Question

The mean of the following distribution is 18. Find the frequency f of the class 19 – 21.

Class	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Frequency	3	6	9	13	f	5	4

Answer 1

Hint: We will first recreate the table using the data given above in the table with adding two more columns or rows of data with the mid values and multiplication of both f and x. Then, we will use the formula for mean and thus get the answer.

Complete step-by-step answer:
Let us first of all recreate the table using the data given in the table above and adding two more columns with the mid values and with the multiplication of both the data given to us already in the question.
Let us first find the mid – points of all the intervals.
11 – 13: the mid – point will be \[\dfrac{{11 + 13}}{2} = 12\]
13 – 15: the mid – point will be \[\dfrac{{13 + 15}}{2} = 14\]
15 – 17: the mid – point will be \[\dfrac{{15 + 17}}{2} = 16\]
17 – 19: the mid – point will be \[\dfrac{{17 + 19}}{2} = 18\]
19 – 21: the mid – point will be \[\dfrac{{19 + 21}}{2} = 20\]
21 – 23: the mid – point will be \[\dfrac{{21 + 23}}{2} = 22\]
23 – 25: the mid – point will be \[\dfrac{{23 + 25}}{2} = 24\]
So, then we will get the following table with us:-

Class	Mid values $({x_i})$	Frequency $({f_i})$	${f_i}{x_i}$
11 – 13	12	3	36
13 – 15	14	6	84
15 – 17	16	9	144
17 – 19	18	13	234
19 – 21	20	f	20f
21 – 23	22	5	110
23 – 25	24	4	96
	TOTAL	40 + f	704 + 20f

Now, we know that we have a formula for mean which is given by:-
\[ \Rightarrow \bar x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\]
Now, since we calculated in the table that $\sum {f = 40 + f} $ (Because 3 + 6 + 9 + 13 + f + 5 + 4 = 40 + f) and \[\sum {{f_i}{x_i}} = 704 + 20f\] (Because 36 + 84 + 144 + 234 + 20f + 110 + 96 = 704 + 20f). Therefore, let us put these values in the formula mentioned above. We will get the following expression:-
\[ \Rightarrow \bar x = \dfrac{{704 + 20f}}{{40 + f}}\]
Since, we are given in the question that the mean is 18.
\[\therefore 18 = \dfrac{{704 + 20f}}{{40 + f}}\]
Cross – multiplying the above expression, we will get:-
\[ \Rightarrow 18(40 + f) = 704 + 20f\]
Simplifying the LHS of the above expression, we will get:-
\[ \Rightarrow 720 + 18f = 704 + 20f\]
Rearranging the terms to get:-
\[ \Rightarrow 20f - 18f = 720 - 704\]
Simplifying it:-
\[ \Rightarrow 2f = 16\]
Simplifying it further:-
\[ \Rightarrow f = 8\]

$\therefore $ The required value of f is 8.

Note: The students must know that mean refers to the average.
The students must note that we did not use any different method than regular in this question because we did not require it, that would have complicated things more than making them easy. We must use our brains to see if we need to modify the question to solve or it is not required.
The students must note that we need as many equations as many unknown variables we have got in the equation. Like here, we got one unknown variable f and for that, we created one equation using the given mean and thus we found the answer.