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

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)100110100 - 110110120110 - 120120130120 - 130130140130 - 140140150140 - 150150160150 - 160
No. of workers373738384040333328282424
Explanation

Solution

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+(N2C)f×h = l + \dfrac{{\left( {\dfrac{N}{2} - C} \right)}}{f} \times h, where ll is the lower limit of the median class interval, NN is the total frequency, CC is the cumulative frequency preceding the median class frequency, ff is the frequency of the median class interval and hh 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 workersCumulative frequency
100110100 - 11037373737
110120110 - 120383838+37=7538 + 37 = 75
120130120 - 130404040+75=11540 + 75 = 115
130140130 - 140333333+115=14833 + 115 = 148
140150140 - 150282828+148=17628 + 148 = 176
150160150 - 160242424+176=20024 + 176 = 200
N=f=200N = \sum f = 200

To find the median class interval, let us check the value of N2\dfrac{N}{2}. Here, N2=2002=100\dfrac{N}{2} = \dfrac{{200}}{2} = 100.
Now, the class-interval containing the cumulative frequency 100100 is 120130120 - 130.
The lower limit of this class-interval is l=120l = 120
The frequency of the median class interval is f=40f = 40
The cumulative frequency preceding the median class frequency is C=75C = 75
The class width is h=10h = 10
Now, we will substitute all of these values in the formula Median =l+(N2C)f×h = l + \dfrac{{\left( {\dfrac{N}{2} - C} \right)}}{f} \times h. Therefore, we get
Median =120+(200275)40×10 = 120 + \dfrac{{\left( {\dfrac{{200}}{2} - 75} \right)}}{{40}} \times 10
Simplifying the expression, we get
\Rightarrow Median=120+2540×10 = 120 + \dfrac{{25}}{{40}} \times 10
Multiplying the terms and taking LCM, we get
\Rightarrow Median =120+254=480+254 = 120 + \dfrac{{25}}{4} = \dfrac{{480 + 25}}{4}
Adding the terms in the numerator, we get
\Rightarrow Median =5054 = \dfrac{{505}}{4}
Dividing 505 by 4, we get

\Rightarrow Median =126.25 = 126.25

Note:
While finding the value of N2\dfrac{N}{2}, the cumulative frequency 100100 lies in the class-interval 120130120 - 130, although the cumulative frequency of that class-interval is 115115. We select this class because all the frequencies preceding to this class cumulate up to only 7575. All frequencies above 7575 and below 115115 will lie in the class-interval 120130120 - 130. The frequency of a class is different from the cumulative frequency.