Question

The table below shows the daily expenditure on food of $25$ households in a locality.

Daily expenditure	$100-150$	$150-200$	$200-250$	$250-300$	$300-350$
No of households	$4$	$5$	$12$	$2$	$2$

Find the mean daily expenditure on food by a suitable method.

Answer 1

In this question, we have the set of given observations.we have to find out the mean.
The mean (or average) of observations, is the sum of the values of all the observations divided by the total number of observations.
If ${x_1},{x_2},{x_3},......,{x_n}$ are observations with respective frequencies ${f_1},{f_2},{f_3},........,{f_n}$ then this means observation ${x_1}$ occurs ${f_1}$ times, ${x_2}$ occurs ${f_2}$ times, and so on.
Now, the sum of the values of all the observations = ${f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}$, and sum of the number of observations = ${f_1} + {f_2} + {f_3} + ........ + {f_n}$
So, the mean x of the data is given by
$x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}$
Or
$x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$

Complete step-by-step solution:
We have daily expenditure on food of $25$ households in a locality.
We need to find out the mean daily expenditure on food.To find the mean first we need to calculate the midpoint $\left( {{x_i}} \right)$ of the respective expenditures and ${f_i}{x_i}$ value for each case then we can put it in the mean formula.

Daily expenditure(in C)	Number of households$\left( {{f_i}} \right)$	Mid-point$\left( {{x_i}} \right)$	${f_i}{x_i}$
$100 - 150$	$4$	$\dfrac{{100 + 150}}{2} = \dfrac{{250}}{2} = 125$	$125 \times 4 = 500$
$150 - 200$	$5$	$\dfrac{{150 + 200}}{2} = \dfrac{{350}}{2} = 175$	$175 \times 5 = 875$
$200 - 250$	$12$	$\dfrac{{200 + 250}}{2} = \dfrac{{450}}{2} = 225$	$225 \times 12 = 2700$
$250 - 300$	$2$	$\dfrac{{250 + 300}}{2} = \dfrac{{550}}{2} = 275$	$275 \times 2 = 550$
$300 - 350$	$2$	$\dfrac{{300 + 350}}{2} = \dfrac{{650}}{2} = 325$	$325 \times 2 = 650$

Now, the mean x of the data is given by,
$x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}$
Or
$x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$
That is, $x = \dfrac{{500 + 875 + 2700 + 550 + 650}}{{4 + 5 + 12 + 2 + 2}} = \dfrac{{5275}}{{25}} = 211$

Hence the mean daily expenditure on food is $211$.

Note: In statistics, the mean is one of the three measures of central tendency, which are single numbers that try to pinpoint the central location within a data set. The mean, or average, is used most commonly, but it is important to differentiate it from the two other measures: median and mode. The median is the middle number when the numbers are listed in ascending order, while the mode is the most frequently occurring number.