The median of the following frequency distribution is $49$ and the total frequency is $100$. Find the missing frequencies $x$ and $y$.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $2$ $6$ $8$ $x$ $20$ $18$ $y$ $14$
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
| Frequency | $2$ | $6$ | $8$ | $x$ | $20$ | $18$ | $y$ | $14$ |
(A) Given total frequency $N = 100$,so $2 + 6 + 8 + x + 20 + 18 + y + 14 = 100$.
$68 + x + y = 100 \implies x + y = 32$ (Equation $1$).
Since the median is $49$,the median class is $40-50$. Here $l = 40$,$f = 20$,$cf = 2 + 6 + 8 + x = 16 + x$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$49 = 40 + \left( \frac{50 - (16 + x)}{20} \right) \times 10$.
$9 = \frac{34 - x}{2} \implies 18 = 34 - x \implies x = 16$.
Substituting $x = 16$ in Equation $1$: $16 + y = 32 \implies y = 16$.
Thus,$x = 16$ and $y = 16$.
$68 + x + y = 100 \implies x + y = 32$ (Equation $1$).
Since the median is $49$,the median class is $40-50$. Here $l = 40$,$f = 20$,$cf = 2 + 6 + 8 + x = 16 + x$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$49 = 40 + \left( \frac{50 - (16 + x)}{20} \right) \times 10$.
$9 = \frac{34 - x}{2} \implies 18 = 34 - x \implies x = 16$.
Substituting $x = 16$ in Equation $1$: $16 + y = 32 \implies y = 16$.
Thus,$x = 16$ and $y = 16$.
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| Below $50$ | $75$ |
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