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

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

Class11 – 1313 – 1515 – 1717 – 1919 – 2121 – 2323 – 25
Frequency36913f54
Explanation

Solution

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 11+132=12\dfrac{{11 + 13}}{2} = 12
13 – 15: the mid – point will be 13+152=14\dfrac{{13 + 15}}{2} = 14
15 – 17: the mid – point will be 15+172=16\dfrac{{15 + 17}}{2} = 16
17 – 19: the mid – point will be 17+192=18\dfrac{{17 + 19}}{2} = 18
19 – 21: the mid – point will be 19+212=20\dfrac{{19 + 21}}{2} = 20
21 – 23: the mid – point will be 21+232=22\dfrac{{21 + 23}}{2} = 22
23 – 25: the mid – point will be 23+252=24\dfrac{{23 + 25}}{2} = 24
So, then we will get the following table with us:-

ClassMid values (xi)({x_i})Frequency (fi)({f_i})fixi{f_i}{x_i}
11 – 1312336
13 – 1514684
15 – 17169144
17 – 191813234
19 – 2120f20f
21 – 23225110
23 – 2524496
TOTAL40 + f704 + 20f

Now, we know that we have a formula for mean which is given by:-
xˉ=fixifi\Rightarrow \bar x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}
Now, since we calculated in the table that f=40+f\sum {f = 40 + f} (Because 3 + 6 + 9 + 13 + f + 5 + 4 = 40 + f) and fixi=704+20f\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:-
xˉ=704+20f40+f\Rightarrow \bar x = \dfrac{{704 + 20f}}{{40 + f}}
Since, we are given in the question that the mean is 18.
18=704+20f40+f\therefore 18 = \dfrac{{704 + 20f}}{{40 + f}}
Cross – multiplying the above expression, we will get:-
18(40+f)=704+20f\Rightarrow 18(40 + f) = 704 + 20f
Simplifying the LHS of the above expression, we will get:-
720+18f=704+20f\Rightarrow 720 + 18f = 704 + 20f
Rearranging the terms to get:-
20f18f=720704\Rightarrow 20f - 18f = 720 - 704
Simplifying it:-
2f=16\Rightarrow 2f = 16
Simplifying it further:-
f=8\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.