Question

Find the mean of the following distribution:
\[ {\text{Class: }}3 - 55 - 77 - 99 - 1111 - 13 \\
  {\text{Frequency: }}5{\text{ }}10{\text{ }}10{\text{ }}7{\text{ 8}} \\
\].

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 \[{x_1},{x_2},....................{\text{ }}{x_n}\] are observations with respective frequencies\[{y_1},{y_2},....................{\text{ }}{y_n}\], then this mean observation \[{x_1}\] occurs \[{y_1}\] times, \[{x_2}\] occurs \[{y_2}\] times, and so on.
Now, the sum of the values of all the observations \[ = {y_1}{x_1} + {y_2}{x_2} + ............{y_n}{x_n}\] and number of observations \[ = {y_1} + {y_2} + ............{\text{ + }}{y_n}\] .
So, the mean $\overline x $ of the data is given by
$\overline x = \dfrac{{{y_1}{x_1} + {y_2}{x_2} + ............{y_n}{x_n}}}{{{y_1} + {y_2} + ............{\text{ + }}{y_n}}}$
Recall that we can write this is short form by using the week letter $\sum {\left( {{\text{Capital Sigma}}} \right)} $ which means summations that is \[\overline x = \dfrac{{{y_i}{x_i}}}{{\sum\limits_{1 = 1}^n {{y_i}} }}\] where it is understood that (i) varies from $1$ to $n$.

Complete step-by-step answer:
Step1:

Class Interval	Frequency\[\left( {{y_i}} \right)\]	Class mark\[\left( {{x_i}} \right)\]	\[{y_i}{x_i}\]
$3 - 5$	$5$	$4$	$20$
$5 - 7$	$10$	$6$	$60$
$7 - 9$	$10$	$8$	$80$
$9 - 11$	$7$	$10$	$70$
$11 - 13$	$8$	$12$	$96$
	$40$		$\sum {{y_i}} {x_i} = 326$

It is assumed that the frequency of each class interval is centered around its mid point. So the midpoint (Class Marks) of each class can be chosen to represent the observation in the class. Class mark of class by finding the average of its upper and lower limits that is.
${\text{Class Mark = }}\dfrac{{{\text{Upper class limit + Lower class mark}}}}{2}$
${\text{Interval }}3 - 5{\text{ }} \Rightarrow {\text{ }}\dfrac{{3 + 5}}{2}{\text{ }} \Rightarrow {\text{ }}\dfrac{8}{2}{\text{ = }}4$ . So $4$ is
Class mark of interval $3 - 5$ and so on.

Step2:
We have the sigma of frequencies is $40$ and sigma of ${y_i}{x_i}$ is $326$ .
So, the mean of the given data is the given by \[\overline x = \dfrac{{\sum {{y_i}{x_i}} }}{{\sum {{y_i}} }} = \dfrac{{326}}{{40}}\]
After solving means $326$ divided by $40$, we get
$ \Rightarrow {\text{ }}\overline x = 8.15$
The mean value is $8.15$

Note: If ${x_i}{\text{ and }}{y_i}$ are numerically large numbers then we can go for the assumed mean method of step-deviation method. If the class size are unequal and ${x_i}$ are large numerically, we can still apply the step-deviation method for finding the mean by taking $h$ to be a suitable divisor of all the dis.