The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in ₹) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
To find the mean daily expenditure, we use the Step-Deviation Method because the class size is uniform.
Step 1: Construct the frequency table. Let the assumed mean (a) be 225 (midpoint of the third class interval). The class size (h) is 50. We calculate the class marks (x_i), deviations (d_i), and step-deviations (u_i).
| Daily expenditure (₹) | Number of households (f_i) | Class mark (x_i) | d_i = x_i - 225 | u_i = d_i/50 | f_i u_i |
|---|---|---|---|---|---|
| 100 - 150 | 4 | 125 | -100 | -2 | -8 |
| 150 - 200 | 5 | 175 | -50 | -1 | -5 |
| 200 - 250 | 12 | 225 | 0 | 0 | 0 |
| 250 - 300 | 2 | 275 | 50 | 1 | 2 |
| 300 - 350 | 2 | 325 | 100 | 2 | 4 |
| Total | ∑f_i = 25 | ∑f_i u_i = -7 |
Step 2: Apply the formula for Mean. Mean = a + (∑f_i u_i / ∑f_i) × h
Step 3: Substitute the values and calculate. Mean = 225 + (-7 / 25) × 50 Mean = 225 + (-) Mean = 225 - 14 Mean = 211
The mean daily expenditure on food is ₹ 211.
Explanation
The question asks for the mean using a suitable method. Since the class intervals have a uniform width of 50, the Step-Deviation Method is the most efficient approach. It simplifies calculations by converting deviations into step-deviations (u_i), resulting in smaller integer values. The assumed mean is chosen at the central class mark (225) to balance positive and negative deviations.
Solution Steps
Step 1: Identify class size h = 50 and choose assumed mean a = 225.
Step 2: Calculate step-deviations u_i and find the sums: ∑f_i = 25 and ∑f_i u_i = -7.
Step 3: Apply the Mean formula: Mean = a + (∑f_i u_i / ∑f_i) × h.
Step 4: Calculate Mean = 225 + (-7/25) × 50 = 225 - 14 = 211.