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Q6:

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

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

Solution :

Given: The frequency distribution of daily expenditure on food for 25 households.

Daily Expenditure (in ₹) 100-150 150-200 200-250 250-300 300-350
Number of Households ($f_i$) 4 5 12 2 2

To Find: The mean daily expenditure on food using a suitable method.

Methodology: Since the class intervals and frequencies are provided, we will use the Step-Deviation Method to calculate the mean, as it simplifies the arithmetic calculations significantly.

Step 1: Construct the Frequency Distribution Table

Let the assumed mean ($a$) be the midpoint of the central class interval. Here, the class intervals are $100-150, 150-200, 200-250, 250-300, 300-350$.

The class mark ($x_i$) is calculated as: $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$.

The class size ($h$) is $150 - 100 = 50$.

Let assumed mean $a = 225$.

The deviation $u_i = \frac{x_i - a}{h}$.

Class Interval Frequency ($f_i$) Class Mark ($x_i$) $u_i = \frac{x_i - 225}{50}$ $f_i u_i$
100-150 4 125 -2 -8
150-200 5 175 -1 -5
200-250 12 225 0 0
250-300 2 275 1 2
300-350 2 325 2 4
Total $\sum f_i = 25$ - - $\sum f_i u_i = -7$

Step 2: Apply the Step-Deviation Formula

The formula for the mean ($\bar{x}$) using the step-deviation method is:

$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

Step 3: Substitute the Values

Given: $a = 225$, $\sum f_i u_i = -7$, $\sum f_i = 25$, $h = 50$.

$\bar{x} = 225 + \left( \frac{-7}{25} \right) \times 50$

[Simplifying the fraction: $\frac{50}{25} = 2$]

$\bar{x} = 225 + (-7 \times 2)$

$\bar{x} = 225 - 14$

$\bar{x} = 211$

Final Answer: The mean daily expenditure on food is ₹ 211.


More Questions from Class 10 Mathematics Statistics EXERCISE 13.1


CBSE Solutions for Class 10 Mathematics Statistics


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