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 |

To find the values of $x$ and $y$, we need to construct the cumulative frequency table and use the given median and total frequency.

Step 1: Construct the Cumulative Frequency Table

Step 2: Use the Total Frequency Condition \nGiven that the total frequency $N = 100$.\nFrom the table, the sum of frequencies is $76 + x + y$.\nTherefore, $$76 + x + y = 100$$ $$x + y = 100 - 76$$ $$x + y = 24 \quad \text{--- (Equation 1)}$$

Step 3: Identify the Median Class \nGiven that the median is 525. This value lies in the class interval $500 - 600$.\nTherefore, the median class is $500 - 600$. \nFrom the median class:

• Lower limit ($l$) = 500
• Class size ($h$) = 100
• Frequency of the median class ($f$) = 20
• Cumulative frequency of the class preceding the median class ($cf$) = $36 + x$
• $\frac{N}{2} = \frac{100}{2} = 50$

Step 4: Apply the Median Formula \nThe formula for median is: $$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$$ \nSubstitute the known values into the formula: $$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$$ \nSimplify the equation: $$525 - 500 = \left( \frac{50 - 36 - x}{20} \right) \times 100$$ $$25 = (14 - x) \times 5$$ $$25 = 70 - 5x$$ $$5x = 70 - 25$$ $$5x = 45$$ $$x = \frac{45}{5} = 9$$

Step 5: Find the Value of $y$ \nSubstitute $x = 9$ into Equation 1: $$9 + y = 24$$ $$y = 24 - 9$$ $$y = 15$$

Conclusion:\nThe values are $x = 9$ and $y = 15$.