Find the missing frequencies in the following frequency distribution whose mean is 34.

We know that,

Mean $=\frac{\sum x_{i} f_{i}}{\sum f_{i}}$

For the following data:

Mean $=\frac{(10 \times 4)+\left(20 \times f_{1}\right)+(30 \times 8)+\left(40 \times f_{2}\right)+(50 \times 3)+(60 \times 4)}{35}$

$\Rightarrow 34=\frac{40+20 f_{1}+240+40 f_{2}+150+240}{35}$

$\Rightarrow 34(35)=670+20 f_{1}+40 f_{2}$

$\Rightarrow 1190-670=20 f_{1}+40 f_{2}$

$\Rightarrow 20 f_{1}+40 f_{2}=520$

$\Rightarrow 20\left(f_{1}+2 f_{2}\right)=520$

$\Rightarrow f_{1}+2 f_{2}=\frac{520}{20}$

$\Rightarrow f_{1}+2 f_{2}=26$

$\Rightarrow f_{1}=26-2 f_{2}$           $\cdots(1)$

Also, $4+f_{1}+8+f_{2}+3+4=35$

$\Rightarrow 19+f_{1}+f_{2}=35$

$\Rightarrow f_{1}+f_{2}=35-19$

$\Rightarrow f_{1}+f_{2}=16$

$\Rightarrow 26-2 f_{2}+f_{2}=16 \quad$ (from (1))

$\Rightarrow 26-f_{2}=16$

$\Rightarrow 26-16=f_{2}$

$\Rightarrow f_{2}=10$

Putting the value of f2 in (1), we get

$f_{1}=26-2(10)=6$

Hence, the value of f1 and  f2 is 6 and 10, respectively.