Question

The marks distribution of 30 students in a mathematics examination are as follows:

Class interval of marks	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
No. of students	2	3	7	6	6	6

Find the mean by assumed mean method and find also the mode of given data.

Answer 1

To solve this problem, we should know the assumed mean method. In the assumed mean method, we will assume a certain number within the data given as the mean and is denoted by a. We will calculate the deviation of different classes from the assumed mean and we will calculate the weighted average of the deviations with the weights being the frequencies and the average is added to the assumed mean. The formula to calculate mode is $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$.

Complete step-by-step answer:
We can write the formula for the assumed mean method as
If a is the assumed mean ${f_i}$ denotes the frequency of the ${i^{th}}$ class which is having a deviation of ${d_i}$ from the assumed mean, the formula for the mean is,
$\bar x = a + \dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}$.
Whenever we are given the classes as an interval, we should take the middle value as the representative of the class. That is, for a class of 100-120, we take the class representative value as 110. Likewise, by restructuring the data, we get

Class Interval	Frequency (${f_i}$)	Mid-value (${x_i}$)
10 – 25	2	17.5
25 – 40	3	32.5
40 – 55	7	47.5
55 – 70	6	62.5
70 – 85	6	77.5
85 – 100	6	92.5

Let us assume the assumed mean as $a = 47.5$. We can write the deviations of different classes as

Class Interval	Frequency (${f_i}$)	Mid-value (${x_i}$)	Deviation${d_i} = {x_i} - 47.5$	${f_i}{d_i}$
10 – 25	2	17.5	-30	-60
25 – 40	3	32.5	-15	-45
40 – 55	7	47.5	0	0
55 – 70	6	62.5	15	90
70 – 85	6	77.5	30	180
85 – 100	6	92.5	45	270
$\sum {{f_i}} = 30$			$\sum {{f_i}{x_i}} = 435$

Using the assumed mean formula, we get,
$\Rightarrow \bar x = 47.5 + \dfrac{{435}}{{30}}$
Divide the numerator by denominator,
$\Rightarrow \bar x = 47.5 + 14.5$
Add the terms,
$\therefore \bar x = 62$
Hence the mean is 62.
Finding Mode:
The formula of mode is,
Mode $= l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h$
The modal class is the interval with the highest frequency.
$\Rightarrow$Modal class $= 40 - 55$
The lower limit of the modal class is,
$\Rightarrow l = 40$
The class-interval is,
$\Rightarrow h = 55 - 40 = 15$
The frequency of the modal class is,
$\Rightarrow {f_1} = 7$
The frequency of the class before the modal class is,
$\Rightarrow {f_0} = 3$
The frequency of the class after modal class is,
$\Rightarrow {f_2} = 6$
Substitute these values in the mode formula,
$\Rightarrow$ Mode $= 40 + \dfrac{{7 - 3}}{{2\left( 7 \right) - 3 - 6}} \times 15$
Simplify the terms,
$\Rightarrow$ Mode $= 40 + \dfrac{4}{{14 - 9}} \times 15$
Subtract the values in the denominator and multiply the terms in the numerator,
$\Rightarrow$ Mode $= 40 + \dfrac{{60}}{5}$
Divide the numerator by denominator,
$\Rightarrow$ Mode $= 40 + 12$
Add the terms,
$\therefore$ Mode $= 52$
Hence, the mode is 52.

Note: Mean is given by the sum of all the observations divided by the number of observations.
Range = Highest Observation – Lowest Observation
The students must also note that if one entry would have been exactly equal to the mean which is 73, then that entry would not be counted as the student getting more than mean marks, because that is just equal to the mean marks, not greater than it.
The students should also note that the average and mean represent the same thing.