Find the missing frequencies in the following frequency distribution whose mean is 34.
$\begin{array}{cccccccc}\text{x}& 10& 20& 30& 40& 50& 60& \text{Total}\\ \text{f}& 4& f_1& 8& f_2& 3& 4& 35\end{array}$
(3 Marks)

We know that,
Mean =$\frac{\sum xifi}{\sum fi}$
(0.5 Marks)

Mean = $\frac{(10\times 4+20\times f1+30\times 8+40\times f2+50\times 3+60\times 4)}{35}$
⇒$34=\frac{(40+20f_1+240+40f_2+150+240)}{35}$
⇒$34\times 35=670+20f_1+40f_2$

⇒$1190-670=20f_1+40f_2$
⇒$20f_1+40f_2=520$
⇒$20(f_1+2f_2)=520$
⇒$f_1+2f_2=\frac{520}{20}$

⇒$f_1+2f_2=26$

⇒$f_1=26-2f_2$ ......................(1)

(1 Mark)

Also, $4+f_1+8+f_2+3+4=35$
⇒ $19+f_1+f_2=35$
⇒ $f_1+f_2=35-19$
⇒$f_1+f_2=16$
⇒ $26-2f_2+f_2=16$
(from (1))
⇒ $26-f_2=16$
⇒ $26-16=f_2$
⇒ $f_2=10$ (0.5 Marks)
Putting the value of $f_2$ in (1), we get

$f_1=26-2\left(10\right)=6$
Hence, the value of $f_1$ and $f_2$ is 6 and 10, respectively.

(1 Mark)