Question 8
The mean of the following frequency distribution is 50 but the frequencies $f_{1}$ and $f_{2}$ in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120.
$\begin{matrix} C l a s s & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 F r e q u e n c y & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}$

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Solution

First, we calculate the class mark of given data
$\begin{matrix} C l a s s & F r e q u e n c y (f_{i}) & C l a s s m a r k s (x_{i}) & u_{i} = \frac{x_{i} - a}{h} & f_{i} u_{i} 0 - 20 & 17 & 10 & - 2 & - 34 20 - 40 & f_{1} & 30 & - 1 & - f_{1} 40 - 60 & 32 & a = 50 & 0 & 0 60 - 80 & f_{2} & 70 & 1 & f_{2} 80 - 100 & 19 & 90 & 2 & 38 \sum f_{i} = 68 + f_{1} + f_{2} & \sum f_{i} u_{i} = 4 + f_{2} - f_{1} \end{matrix}$
Given that, sum of all frequencies=120
$\Rightarrow \sum f_{i} = 68 + f_{1} + f_{2} = 120 \Rightarrow f_{1} + f_{2} = 52 (i)$
Here, (assumed mean) a=50
and (class width) h=20
By step deviation method,
$M e a n = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h \Rightarrow 50 = 50 + \frac{(4 + f_{2} - f_{1})}{120} \times 20 \Rightarrow 4 + f_{2} + f_{1} = 0 \Rightarrow - f_{2} + f_{1} = 4 . . . . (i i) O n a d d i n g E q s . (i) a n d (i i), w e g e t 2 f_{1} = 56 \Rightarrow f_{1} = 28$
Put the value of $f_{1}$ in Eq.(i), we get
$f_{2} = 52 - 28 \Rightarrow f_{2} = 24$
Hence, $f_{1} = 28 a n d f_{2} = 24.$

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Question 8
The mean of the following frequency distribution is 50 but the frequencies $f_{1}$ and $f_{2}$ in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120.
$\begin{matrix} C l a s s & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 F r e q u e n c y & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}$