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Question: The following data gives the distribution of total monthly household expenditure of 200 families in ...

The following data gives the distribution of total monthly household expenditure of 200 families in a village. Find the modal monthly expenditure of the families. Also, find the mean 150020001500 - 2000 monthly expenditure.

Expenditure (in Rs)Number of families
100015001000 - 15002424
150020001500 - 20004040
200025002000 - 25003333
250030002500 - 30002828
300035003000 - 35003030
350040003500 - 40002222
400045004000 - 45001616
450050004500 - 500077
Explanation

Solution

In this question, we have been asked to find the mode and mean. Starting with mode, mark the interval with the highest frequency as f1{f_1}, the interval before that as f0{f_0} and after the modal class as f2{f_2}. After recognising the modal class, simply put the values in the formula to find the answer.
Next, find mean. First step is to write the midpoint of the class interval as xi{x_i} Then, multiply the frequency with the midpoint to find fixi{f_i}{x_i}. Find the sum of fixi{f_i}{x_i} and fi{f_i}, and put in the formula to find the mean.

Formula used: 1) Mode = l+f1f02f1f0f2×hl + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h, where l=l = lower limit of modal class, f1={f_1} = frequency of the modal class, f0={f_0} = frequency of class before modal class, f2={f_2} = frequency of class after modal class and h=h = class interval.
2) Mean = fixifi\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}

Complete step-by-step solution:
We are given a distribution table of monthly expenditure of 200 families. We have been asked to find the mode and mean.
Let us first start with the mode.

Expenditure (in Rs)Number of families
100015001000 - 15002424=f0{f_0}
150020001500 - 20004040=f1{f_1}
200025002000 - 25003333=f2{f_2}
250030002500 - 30002828
300035003000 - 35003030
350040003500 - 40002222
400045004000 - 45001616
450050004500 - 500077

Since 150020001500 - 2000 has the highest frequency, it is our modal class. The class before modal class will be fo{f_o} and the class after that will be f2{f_2}.
By looking at the table, l=1500l = 1500, h=500h = 500.
Now, let us put the values in the formula l+f1f02f1f0f2×hl + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h.
\Rightarrow Mode = 1500+40242×402433×5001500 + \dfrac{{40 - 24}}{{2 \times 40 - 24 - 33}} \times 500
Simplifying the equation,
\Rightarrow Mode = 1500+168057×5001500 + \dfrac{{16}}{{80 - 57}} \times 500
\Rightarrow Mode = 1500+1623×5001500 + \dfrac{{16}}{{23}} \times 500
\Rightarrow Mode = 1500+347.82=1847.821500 + 347.82 = 1847.82
Therefore, modal monthly expenditure of families is 1847.821847.82Rs.
Now, let us move towards finding mean. First step is to write the midpoint of the class interval as xi{x_i} Then, multiply the frequency with the midpoint to find fixi{f_i}{x_i}. Find the sum of fixi{f_i}{x_i} and fi{f_i} and put in the formula to find the mean.

Expenditure (in Rs)Number of families(fi)({f_i})Mid-point(xi)({x_i})fixi{f_i}{x_i}
100015001000 - 150024241250125030,00030,000
150020001500 - 200040401750175070,00070,000
200025002000 - 250033332250225074,25074,250
250030002500 - 300028282750275077,00077,000
300035003000 - 350030303250325097,50097,500
350040003500 - 400022223750375082,50082,500
400045004000 - 450016164250425068,00068,000
450050004500 - 5000774750475033,25033,250
fi=200\sum {{f_i} = 200} fixi=5,32,500\sum {{f_i}{x_i} = 5,32,500}

Mean = fixifi\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}
Putting the values,
Xˉ=532500200=2662.5\Rightarrow \bar X = \dfrac{{532500}}{{200}} = 2662.5

\therefore The mean monthly expenditure is 2662.52662.5 Rs.

Note: We have to remember that, the mean is the mathematical average of a set of two or more numbers. The arithmetic mean and the geometric mean are two types of mean that can be calculated. Summing the numbers in a set and dividing by the total number gives you the arithmetic mean. The students can use any other formula to find the mean – assumed mean method or shortcut method as per their choice. The formula used here is a direct and shorter method but it involves a lot of calculation.