The frequency distribution of marks obtained by $400$ students is given below. If the mean of the frequency distribution is $41.2$,find the missing frequencies $x$ and $y$.

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Number of students	$26$	$26$	$x$	$110$	$84$	$y$	$36$	$32$

(N/A) The total number of students is $N = 400$.
Therefore,$26 + 26 + x + 110 + 84 + y + 36 + 32 = 400$.
$314 + x + y = 400 \implies x + y = 86$ ---$(i)$
The mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 41.2$.
The class marks $(x_i)$ are: $5, 15, 25, 35, 45, 55, 65, 75$.
The sum of $f_i x_i$ is: $(26 \times 5) + (26 \times 15) + (x \times 25) + (110 \times 35) + (84 \times 45) + (y \times 55) + (36 \times 65) + (32 \times 75)$.
$= 130 + 390 + 25x + 3850 + 3780 + 55y + 2340 + 2400 = 12890 + 25x + 55y$.
Thus,$\frac{12890 + 25x + 55y}{400} = 41.2$.
$12890 + 25x + 55y = 16480 \implies 25x + 55y = 3590$.
Dividing by $5$,we get $5x + 11y = 718$ ---(ii).
From equation $(i)$,$x = 86 - y$. Substituting this into $(ii)$:
$5(86 - y) + 11y = 718 \implies 430 - 5y + 11y = 718$.
$6y = 288 \implies y = 48$.
Therefore,$x = 86 - 48 = 38$.
Answer: $x = 38, y = 48$.