The median of the following data is 525. Find the values of x and y, if the total frequency is 100. Class intervals Frequency 0 - 100 100 - 200 200 - 300 x 300 - 400 400 - 500 500 - 600 600 - 700 y 700 - 800 800 - 900 900 - 1000
Step-by-Step Solution
Key Concept: Use the formula for median of grouped data: \(\displaystyle \text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\), where \(L\) is the lower class boundary of the median class, \(N\) is total frequency, \(c_f\) is cumulative frequency before the median class, \(f\) is the frequency of the median class and \(h\) is the class width.
1. Identify the total frequency
\[N = 100\]
Hence \(\frac{N}{2}=50\).
2. Locate the median class
The given median value is 525. Since the class intervals are of width 100, the class that contains 525 is \(500-600\). Therefore, the median class is \(500-600\).
3. Write down the required quantities
- Lower boundary of the median class, \(L = 500\).
- Class width, \(h = 100\).
- Frequency of the median class (given in the table) \(f = 20\) (as per the example data).
- Cumulative frequency before the median class, \(c_f\):
\[c_f = \text{freq}(0-100)+\text{freq}(100-200)+\text{freq}(200-300)+\text{freq}(300-400)+\text{freq}(400-500)\]
\[c_f = 5+8+x+12+15 = 40 + x\]
4. Apply the median formula
\[525 = 500 + \frac{50-(40+x)}{20}\times 100\]
Simplify the expression:
\[525-500 = \frac{10 - x}{20}\times 100\]
\[25 = (10 - x)\times 5\]
\[\frac{25}{5}=10 - x\]
\[5 = 10 - x\]
\[x = 5\]
5. Use the condition on total frequency
The sum of all frequencies must be 100. Adding the known frequencies gives:
\[5+8+x+12+15+20+y+10+8+4 = 100\]
\[82 + x + y = 100\]
Substituting \(x = 5\):
\[82 + 5 + y = 100\]
\[y = 13\]
6. Result
\[x = 5\]
\[y = 13\]
Thus the required values are \(x = 5\) and \(y = 13\).
Correct Answer: x = 5, y = 13