
The following table gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
Expenditure (in Rs.) Frequency 1000-1500 24 1500-2000 40 2000-2500 33 2500-3000 28 3000-3500 30 3500-4000 22 4000-4500 16 4500-5000 7
| Expenditure (in Rs.) | Frequency |
| 1000-1500 | 24 |
| 1500-2000 | 40 |
| 2000-2500 | 33 |
| 2500-3000 | 28 |
| 3000-3500 | 30 |
| 3500-4000 | 22 |
| 4000-4500 | 16 |
| 4500-5000 | 7 |
Answer
538.5k+ views
Hint: First, take the mid values of each class as ${x_i}$ and frequency ${f_i}$. The mean value is equivalent to the fraction between the addition of a product of mid-value with frequency and the total frequency and the formula to calculate mode is $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$.
Complete Step by Step Solution:
We are given the monthly expenditure of families.
Let us assume that ${f_i}$ represents the number of consumers and ${x_i}$ is the difference in the intervals divided by 2.
Let the missing frequency be $x$.
The frequency distribution table for the given data is as follows:
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{532500}}{{200}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 2662.5$
Hence the mean expenditure is Rs. 2662.5.
The formula of mode is,
Mode $ = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$
The modal class is the interval with the highest frequency.
$ \Rightarrow $Modal class $ = 1500 - 2000$
The lower limit of the modal class is,
$ \Rightarrow l = 1500$
The class-interval is,
$ \Rightarrow h = 1500 - 1000 = 500$
The frequency of the modal class is,
$ \Rightarrow {f_1} = 40$
The frequency of the class before the modal class is,
$ \Rightarrow {f_0} = 24$
The frequency of the class after modal class is,
$ \Rightarrow {f_2} = 33$
Substitute these values in the mode formula,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{40 - 24}}{{2\left( {40} \right) - 24 - 33}} \times 500$
Simplify the terms,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{16}}{{80 - 57}} \times 500$
Subtract the values in the denominator and multiply the terms in the numerator,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{8000}}{{23}}$
Divide the numerator by denominator,
$ \Rightarrow $ Mode $ = 1500 + 347.83$
Add the terms,
$\therefore $ Mode $ = 1847.83$
Hence, the mode is Rs. 1847.83.
Note: In solving these types of questions, students should know the formulae of mean and mode. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.
Complete Step by Step Solution:
We are given the monthly expenditure of families.
Let us assume that ${f_i}$ represents the number of consumers and ${x_i}$ is the difference in the intervals divided by 2.
Let the missing frequency be $x$.
The frequency distribution table for the given data is as follows:
| Class | Frequency (${f_i}$) | Mid-value (${x_i}$) | ${f_i}{x_i}$ |
| 1000-1500 | 24 | 1250 | 30000 |
| 1500-2000 | 40 | 1750 | 70000 |
| 2000-2500 | 33 | 2250 | 74250 |
| 2500-3000 | 28 | 2750 | 77000 |
| 3000-3500 | 30 | 3250 | 97500 |
| 3500-4000 | 22 | 3750 | 82500 |
| 4000-4500 | 16 | 4250 | 68000 |
| 4500-5000 | 7 | 4750 | 33250 |
| Total | $\sum {{f_i}} = 200$ | $\sum {{f_i}{x_i}} = 532500$ |
We know that the general formula to find the mean value is,
Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$
Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.
$ \Rightarrow $ Mean $ = \dfrac{{532500}}{{200}}$
Divide numerator by the denominator,
$\therefore $ Mean $ = 2662.5$
Hence the mean expenditure is Rs. 2662.5.
The formula of mode is,
Mode $ = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$
The modal class is the interval with the highest frequency.
$ \Rightarrow $Modal class $ = 1500 - 2000$
The lower limit of the modal class is,
$ \Rightarrow l = 1500$
The class-interval is,
$ \Rightarrow h = 1500 - 1000 = 500$
The frequency of the modal class is,
$ \Rightarrow {f_1} = 40$
The frequency of the class before the modal class is,
$ \Rightarrow {f_0} = 24$
The frequency of the class after modal class is,
$ \Rightarrow {f_2} = 33$
Substitute these values in the mode formula,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{40 - 24}}{{2\left( {40} \right) - 24 - 33}} \times 500$
Simplify the terms,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{16}}{{80 - 57}} \times 500$
Subtract the values in the denominator and multiply the terms in the numerator,
$ \Rightarrow $ Mode $ = 1500 + \dfrac{{8000}}{{23}}$
Divide the numerator by denominator,
$ \Rightarrow $ Mode $ = 1500 + 347.83$
Add the terms,
$\therefore $ Mode $ = 1847.83$
Hence, the mode is Rs. 1847.83.
Note: In solving these types of questions, students should know the formulae of mean and mode. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.
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