If the mean marks of $100$ students is $54$, find the values of $p$ and $q$:
| Marks | $0 - 20$ | $20 - 40$ | $40 - 60$ | $60 - 80$ | $80 - 100$ |
|---|---|---|---|---|---|
| Number of students | $16$ | $p$ | $24$ | $26$ | $q$ |
| Marks (Class Interval) | Mid Value $\left(x_i\right)$ | Number of students $\left(\text{Frequency } f_i\right)$ | $f_i x_i$ |
|---|---|---|---|
| $0 - 20$ | $10$ | $16$ | $160$ |
| $20 - 40$ | $30$ | $p$ | $30 p$ |
| $40 - 60$ | $50$ | $24$ | $1200$ |
| $60 - 80$ | $70$ | $26$ | $1820$ |
| $80 - 100$ | $90$ | $q$ | $90 q$ |
Total number of students $= \Sigma f_i = 66 + p + q = 100$ $\;\;\;$ [given]
$\implies$ $p + q = 34$ $\implies$ $p = 34 - q$ $\;\;\; \cdots \; (1)$
$\begin{aligned} \Sigma f_i x_i & = 3180 + 30p + 90q \\\\ & = 3180 + 30 \left(p + 3q\right) \\\\ & = 3180 + 30 \left(34 - q + 3q\right) \;\;\; \left[\text{by equation (1)}\right] \\\\ & = 3180 + 30 \left(34 + 2q\right) \\\\ & = 3180 + 60 \left(17 + q\right) \end{aligned}$
$\text{Mean} = \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{3180 + 60 \left(17 + q\right)}{100} = 54 \;\;\;$ [given]
i.e. $\;$ $3180 + 60 \left(17 + q\right) = 5400$
i.e. $\;$ $60 \left(17 + q\right) = 2220$
i.e. $\;$ $17 + q = 37$ $\implies$ $q = 20$
Substituting the value of $q$ in equation $(1)$ gives
$p = 34 - 20 = 14$
$\therefore \;$ $p = 14, \; q = 20$