Question

The following table gives the daily income of 50 workers of a factory:

Daily income (in Rs.)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

Find the mean, median, and mode of the above data.

Answer 1

Hint: First we will calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency. After that calculate the median by the formula $l + \dfrac{{\dfrac{N}{2} - cf}}{f} \times h$. Then use the formula of mode $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}$ to calculate the mode.

Complete step by step solution:
We are given that the daily income is the confidence interval C.I.
Let us assume that ${f_i}$ represents the number of workers and ${x_i}$ is the mid-value of the interval.
We will now form a table to find the value of the product ${f_i}{x_i}$ for the mean.

C.I.	${x_i}$	${f_i}$	${f_i}{x_i}$
100 – 120	110	12	1320
120 – 140	130	14	1820
140 – 160	150	8	1200
160 – 180	170	6	1020
180 – 200	190	10	1900
Total		$\sum {{f_i}} = 50$	$\sum {{f_i}{x_i}} = 7260$

We know that the formula to calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency.
Substitute the values to find the value of mean in the above formula, we get,
$ \Rightarrow $ Mean $ = \dfrac{{7260}}{{50}}$
Divide the numerator by the denominator,
$ \Rightarrow $ Mean $ = 145.2$
Hence, the mean is 145.2.

We will now find the value of $cf$ from the above table for median and mode.
We know from the above table that the value of $n$ is 50.

C.I.	${x_i}$	${f_i}$	$cf$
100 – 120	110	12	12
120 – 140	130	14	26
140 – 160	150	8	34
160 – 180	170	6	40
180 – 200	190	10	50

The value of $\dfrac{n}{2}$ is,
$ \Rightarrow \dfrac{{50}}{2} = 25$
So, the median class is 120 – 140.
The lowest value of the median class is,
$ \Rightarrow l = 120$
The frequency of the median class is,
$ \Rightarrow f = 14$
The difference of interval is,
$ \Rightarrow h = 20$
The cumulative frequency above the median class is,
$ \Rightarrow cf = 12$
Substitute these values in the median formula,
$ \Rightarrow $ Median $ = 120 + \dfrac{{25 - 12}}{{14}} \times 20$
Simplify the terms,
$ \Rightarrow $ Median $ = 120 + \dfrac{{13}}{{14}} \times 20$
Divide the numerator by denominator and multiply with 20,
$ \Rightarrow $ Median $ = 120 + 18.57$
Add the terms,
$ \Rightarrow $ Median $ = 138.57$
Hence, the median is 138.57.

Now the modal class is the class where ${f_i}$ is the highest, thus the modal class from the above table is,
$ \Rightarrow 120 - 140$
Then in the mode class, we have
$ \Rightarrow l = 120$
$ \Rightarrow {f_0} = 12$
$ \Rightarrow {f_1} = 14$
$ \Rightarrow {f_2} = 8$
$h = 20$
Substitute the values in mode formula,
$ \Rightarrow $ Mode $ = 120 + \dfrac{{14 - 12}}{{2\left( {14} \right) - 12 - 8}} \times 20$
Simplify the terms,
$ \Rightarrow $ Mode $ = 120 + \dfrac{2}{8} \times 20$
Multiply the numerator and then divide by denominator,
$ \Rightarrow $ Mode $ = 120 + 5$
Add the terms,
$ \Rightarrow $ Mode $ = 125$
Hence, the mode is 125.

Note: The only difficulty where students will face is selecting the median and modal classes. After selecting the classes taking the frequencies and cumulative frequencies for substituting in the formula is important. Due to confusion students will take frequencies of median and modal classes instead of preceding and succeeding frequencies. The selection needs to be taken care of.