The median of the following frequency distribution is $525$ and the total frequency is $100$. Find the missing frequencies $x$ and $y$.
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ $600-700$ $700-800$ $800-900$ $900-1000$ Frequency $2$ $5$ $x$ $12$ $17$ $20$ $y$ $9$ $7$ $4$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ | $600-700$ | $700-800$ | $800-900$ | $900-1000$ |
| Frequency | $2$ | $5$ | $x$ | $12$ | $17$ | $20$ | $y$ | $9$ | $7$ | $4$ |
(A) Given,total frequency $N = 100$.
Sum of frequencies: $2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$
$76 + x + y = 100 \implies x + y = 24$ --- $(1)$
Since the median is $525$,the median class is $500-600$.
Here,$l = 500$,$f = 20$,$cf = (2 + 5 + x + 12 + 17) = 36 + x$,$h = 100$,and $N/2 = 50$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 9$
Substituting $x = 9$ in $(1)$: $9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.
Sum of frequencies: $2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$
$76 + x + y = 100 \implies x + y = 24$ --- $(1)$
Since the median is $525$,the median class is $500-600$.
Here,$l = 500$,$f = 20$,$cf = (2 + 5 + x + 12 + 17) = 36 + x$,$h = 100$,and $N/2 = 50$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 9$
Substituting $x = 9$ in $(1)$: $9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.
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