The median of the following frequency distribution is $50$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ Frequency $5$ $x$ $20$ $y$ $2$
| Class | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ |
| Frequency | $5$ | $x$ | $20$ | $y$ | $2$ |
(A) Given total frequency $N = 60$.
Sum of frequencies: $5 + x + 20 + y + 2 = 60 \implies x + y + 27 = 60 \implies x + y = 33$ (Equation $1$).
Since the median is $50$,the median class is $40-60$. Here,$l = 40$,$h = 20$,$f = 20$,$cf = 5 + x$,and $N/2 = 30$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$50 = 40 + \left( \frac{30 - (5 + x)}{20} \right) \times 20$.
$50 = 40 + (25 - x)$.
$50 = 65 - x \implies x = 15$.
Substituting $x = 15$ in Equation $1$: $15 + y = 33 \implies y = 18$.
Thus,$x = 15$ and $y = 18$.
Sum of frequencies: $5 + x + 20 + y + 2 = 60 \implies x + y + 27 = 60 \implies x + y = 33$ (Equation $1$).
Since the median is $50$,the median class is $40-60$. Here,$l = 40$,$h = 20$,$f = 20$,$cf = 5 + x$,and $N/2 = 30$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$50 = 40 + \left( \frac{30 - (5 + x)}{20} \right) \times 20$.
$50 = 40 + (25 - x)$.
$50 = 65 - x \implies x = 15$.
Substituting $x = 15$ in Equation $1$: $15 + y = 33 \implies y = 18$.
Thus,$x = 15$ and $y = 18$.
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