Question

Using an appropriate method, find the mean of following frequency distribution:

Class	84 - 90	90 - 96	96 - 102	102-108	108 - 114	114 - 120
Frequency	8	10	16	23	12	11

Which method did you use, and why?

Answer 1

Hint: The mean (or average) of observations, as we know is the sum of the values of all the observations divided by the total number of observations. If $$x_1,x_2,........,x_n$$ are observation with respective frequencies $$f_1,f_2,........,f_n$$, then their mean observation $$x_1$$ occurs $j_1$ time, $$x_2$$ occurs $j_2$ time and so on.
So, the mean $\bar{x}$ of the data is give by
$$\bar{x}={\displaystyle \frac{f_1{x}_1+f_2{x}_2+.......+f_n{x}_n}{f_1+f_2+.......+f_n}}$$
Recall that we can write this in short form by using the Greek letter $\sum \left(CapitalSigma\right)$ which means summation.
$$\bar{x}=\sum _{i=1}^n{\displaystyle \frac{f_i{x}_i}{\sum _{i=1}^n{f}_1}}$$ Varies from $1$ to $n$.

Complete step-by-step answer:

class Interval	Frequency($$f_i$$)	Class mark($$x_1$$)	$$f_i{x}_i$$
$84-90$	$8$	$87$	$696$
$90-96$	$10$	$93$	$930$
$96-102$	$16$	$99$	$1584$
$102-108$	$$23$$	$$105$$	$$2415$$
$108-114$	$$12$$	$$111$$	$$1332$$
$114-120$	$$11$$	$$117$$	$$1287$$
	$$\mathbf{80}$$		$$\sum f_i{x}_i=\mathbf{8244}$$

It is assumed that the frequency of each class – Internal is centered around its main point. So the midpoint (class marks) of each class can be chosen to represent the observation in the case.
Class mark = ${\displaystyle \frac{UpperClassLimit+LowerClassLimit}{2}}$
We have the sigma of frequencies is $80$and sigma is $f_1{x}_1$ is $8244$. So, the mean $\bar{x}$ of the given data is the give by
$$\bar{x}={\displaystyle \frac{\sum f_i{x}_i}{\sum f_i}}={\displaystyle \frac{8244}{80}}=103.05$$
The mean value is $103.05$
The choice of method to be used depends on the numerical value of $x_i$ and $f_i$.
 If $x_i$ and $f_i$ are sufficiently small, then the direct method is an appropriate choice.

Note: If $x_i$ and $f_i$ are numerically large numbers, then we can go for the assumed mean method of step – deviation method. If the class size is unequal and $x_i$ are large. Numerically, we can still apply the step-deviation method for te mean by taking h to be a suitable divisor of all the