\]
\[130 - 140\]	\[33\]	\[33 + 115 = 148\]
\[140 - 150\]	\[28\]	\[28 + 148 = 176\]
\[150 - 160\]	\[24\]	\[24 + 176 = 200\]
	\[N = \sum f = 200\]

To find the median class interval, let us check the value of \[\dfrac{N}{2}\]. Here, \[\dfrac{N}{2} = \dfrac{{200}}{2} = 100\].
Now, the class-interval containing the cumulative frequency \[100\] is \[120 - 130\].
The lower limit of this class-interval is \[l = 120\]
The frequency of the median class interval is \[f = 40\]
The cumulative frequency preceding the median class frequency is \[C = 75\]
The class width is \[h = 10\]
Now, we will substitute all of these values in the formula Median \[ = l + \dfrac{{\left( {\dfrac{N}{2} - C} \right)}}{f} \times h\]. Therefore, we get
Median \[ = 120 + \dfrac{{\left( {\dfrac{{200}}{2} - 75} \right)}}{{40}} \times 10\]
Simplifying the expression, we get
\[ \Rightarrow \] Median\[ = 120 + \dfrac{{25}}{{40}} \times 10\]
Multiplying the terms and taking LCM, we get
\[ \Rightarrow \] Median \[ = 120 + \dfrac{{25}}{4} = \dfrac{{480 + 25}}{4}\]
Adding the terms in the numerator, we get
\[ \Rightarrow \] Median \[ = \dfrac{{505}}{4}\]
Dividing 505 by 4, we get

\[ \Rightarrow \] Median \[ = 126.25\]

Note:
While finding the value of \[\dfrac{N}{2}\], the cumulative frequency \[100\] lies in the class-interval \[120 - 130\], although the cumulative frequency of that class-interval is \[115\]. We select this class because all the frequencies preceding to this class cumulate up to only \[75\]. All frequencies above \[75\] and below \[115\] will lie in the class-interval \[120 - 130\]. The frequency of a class is different from the cumulative frequency.