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Question: Find the missing frequencies in the following frequency distribution if it is known that the mean of...

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.

X:1030507090
F:17f1{f_1}32f2{f_2}19
Explanation

Solution

To find the missing values of frequencies we will generate two equations in terms of frequency. One of the equations by applying the condition of the sum of the frequency and the other by applying the mean of the grouped data. We will get two equations and two variables just solve the two generated equations and then find the value of missing frequencies.

Complete Step by Step Solution:
Given the sum of frequency =120
So, adding all the given frequencies we will get,
17+f1+32+f2+19=120\Rightarrow 17 + {f_1} + 32 + {f_2} + 19 = 120
Add the terms on the left side,
f1+f2+68=120\Rightarrow {f_1} + {f_2} + 68 = 120
Move f2{f_2} and constant part on the right side and subtract,
f1=52f2\Rightarrow {f_1} = 52 - {f_2} ….. (1)
The mean of the grouped data is given as 50 so we will calculate the mean and equate it with the given value of mean such that we will get another equation.
We know mean is defined as,
Mean =xifixi = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}
On putting the given values in the above formula, we get
50=(10×17)+(30×f1)+(50×32)+(70×f2)+(90×19)120\Rightarrow 50 = \dfrac{{\left( {10 \times 17} \right) + \left( {30 \times {f_1}} \right) + \left( {50 \times 32} \right) + \left( {70 \times {f_2}} \right) + \left( {90 \times 19} \right)}}{{120}}
Multiply the terms in the numerator,
50=170+30f1+1600+70f2+1710120\Rightarrow 50 = \dfrac{{170 + 30{f_1} + 1600 + 70{f_2} + 1710}}{{120}}​​
Cross-multiply the terms and add the terms,
3480+30f1+70f2=6000\Rightarrow 3480 + 30{f_1} + 70{f_2} = 6000
Move constant part on the right side and subtract,
30f1+70f2=2520\Rightarrow 30{f_1} + 70{f_2} = 2520
Divide both sides by 10,
3f1+7f2=252\Rightarrow 3{f_1} + 7{f_2} = 252
Substitute the values from equation (1),
3(52f2)+7f2=252\Rightarrow 3\left( {52 - {f_2}} \right) + 7{f_2} = 252
Simplify the terms,
1563f2+7f2=252\Rightarrow 156 - 3{f_2} + 7{f_2} = 252
Add the like terms and move the constant to the right side and subtract,
4f2=96\Rightarrow 4{f_2} = 96
Divide both sides by 4,
f2=24\Rightarrow {f_2} = 24
Substitute the value of f2{f_2} in equation (1),
f1=5224\Rightarrow {f_1} = 52 - 24
Simplify the terms,
f1=28\Rightarrow {f_1} = 28

Hence the value of f1{f_1} and f2{f_2} are 28 and 24.

Note: This is solved by a simple mean calculation method. It can be solved by assuming a mean method also. Generally, there must at least be two equations to find two unknowns.
In the mean formula, while computing fx\sum {fx} , don’t take the sum of ff and xx separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data.