Question

Find the mean marks from the following data :

Marks	Number of students
Below 10	5
Below 20	9
Below 30	17
Below 40	29
Below 50	45
Below 60	60
Below 70	70
Below 80	78
Below 90	83
Below 100	85

A. 37.12 marks
B. 41.5 marks
C. 44.26 marks
D. 48.4 marks

Answer 1

Hint: To find the mean marks from the given data, we have to convert it into frequency distribution. We can write below 10 as 0-10, below 20 as 10-20 and so on. The corresponding frequencies ( ${{f}_{i}}$) can be found by subtracting the previous number of students from the current number of students. That is, for example, in the class range 10-20, we can find the frequency by doing $9-5=4$ . We have to find the midpoint $\left( {{x}_{i}} \right)$ using the formula $\text{Midpoint}=\dfrac{\text{Lower limit+Upper limit}}{2}$ . We will multiply ${{f}_{i}}\text{ and }{{x}_{i}}$ to get ${{f}_{i}}{{x}_{i}}$ . Now, find the sum of ${{f}_{i}}\text{ }$ separately and ${{f}_{i}}{{x}_{i}}$ . Mean is given by $\text{Mean}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}$ . Substitute the values to get the correct option.

Complete step-by-step answer:
We have to find the mean marks from the given data. Here, the data is given in less than type. So, let us convert it into frequency distribution.
We can write below 10 as 0-10, below 20 as 10-20 and so on. The corresponding frequencies can be found by subtracting the previous number of students from the current number of students. That is, for example, in the class range 10-20, we can find the frequency by doing $9-5=4$ . Now, let us tabulate this. Frequency distribution is shown below.

Marks	Number of students	Class Marks	Frequency( ${{f}_{i}}$)
Below 10	5	0-10	5
Below 20	9	10-20	$9-5=4$
Below 30	17	20-30	$17-9=8$
Below 40	29	30-40	$29-17=12$
Below 50	45	40-50	$45-29=16$
Below 60	60	50-60	$60-45=15$
Below 70	70	60-70	$70-60=10$
Below 80	78	70-80	$78-70=8$
Below 90	83	80-90	$83-78=5$
Below 100	85	90-100	$85-83=2$

Now, let us find the midpoint. This is done by using the formula
$\text{Midpoint}=\dfrac{\text{Lower limit+Upper limit}}{2}$
For example, for the interval 10-20, we can compute midpoint as
$\text{Midpoint}=\dfrac{10+20}{2}=\dfrac{30}{2}=15$
Similarly, we can do the same and tabulate these. We will get

Class Marks	Frequency ( ${{f}_{i}}$)	Midpoint$\left( {{x}_{i}} \right)$
0-10	5	5
10-20	4	15
20-30	8	25
30-40	12	35
40-50	16	45
50-60	15	55
60-70	10	65
70-80	8	75
80-90	5	85
90-100	2	95

Now, let us multiply frequency and midpoint. This is denoted as ${{f}_{i}}{{x}_{i}}$ . Let us also find the sum of frequencies and sum of ${{f}_{i}}{{x}_{i}}$ .

Class Marks	Frequency ( ${{f}_{i}}$)	Midpoint$\left( {{x}_{i}} \right)$	${{f}_{i}}{{x}_{i}}$
0-10	5	5	25
10-20	4	15	60
20-30	8	25	200
30-40	12	35	420
40-50	16	45	720
50-60	15	55	825
60-70	10	65	650
70-80	8	75	600
80-90	5	85	425
90-100	2	95	190
	$\sum{{{f}_{i}}}=85$		$\sum{{{f}_{i}}{{x}_{i}}}=4115$

We know that mean is given by
$\text{Mean}=\dfrac{\sum{{{f}_{i}}}{{x}_{i}}}{\sum{{{f}_{i}}}}$
Let us substitute the values. We will get
$\text{Mean}=\dfrac{4115}{85}=48.4$
Hence, the correct option is D.

So, the correct answer is “Option D”.

Note: You may make error when writing the formula for mean as $\text{Mean}=\dfrac{\sum{{{f}_{i}}}}{\sum{{{f}_{i}}{{x}_{i}}}}$ . The method used in the solution to find the mean is the direct method. We can also find the mean using the assumed mean method or step deviation method.