Question 3

We need to find the missing frequency 'f' using the given mean and frequency distribution.

Step 1 — Calculate Class Marks and Deviations

Let's first find the class mark ($x_i$) for each interval. We use the formula: $x_i = (\text{Upper limit} + \text{Lower limit}) / 2$. We will take the assumed mean ($a$) as 18, which is the class mark of the class 17-19. Now, we calculate the deviation ($d_i = x_i - a$) and the product ($f_i d_i$).

| Daily pocket allowance (in ₹) | Number of children ($f_i$) | Class mark ($x_i$) | $d_i = x_i - 18$ | $f_i d_i$ | | :---------------------------- | :------------------------- | :----------------- | :---------------- | :-------- | | 11 - 13 | 7 | 12 | -6 | -42 | | 13 - 15 | 6 | 14 | -4 | -24 | | 15 - 17 | 9 | 16 | -2 | -18 | | 17 - 19 | 13 | 18 | 0 | 0 | | 19 - 21 | f | 20 | 2 | 2f | | 21 - 23 | 5 | 22 | 4 | 20 | | 23 - 25 | 4 | 24 | 6 | 24 | | Total | $\sum f_i$ | | | $\sum f_i d_i$ |

Let's sum the frequencies ($f_i$).

$\sum f_i = 7 + 6 + 9 + 13 + f + 5 + 4$

$= 44 + f$

Now, let's sum the products ($f_i d_i$).

$\sum f_i d_i = -42 - 24 - 18 + 0 + 2f + 20 + 24$

$= -84 + 2f + 44$

$= 2f - 40$

Step 2 — Apply Mean Formula

The formula for the mean using the assumed mean method is: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$. We are given that the mean pocket allowance ($\bar{x}$) is ₹18. We substitute the values into the formula.

$18 = 18 + \frac{2f - 40}{44 + f}$

Subtract 18 from both sides.

$0 = \frac{2f - 40}{44 + f}$

To make the fraction zero, the numerator must be zero.

$2f - 40 = 0$

Add 40 to both sides.

$2f = 40$

Divide by 2.

$f = 20$

$\boxed{f = 20}$

Answer

The value of the missing frequency f is 20.