CBSE · Class 10 · Mathematics · StatisticsThe 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 |
Step-by-Step Solution
To find the values of $x$ and $y$, we construct the cumulative frequency table:
| Class Interval | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0-100 | 2 | 2 |
| 100-200 | 5 | 7 |
| 200-300 | $x$ | $7 + x$ |
| 300-400 | 12 | $19 + x$ |
| 400-500 | 17 | $36 + x$ |
| 500-600 | 20 | $56 + x$ |
| 600-700 | $y$ | $56 + x + y$ |
| 700-800 | 9 | $65 + x + y$ |
| 800-900 | 7 | $72 + x + y$ |
| 900-1000 | 4 | $76 + x + y$ |
| \nGiven, total frequency $N = 100$\nSo, $76 + x + y = 100$ | ||
| $x + y = 24$ -----(1) | ||
| \nGiven, Median = 525, which lies in the class interval 500-600.\nTherefore, |
- Lower limit of median class ($l$) = 500
- Frequency of median class ($f$) = 20
- Cumulative frequency of preceding class ($cf$) = $36 + x$
- Class size ($h$) = 100 \nThe median formula is: $$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$$ \nSubstitute the values: $$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$$ $$525 - 500 = (50 - 36 - x) \times 5$$ $$25 = (14 - x) \times 5$$ $$5 = 14 - x$$ $$x = 14 - 5 = 9$$ \nSubstitute $x = 9$ in equation (1): $$9 + y = 24$$ $$y = 24 - 9 = 15$$ \nThus, the values are $x = 9$ and $y = 15$.
💡 Study Guide: This question tests core syllabus concepts from Statistics. For formulas, key summaries, and mock exam reference guides, read the full Statistics Revision Notes.