Question

In a higher secondary school, an inter-school quiz competition was held among 110 students. The marks obtained by 110 students are given below:

Marks	30-35	35-40	40-45	45-50	50-55	55-60	60-65
Number of Students	14	16	28	23	18	8	3

Find the mean.

Answer 1

Hint: We will make a new row in the above table which will contain the class mark of all the classes. We will calculate the classmark for a class using the formula $\text{class mark}=\dfrac{\text{upper limit}+\text{lower limit}}{2}$
Then we will find the sum of marks of all students by multiplying the class mark to its frequency and adding them. Then we will use the formula for mean, which is $\text{mean}=\dfrac{\text{sum of all observations}}{\text{number of observations}}$.

Complete step by step solution:
The first class is 30-35. Now, we will calculate the classmark for this class. The classmark is calculated by the following formula
$\text{class mark}=\dfrac{\text{upper limit}+\text{lower limit}}{2}$
So, we have the first-class mark as $\dfrac{30+35}{2}=\dfrac{65}{2}=32.5$. Similarly, for the class 35-40, we get the classmark as $\dfrac{35+40}{2}=\dfrac{75}{2}=37.5$. We can calculate the rest of the classmark in a similar manner. So, we get the following table,

Marks	30-35	35-40	40-45	45-50	50-55	55-60	60-65
Classmark	32.5	37.5	42.5	47.5	52.5	57.5	62.5
Number of Students	14	16	28	23	18	8	3

Now, we have to find the sum of all the marks obtained by the 110 students. For this we will multiply the class mark with its frequencies and add them all together. So, we have the following expression for the sum of all observations,
$\text{sum}=\left( 32.5\times 14 \right)+\left( 37.5\times 16 \right)+\left( 42.5\times 28 \right)+\left( 47.5\times 23 \right)+\left( 52.5\times 18 \right)+\left( 57.5\times 8 \right)+\left( 62.5\times 3 \right)$
Simplifying the above expression, we get
$\begin{align}
  & \text{sum of all observations}=455+600+1190+1092.5+945+460+187.5 \\
 & \therefore \text{sum of all observations}=4930 \\
\end{align}$
Now, the formula for mean is given by
$\text{mean}=\dfrac{\text{sum of all observations}}{\text{number of observations}}$
Substituting the sum of all observations and the number of observations, we get
$\begin{align}
  & \text{mean}=\dfrac{4930}{110} \\
 & \therefore \text{mean}=44.818 \\
\end{align}$
Hence, the mean of the marks is 44.818.

Note:
 Since the classes given in the table are continuous, we could directly use the upper limits and lower limits to find the class mark. If the classes are not continuous, then we first find the boundaries of the classes. These can be calculated by taking the average of the lower limit of the class and the upper limit of the previous class. After that, we use the same formula to find the class mark.