Question

The mean of the following table is 57.6.But the frequencies \[{f_1}\] and \[{f_2}\] are missing. Find the value of missing frequencies \[{f_1}\] and \[{f_2}\]

Class interval	\[0 - 20\]	\[20 - 40\]	\[40 - 60\]	\[60 - 80\]	\[80 - 100\]	\[100 - 120\]	Total
Frequency	7	\[{f_1}\]	12	\[{f_2}\]	8	5	50

Answer 1

Hint: Here, we must know that the mean of a number of observations is the sum of the values of all the observations divided by the total number of observations. As we know that the mean of any data is given by this formula
\[\overline x = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4}............................. + {x_n}}}{n}\]
Given class intervals and their frequencies would lead us to the calculation of mean using the standard formula where an additional term will add known as xi as follows:
\[\overline x = \dfrac{{\sum {fx} }}{{\sum f }}\]

Complete step-by-step solution:
From the given values and to apply the formula lets first draw a table for computing \[{f_i}\]\[{X_i}\] term easily:

Class interval	\[{f_i}\]	\[{X_i}\]	\[{f_i}\]\[{X_i}\]
\[0 - 20\]	7	10	70
\[20 - 40\]	\[{f_1}\]	30	30 \[{f_1}\]
\[40 - 60\]	12	50	600
\[60 - 80\]	\[{f_2}\]	70	70\[{f_2}\]
\[80 - 100\]	8	90	720
\[100 - 120\]	5	110	550
	50(Given)		Total $ = 1940 + 30{f_1} + 70{f_2}$

The Mean of data given is =\[\overline x = 57.6\]
Applying the standard formula we will get the required calculation as follows:
$\Rightarrow$\[\overline x = \dfrac{{(70 + 30{f_1} + 600 + 70{f_2} + 720 + 550)}}{{50}} = 57.6\]
Multiplying 50 on both sides we get;
$\Rightarrow$\[\overline x = (70 + 30{f_1} + 600 + 70{f_2} + 720 + 550) = 57.6 \times 50\]
Simplifying equation further we get;
$\Rightarrow$\[30{f_1} + 70{f_2} = 2880 - 1940 = 940\]
After calculation we get an equation in terms of \[{f_1}\] and \[{f_2}\] as such:
$\Rightarrow$\[30{f_1} + 70{f_2} = 940\; - - - - \left( 1 \right)\]
Also we must know that the sum of all the frequencies is given to be 50, thus we have following summation;
$\Rightarrow$\[7 + {f_1} + 12 + {f_2} + 8 + 5 = 50\]
On simplification of the above equation;
$\Rightarrow$\[32 + {f_1} + {f_2} = 50\]
Here we get another equation in terms of f1 and f2 as such:
$\Rightarrow$\[{f_1} + {f_2} = 18\; - - - - \left( 2 \right)\]
Using equation 1 and 2 we get two equations with two variables thus in order to solve them,
Multiply equation (2) with 30 and subtract equation (2) from equation (1) as shown:
$\Rightarrow$\[30{f_1} + 70{f_2} = 940 - - - - \left( 3 \right)\]
$\Rightarrow$\[30{f_1} + 30{f_2} = 540 - - - - \left( 4 \right)\]
After subtracting equation (4) from (3) we get the values as:
$\Rightarrow$\[40{f_2} = 400\]
$\Rightarrow$\[{f_2} = \dfrac{{400}}{{40}} = 10\]
Put \[{f_2}\] in equation (2) to get the following result,
$\Rightarrow$\[{f_1} + {f_2} = 18\]
Putting \[{f_2} = 10\]
$\Rightarrow$\[{f_1} + 10 = 18\]
$\Rightarrow$\[{f_1} = 8\]

Therefore the value of $f_1$ is 8 and the value of $f_2$ is 10.

Note: Other methods like harmonic mean and geometric mean are also applicable. It is the arithmetic mean method in which class marks would serve as representative of the whole class and are represented by \[{X_i}\].