(X=8, Y=7)
| Class | Frequency $(f)$ | Cumulative frequency $(cf)$ |
| $0-10$ | $5$ | $5$ |
| $10-20$ | $x$ | $5+x$ |
| $20-30$ | $20$ | $25+x$ |
| $30-40$ | $15$ | $40+x$ |
| $40-50$ | $y$ | $40+x+y$ |
| $50-60$ | $5$ | $45+x+y$ |
Given total frequency $n = 60$.
Therefore,$45 + x + y = 60 \implies x + y = 15$ (Equation $1$).
Since $n = 60$,$\frac{n}{2} = 30$.
The median $28.5$ lies in the class $20-30$. Thus,the median class is $20-30$.
Here,$l = 20$,$cf = 5 + x$,$f = 20$,and class size $h = 10$.
Using the median formula: $Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$
$8.5 = \frac{25 - x}{2}$
$17 = 25 - x \implies x = 8$.
Substituting $x = 8$ in Equation $1$: $8 + y = 15 \implies y = 7$.
The missing frequencies are $x = 8$ and $y = 7$.