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Q.
If the mean of the following table is 56.25 , but the frequencies $f_1$ and $f_2$ are missing, then find the missing frequencies.
Statistics
Solution:
$\Sigma f=100 $
$\therefore f_1+f_2=40$
Mid values of class are 7.5, 22.5, 37.5, 52.5, 67.5 and 82.5
$\therefore \operatorname{Mean}=\Sigma f x / \Sigma f$
$[(7.5)(5)+(22.5)(10)+(37.5)(f_1)+(52.5)(25)+(67.5).$ $.f_2+(82.5) 20] / 100$
$=56.25$
$\therefore 3225+37.5 f_1+67.5 f_2=5625$
$\Rightarrow 37.5 f_1+67.5 f_2=2400$
Solving (1) and (2), we get $f_1=10$ and $f_2=30$
