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:
We will use the step-deviation method to solve this question because the data given is large and will be convenient to apply if all the dᵢ have a common factor.
When the numerical values of xᵢ and fᵢ are large, finding the product of xᵢ and fᵢ becomes tedious. We can change each xᵢ to a smaller number so that our calculations become easy. Now we have to subtract a fixed number from each of these xᵢ.
- The first step is to select one among the xᵢ as the assumed mean which is named as ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xᵢ which lies in the centre of x₁, x₂, . . ., xₙ.
- The next step is to find the difference ‘ dᵢ ’ between a and each of the xᵢ, that is, the deviation of ‘ a ’ from each of the xᵢ. i.e., dᵢ = xᵢ - a
- The third step is to find ‘ uᵢ ’ by dividing dᵢ by the class size h-for each of the xᵢ. i.e., uᵢ= dᵢ/h
- The next step is to find the product of uᵢ with the corresponding fᵢ, and take the sum of all the fᵢuᵢ.
The step-deviation method will be convenient to apply if all the dᵢ have a common factor. Now use the values in the formula below.
Mean, (x) = a + (Σfᵢuᵢ/Σfᵢ) × h
We know that, Class mark, xᵢ = (Upper class limit + Lower class limit) / 2
Class size, h = 50
Taking assumed mean, a = 225
From the table, we obtain
Σfᵢ = 25
Σfᵢuᵢ = -7
Mean, (x) = a + (Σfᵢuᵢ/Σfᵢ) × h
= 225 + (- 7/25) × 50
= 225 - 14
= 211
Thus, the mean daily expenditure on food is ₹ 211.
☛ Check: NCERT Solutions Class 10 Maths Chapter 14
Video Solution:
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.
NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 Question 6
Summary:
The table below shows the daily expenditure on food of 25 households in a locality. The mean daily expenditure on food of 25 households in a locality is ₹ 211 using step deviation method.
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