Statistics | Exercise 13.1

Question 6

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

Understand the Question

To find the mean daily expenditure for grouped data with large numerical values and equal class intervals, the Step-Deviation Method is the most efficient.
The class mark ( $x_i$ ) is the midpoint of each class interval: $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$
Using an assumed mean $a$ and class size $h$ , we compute $u_i = \dfrac{x_i - a}{h}$ and then calculate the mean using: $\bar{x} = a + \left(\dfrac{\sum f_i u_i}{\sum f_i}\right) \times h$

Step 1 · Prepare the Step-Deviation Table

Let Assumed Mean $a = 225$ and Class Size $h = 50$ .

\begin{array}{|c|c|c|c|c|c|} \hline \text{Daily expenditure (in ₹)} & \text{Number of households } (f_i) & \text{Class mark } (x_i) & d_i = x_i - 225 & u_i = \dfrac{d_i}{50} & f_i u_i \\ \hline 100 - 150 & 4 & 125 & -100 & -2 & -8 \\ \hline 150 - 200 & 5 & 175 & -50 & -1 & -5 \\ \hline 200 - 250 & 12 & 225 & 0 & 0 & 0 \\ \hline 250 - 300 & 2 & 275 & 50 & 1 & 2 \\ \hline 300 - 350 & 2 & 325 & 100 & 2 & 4 \\ \hline \text{Total} & \sum f_i = 25 & & & & \sum f_i u_i = -7 \\ \hline \end{array}

Step 2 · Calculate the Mean

Using the step-deviation formula $\bar{x} = a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) \times h$

Substitute $a = 225$ , $\sum f_i u_i = -7$ , $\sum f_i = 25$ , and $h = 50$

\begin{aligned} \bar{x} &= 225 + \left( \dfrac{-7}{25} \right) \times 50 \\[0.6em] &= 225 + (-7 \times 2) \\[0.6em] &= 225 - 14 \\[0.6em] &= 211 \end{aligned}

Answer

₹ $211$

Common Mistakes

Negative Sign Error: Neglecting the negative sign in $\sum f_i u_i = -13 + 6 = -7$ , leading to addition instead of subtraction from the assumed mean.
Omitting Class Size ( $h$ ): Forgetting to multiply $\dfrac{\sum f_i u_i}{\sum f_i}$ by $h = 50$ in the final formula.

More questions in Exercise 13.1