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

The mean of the following table is 57.6.But the frequencies f1{f_1} and f2{f_2} are missing. Find the value of missing frequencies f1{f_1} and f2{f_2}

Class interval0−200 - 2020−4020 - 4040−6040 - 6060−8060 - 8080−10080 - 100100−120100 - 120Total
Frequency7f1{f_1}12f2{f_2}8550
Explanation

Solution

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
x‾=x1+x2+x3+x4.............................+xnn\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:
x‾=∑fx∑f\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 intervalfi{f_i}Xi{X_i}{f_i}$$$${X_i}
0−200 - 2071070
20−4020 - 40f1{f_1}3030 f1{f_1}
40−6040 - 601250600
60−8060 - 80f2{f_2}7070f2{f_2}
80−10080 - 100890720
100−120100 - 1205110550
50(Given)Total =1940+30f1+70f2 = 1940 + 30{f_1} + 70{f_2}

The Mean of data given is =x‾=57.6\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$$ Simplifying equation further we get; $\Rightarrow30{f_1} + 70{f_2} = 2880 - 1940 = 940Aftercalculationwegetanequationintermsof After calculation we get an equation in terms of{f_1}andand{f_2} as such: $\Rightarrow$$$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; \Rightarrow32 + {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: \Rightarrow40{f_2} = 400$$ $\Rightarrow{f_2} = \dfrac{{400}}{{40}} = 10Put Put{f_2} in equation (2) to get the following result, $\Rightarrow$$${f_1} + {f_2} = 18
Putting f2=10{f_2} = 10
\Rightarrow$$${f_1} + 10 = 18$$ \Rightarrow$f1=8{f_1} = 8

Therefore the value of f1f_1 is 8 and the value of f2f_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 Xi{X_i}.