Question

Find the mean of the following data using the assumed mean method.

${\text{Class Interval}}$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$
${\text{Frequency}}$	$7$	$12$	$8$	$10$

Answer 1

Hint: Mean is the sum of a collection of numbers divided by the count of numbers divided by the count of numbers in the collection. In this question, we are to use the assumed mean method, in which we generally assume a middle number from the class marks (the mid-value of class interval) in order to ease our calculations, and quicken our process. Then, we find the deviations of the class marks of each interval from the assumed mean. We substitute the values of the known quantities into the formula $\bar x = A + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}$ to get to the required answer.

Complete step by step answer:
We are to find the class marks for each class interval.
So, the class marks, ${x_i} = \dfrac{{{\text{(Upper limit of class interval) + (Lower limit of class interval)}}}}{2}$
For, first interval: $10 - 15$, ${x_i} = \dfrac{{10 + 15}}{2}$.
$ \Rightarrow {x_i} = \dfrac{{25}}{2} = 12.5$
For, second interval: $15 - 20$, ${x_i} = \dfrac{{15 + 20}}{2}$
$ \Rightarrow {x_i} = \dfrac{{35}}{2} = 17.5$
For, third interval: $20 - 25$, ${x_i} = \dfrac{{20 + 25}}{2}$
$ \Rightarrow {x_i} = \dfrac{{45}}{2} = 22.5$
For, fourth interval $25 - 30$, ${x_i} = \dfrac{{25 + 30}}{2}$
$ \Rightarrow {x_i} = \dfrac{{55}}{2} = 27.5$

Now, let the assumed mean be, $A = 17.5$. Then, we have to calculate the deviations from this assumed mean.
Now, ${d_i} = {\text{(Class marks,}}\;{x_i}{\text{) - (Assumed mean, }}A)$
Therefore, for the first interval: $10 - 15$, ${d_i} = 12.5 - 17.5 = - 5$.
For second interval: $15 - 20$, ${d_i} = 17.5 - 17.5 = 0$
For third interval: $20 - 25$, ${d_i} = 22.5 - 17.5 = 5$
For fourth interval: $25 - 30$, ${d_i} = 27.5 - 17.5 = 10$
Putting these values in the table, we get,

${\text{Class Interval}}$	${\text{Class Marks(}}{x_i})$	${d_i} = {x_i} - A$	${\text{Frequency(}}{f_i})$	${d_i} \times {f_i}$
$10 - 15$	$12.5$	$ - 5$	$7$	$( - 5) \times 7 = - 35$
$15 - 20$	$17.5 = A$	$0$	$12$	$0 \times 12 = 0$
$20 - 25$	$22.5$	$5$	$8$	$5 \times 8 = 40$
$25 - 30$	$27.5$	$10$	$10$	$10 \times 10 = 100$
${\text{Total}}$			$\sum {{f_i} = } 7 + 12 + 8 + 10 = 37$	$\sum {{f_i}{d_i} = } - 35 + 0 + 40 + 100 = 105$

Now, by the formula of assumed mean method, the mean of the data, $\bar x = A + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}$
Now, by substituting the values of$A$,$\sum {{f_i}} $, $\sum {{f_i}{d_i}} $, we get,
$\bar x = 17.5 + \dfrac{{105}}{{37}}$
Simplifying the calculations, we get,
$ \Rightarrow \bar x = 17.5 + 2.84$
$ \therefore \bar x = 20.34$

Therefore, the arithmetic mean of the data is $20.34$.

Note: Further easier ways of finding mean are also there, like the step-deviation method, in which ${d_i}$ is further divided by the width of class $h$for each class interval to make the calculations further easier. For step deviation method, the formula of mean is $\bar x = A + \dfrac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }} \times h$ where, ${u_i} = \dfrac{{{d_i}}}{h}$, for each interval. But, one may argue that these processes are more complex and harder to remember. In such a case, the standard formula of mean can be used, which is, $\bar x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$. But, one thing must be kept in mind, the problem should be done by the method as stated in the question and if no method is stated, then any of the three methods can be opted.