Question

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

No. of heart beats per minute	0 - 6	6 - 10	10 – 14	14 – 20	20 – 28	28 – 38	38 – 40
No. of women	11	10	7	4	4	3	1

Answer 1

Hint: We will first create a new table using the data given above to us with the mid value of the intervals, assumed mean and then the product of assumed mean and frequency. After this, we will use formulas of mean to find the required answer.

Complete step-by-step answer:
Let us first find the mid – points of all the intervals.
0 – 6: the mid – point will be \[\dfrac{{0 + 6}}{2} = \dfrac{6}{2} = 3\]
6 – 10: the mid – point will be \[\dfrac{{6 + 10}}{2} = \dfrac{{16}}{2} = 8\]
10 – 14: the mid – point will be \[\dfrac{{10 + 14}}{2} = \dfrac{{24}}{2} = 12\]
14 – 20: the mid – point will be \[\dfrac{{14 + 20}}{2} = \dfrac{{34}}{2} = 17\]
20 – 28: the mid – point will be \[\dfrac{{20 + 28}}{2} = \dfrac{{48}}{2} = 24\]
28 – 38: the mid – point will be \[\dfrac{{28 + 38}}{2} = \dfrac{{66}}{2} = 33\]
38 – 40: the mid – point will be \[\dfrac{{38 + 40}}{2} = \dfrac{{78}}{2} = 39\]
Now, let us create a new table:

Class - Interval	Mid – value $({x_i})$	Frequency $({f_i})$	${f_i}{x_i}$
0 – 6	3	11	33
6 – 10	8	10	80
10 – 14	12	7	84
14 – 20	17	4	68
20 – 28	24	4	96
28 – 38	33	3	99
38 – 40	39	1	39
	Total	40	499

Now, we also have formula of mean which is given by:
$\bar x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$
Now, putting in the values as per our question, we will get:-
$ \Rightarrow \bar x = \dfrac{{499}}{{40}}$
Simplifying the RHS of the above expression will lead us to:-
$ \Rightarrow \bar x = 12.475$
Hence, the mean number of days a student was absent is 12.475

$\therefore $ The required answer is 12.475

Note: The students must know that mean refers to the average. Like if we take the example of the given question, if we pick out a random student, the average numbers of days he / he would have been absent will be 12.475
We can find mean by assumed mean method as well but for that we require the class intervals to be equally spaced but here we do not have that. And, as well this method is very easy here because the calculations are pretty small. Therefore, we have used the normal mean formula in this data.