Question

Calculate the mean of the following distribution using the shortcut method:

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
Number of students	2	6	10	12	9	7	4

Answer 1

Hint: First of all, we are going to make the marks distribution continuous in such a way that the upper limit of any interval is equal to the lower limit of the interval which is written just after that. For e.g., in the interval 21-30, the upper limit of this interval is 30 and the lower limit of the next interval (31-40) is 31. As you can see that the upper limit of any interval is not equal to the lower limit of the next interval so we are going to make them equal to find the mean. After that, we are going to find the middle value for each interval and then multiply each middle value with the number of students written below that interval. Then we add these multiplications followed by division of this addition with the total number of students to get the mean.

Complete step-by-step solution:
The marks distribution table given in the above problem is as follows:

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
Number of students	2	6	10	12	9	7	4

Now, we are going to change the lower and upper limits of the marks intervals given above by subtracting 0.5 from the lower limit and adding 0.5 in the upper limit so after making this conversion the marks intervals will look like:

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
New Marks	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5	60.5-70.5	70.5-80.5
Number of students	2	6	10	12	9	7	4

Now, we are going to use this new marks row for the mean calculation so there is a table below which has new marks and the number of students rows.

New Marks	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5	60.5-70.5	70.5-80.5
Number of students	2	6	10	12	9	7	4

Now, we are going to find the middle values for all the intervals in the marks distribution.
The formula which we are using to evaluate the middle value of the marks interval is given below:
$\dfrac{\text{Lower Limit}+\text{Upper Limit}}{2}$
The middle value for 10.5-20.5 is equal to:
$\begin{align}
  & \dfrac{10.5+20.5}{2} \\
 & =\dfrac{31}{2}=15.5 \\
\end{align}$
Similarly, we can find all the middle values of the marks interval and in the below, we are tabulating the middle value.

Marks	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5	60.5-70.5	70.5-80.5
Middle value	15.5	25.5	35.5	45.5	55.5	65.5	75.5
Frequency	2	6	10	12	9	7	4

We are assigning the general name for middle value as ${{x}_{i}}$ in which if “i” equals 1 then we get 15.5, if “i” equals 2 then 25.5 and so on till “i” equals 7 where the middle value is 75.5. Similarly, we are assigning the name for general frequency as ${{f}_{i}}$.

Marks	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5	60.5-70.5	70.5-80.5
Middle value $\left( {{x}_{i}} \right)$	15.5	25.5	35.5	45.5	55.5	65.5	75.5
Frequency $\left( {{f}_{i}} \right)$	2	6	10	12	9	7	4

Now, we are multiplying ${{f}_{i}}\text{ }and\text{ }{{x}_{i}}$ together and writing it in the tabular form we get,

Marks	10.5-20.5	20.5-30.5	30.5-40.5	40.5-50.5	50.5-60.5	60.5-70.5	70.5-80.5
Middle value $\left( {{x}_{i}} \right)$	15.5	25.5	35.5	45.5	55.5	65.5	75.5
Frequency $\left( {{f}_{i}} \right)$	2	6	10	12	9	7	4
${{f}_{i}}{{x}_{i}}$	31.0	153.0	355.0	546.0	499.5	458.5	302.0

Adding all the values of ${{f}_{i}}{{x}_{i}}$ we get,
\[\begin{align}
  & \sum\limits_{i=1}^{7}{{{f}_{i}}}{{x}_{i}}=31.0+153.0+355.0+546.0+499.5+458.5+302.0 \\
 & \Rightarrow \sum\limits_{i=1}^{7}{{{f}_{i}}}{{x}_{i}}=2345 \\
\end{align}\]
The summation of all the frequencies is equal to:
$\begin{align}
  & \sum\limits_{i=1}^{7}{{{f}_{i}}}=2+6+10+12+9+7+4 \\
 & \Rightarrow \sum\limits_{i=1}^{7}{{{f}_{i}}}=50 \\
\end{align}$
Now, dividing 2345 to the summation of all the frequencies we get,
$\begin{align}
&\dfrac{\sum\limits_{i=1}^{7}{{{f}_{i}}}{{x}_{i}}}{\sum\limits_{i=1}^{7}{{{f}_{i}}}}=\dfrac{2345}{50} \\
& \Rightarrow \dfrac{\sum\limits_{i=1}^{7}{{{f}_{i}}}{{x}_{i}}}{\sum\limits_{i=1}^{7}{{{f}_{i}}}}=46.9 \\
\end{align}$
Hence, we got the mean for the above marks distribution is 46.9.

Note: There is a point to be noted here that we can check if the mean value that we got above is correct or not. The mean value that we got in the above solution is 46.9. And you can find that this value which is 46.9 is lying in one of the marks intervals lies in the marks interval given in the above problem and that interval is 41-50.
Now, if you got the mean as 469.50 then quickly you can understand that this mean value is wrong because this value is not lying in any of the marks intervals given above.