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

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 $\overline x$ of the data is give by
$\overline x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ....... + {f_n}{x_n}}}{{{f_1} + {f_2} + ....... + {f_n}}}$
Recall that we can write this in short form by using the Greek letter $\sum {\left( {CapitalSigma} \right)}$ which means summation.
$\overline x = \sum\limits_{i = 1}^n {\dfrac{{{f_i}{x_i}}}{{\sum\limits_{i = 1}^n {{f_1}} }}}$ 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 = $\dfrac{{UpperClassLimit + LowerClassLimit}}{2}$
We have the sigma of frequencies is $80$and sigma is ${f_1}{x_1}$ is $8244$. So, the mean $\overline x$ of the given data is the give by
$\overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} = \dfrac{{8244}}{{80}} = 103.05$
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