The median of the following frequency distribution is $28.5$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.
Class$0-10$$10-20$$20-30$$30-40$$40-50$$50-60$
Frequency$5$$x$$20$$15$$y$$5$

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(X=8, Y=7)
ClassFrequency $(f)$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 $y = 7$.

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The mean of the following data is $55.5$ and the total frequency is $20$. Find the missing frequencies $x$ and $y$.
Class$30-40$$40-50$$50-60$$60-70$$70-80$
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