Question

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ${\rm{Rs}}.18$. Find the missing frequency $f$.

Daily pocket allowance (in Rs.)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of Children	7	6	9	13	f	5	4

Answer 1

Here, we will find the Class marks and then the sum of frequencies. Then we will find the sum of the product of frequencies and their respective class marks. We will then substitute these values in the formula of Mean, and equate it to 18 to find the missing value of the frequency $f$.

Formula used:
We will use the formula Mean $= \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where ${x_i}$ is the class mark and ${f_i}$ is the frequency.

Complete step-by-step answer:
We are given the mean pocket allowance of children of a locality and we are required to find the missing frequency$f$.
We will find the Class marks of the given class intervals.
Class marks or ${x_i}$ is equal to the sum of lower limit and upper limit divided by 2.
Hence, we will draw the following table:

Daily pocket allowance (in Rs.)	Number of Children$\left( {{f_i}} \right)$	Class Marks$\left( {{x_i}} \right)$	${f_i}{x_i}$
11-13	7	$\dfrac{{11 + 13}}{2} = \dfrac{{24}}{2} = 12$	$7 \times 12 = 84$
13-15	6	$\dfrac{{13 + 15}}{2} = \dfrac{{28}}{2} = 14$	$6 \times 14 = 84$
15-17	9	$\dfrac{{15 + 17}}{2} = \dfrac{{32}}{2} = 16$	$9 \times 16 = 144$
17-19	13	$\dfrac{{17 + 19}}{2} = \dfrac{{36}}{2} = 18$	$13 \times 18 = 234$
19-21	$f$	$\dfrac{{19 + 21}}{2} = \dfrac{{40}}{2} = 20$	$f \times 20 = 20f$
21-23	5	$\dfrac{{21 + 23}}{2} = \dfrac{{44}}{2} = 22$	$5 \times 22 = 110$
23-25	4	$\dfrac{{23 + 25}}{2} = \dfrac{{48}}{2} = 24$	$4 \times 24 = 96$
Total:	$\sum {{f_i} = } 44 + f$		$\sum {{f_i}{x_i} = } 752 + 20f$

We know that the formula of Mean $= \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$.
Substituting $\sum {{f_i} = } 44 + f$ and $\sum {{f_i}{x_i} = } 752 + 20f$ in the formula of mean, we get
Hence, Mean$= \dfrac{{752 + 20f}}{{44 + f}}$…………………………$\left( 1 \right)$
But in the question it is given that Mean$= 18$.
Substituting Mean$= 18$ in the equation $\left( 1 \right)$, we get
$\Rightarrow 18 = \dfrac{{752 + 20f}}{{44 + f}}$
On cross multiplication, we get
$\Rightarrow 18\left( {44 + f} \right) = 752 + 20f$
$\Rightarrow 18 \times 44 + 18f = 752 + 20f$
Multiplying 18 by 44, we get
$\Rightarrow 792 + 18f = 752 + 20f$
Adding and subtracting like terms, we get
$\Rightarrow 2f = 40$
Dividing both sides by 2, we get
$\Rightarrow f = 20$
Hence, the required value of missing frequency $f$ is 20.

Note: Mean is the average of a given set of numbers. It is usually stated as the sum of the observations divided by the total number of observations. In this question as well, we have multiplied the frequencies by the class marks which actually represent the marks of the whole class falling in that particular frequency. By adding all the products of the frequencies and the class marks, we actually find the ‘sum of observations’. And finally, when we divide this by the sum of frequencies, we get our required mean as the sum of frequencies gives us the total number of observations.