Question

A class teacher has the following absentee record of 40 students of a class for a whole term. Find the mean number of days a student was absent.

No. of Days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
No. of Students	11	10	7	4	4	3	1

Answer 1

Hint:
Here, we will find the mid values of the class intervals. Then, we will multiply the mid values with the frequency of the corresponding class interval. Finally, we will use the formula for mean of a continuous series to find the mean number of days a student was absent.

Formula Used: We will use the formula of the mean of a continuous series, \[\dfrac{{\sum {fm} }}{{\sum f }}\].

Complete step by step solution:
We can observe that the given series is an exclusive series, and not an inclusive series.
Therefore, we do not need to convert it to an exclusive series.
First, we will calculate the mid values of the class intervals.
The mid values of the class intervals is the average of the lower limit and the upper limit.
Therefore, we get
Mid value of the class interval 0 – 6 \[ = \dfrac{{0 + 6}}{2} = \dfrac{6}{2} = 3\]
Mid value of the class interval 6 – 10 \[ = \dfrac{{6 + 10}}{2} = \dfrac{{16}}{2} = 8\]
Mid value of the class interval 10 – 14 \[ = \dfrac{{10 + 14}}{2} = \dfrac{{24}}{2} = 12\]
Mid value of the class interval 14 – 20 \[ = \dfrac{{14 + 20}}{2} = \dfrac{{34}}{2} = 17\]
Mid value of the class interval 20 – 28 \[ = \dfrac{{20 + 28}}{2} = \dfrac{{48}}{2} = 24\]
Mid value of the class interval 28 – 38 \[ = \dfrac{{28 + 38}}{2} = \dfrac{{66}}{2} = 33\]
Mid value of the class interval 38 – 40 \[ = \dfrac{{38 + 40}}{2} = \dfrac{{78}}{2} = 39\]
We will add the mid values to a new row in the table.
Thus, we get the table

No. of Days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
No. of Students \[\left( f \right)\]	11	10	7	4	4	3	1
Mid values \[\left( m \right)\]	3	8	12	17	24	33	39

We will now find the mean using the formula \[\dfrac{{\sum {fm} }}{{\sum f }}\].
Multiplying the frequencies by the mid values and making a new row in the table, we get

No. of Days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
No. of Students \[\left( f \right)\]	11	10	7	4	4	3	1
Mid values \[\left( m \right)\]	3	8	12	17	24	33	39
\[fm\]	33	80	84	68	96	99	39

Now, we will find the sum of \[fm\] and the sum of all frequencies.
Therefore, we get
\[\sum {fm} = 33 + 80 + 84 + 68 + 96 + 99 + 39\] and \[\sum {f = 11 + 10 + 7 + 4 + 4 + 3 + 1} \]
Adding the terms in the expressions, we get
\[\sum {fm} = 499\] and \[\sum {f = 40} \]
Substituting \[\sum {fm} = 499\] and \[\sum {f = 40} \] in the formula for mean, we get
Mean number of days a student was absent \[ = \dfrac{{499}}{{40}}\]
Simplifying the expression, we get
Mean number of days a student was absent \[ = 12.475\]

Therefore, we get the mean number of days a student was absent as \[12.475\] days.

Note:
We calculated the mid values of the class intervals. These mid values are called class marks of the class interval. The class marks of the class intervals are the average of the lower limit and the upper limit of a class interval.