Statistics

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

$p = 34 - 20 = 14$

$\therefore \;$ $p = 14, \; q = 20$