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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Similar Questions
The annual profits earned by $30$ shops of a shopping complex in a locality give rise to the following distribution:
Profit (Rs in lakhs) Number of shops (frequency) More than or equal to $5$ $30$ More than or equal to $10$ $28$ More than or equal to $15$ $16$ More than or equal to $20$ $14$ More than or equal to $25$ $10$ More than or equal to $30$ $7$ More than or equal to $35$ $3$
Draw both ogives for the data above. Hence,obtain the median profit.
| Profit (Rs in lakhs) | Number of shops (frequency) |
| More than or equal to $5$ | $30$ |
| More than or equal to $10$ | $28$ |
| More than or equal to $15$ | $16$ |
| More than or equal to $20$ | $14$ |
| More than or equal to $25$ | $10$ |
| More than or equal to $30$ | $7$ |
| More than or equal to $35$ | $3$ |
Draw both ogives for the data above. Hence,obtain the median profit.
Medium
View SolutionFind the median of the following frequency distribution:
Class $65-85$ $85-105$ $105-125$ $125-145$ $145-165$ $165-185$ $185-205$ Frequency $4$ $5$ $13$ $20$ $14$ $8$ $4$
| Class | $65-85$ | $85-105$ | $105-125$ | $125-145$ | $145-165$ | $165-185$ | $185-205$ |
| Frequency | $4$ | $5$ | $13$ | $20$ | $14$ | $8$ | $4$ |
Medium
View SolutionThe median of the following data is $525$. Find the values of $x$ and $y$,if the total frequency is $100$.
Class interval Frequency $0-100$ $2$ $100-200$ $5$ $200-300$ $x$ $300-400$ $12$ $400-500$ $17$ $500-600$ $20$ $600-700$ $y$ $700-800$ $9$ $800-900$ $7$ $900-1000$ $4$
| Class interval | Frequency |
| $0-100$ | $2$ |
| $100-200$ | $5$ |
| $200-300$ | $x$ |
| $300-400$ | $12$ |
| $400-500$ | $17$ |
| $500-600$ | $20$ |
| $600-700$ | $y$ |
| $700-800$ | $9$ |
| $800-900$ | $7$ |
| $900-1000$ | $4$ |
Medium
View SolutionIs it correct to say that an ogive is a graphical representation of a frequency distribution? Give reason.
Easy
View SolutionThe median class of the following frequency distribution is ...........
Class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $6$ $10$ $5$ $6$ $4$ $2$ $2$
| Class | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
| Frequency | $6$ | $10$ | $5$ | $6$ | $4$ | $2$ | $2$ |
Easy
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