# The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs18. Find the missing frequency f

**Solution:**

We will use Assumed Mean Method to solve this question because the data given is large.

When the numerical values of xᵢ and fᵢ are large, then 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 choose one value among xᵢ as the assumed mean and denote it by ‘ a ’. Also, to further reduce our calculation work, we may take ‘ a ’ to be that xᵢ which lies in the center of x₁, x₂, . . ., xₙ. So, we can choose a.
- The second step is to find the difference ‘ dᵢ ’. That is the difference between a and each of the xᵢ.
_{ }The deviation of ‘ a ’ from each of the xᵢ. i.e., dᵢ = xᵢ - a - The third step is to find the product of dᵢ with the corresponding fᵢ
_{.}After that take the sum of all the fᵢdᵢ.

Now use the values in the below formula

Mean, (x) = a + Σfᵢdᵢ/ Σfᵢ

We know that, Class mark, xᵢ = (Upper class limit + Lower class limit )/2

Taking assumed mean, a = 18

From the table, we obtain.

Σfᵢ = 44 + f

Σfᵢdᵢ = 2 f - 40

Mean, (x) = a + Σfᵢdᵢ/Σfᵢ

18 = 18 + [(- 40 + 2f)/(44 + f)]

18 - 18 = (2f - 40)/(44 + f)

2f - 40 = 0

f = 20

Thus, for the given problem the missing frequency f is 20.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 14

**Video Solution:**

## The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f

NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1 Question 3

**Summary:**

The missing frequency f for the given distribution showing the daily pocket allowance of children of a locality mean pocket allowance of Rs 18 is 20.

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