Q34. The population of lions was noted in different regions
across the world in the following table:
| Number of lions |
Number of regions |
| 0 - 100 |
2 |
| 100 - 200 |
5 |
| 200 - 300 |
9 |
| 300 - 400 |
12 |
| 400 - 500 |
X |
| 500 - 600 |
20 |
| 600 - 700 |
15 |
| 700 - 800 |
9 |
| 800 - 900 |
y |
| 900 - 1000 |
2 |
| Total |
100 |
If the median of the given data is 525, find the values of \(x\) and \(y\).
Total \(N = 100\). Median = 525.
Median lies in class 500-600. So \(l = 500, h = 100, f = 20\).
Cumulative Frequency (CF) preceding class = \(2+5+9+12+x = 28+x\).
Formula: \(\text{Median} = l + \left(\frac{\frac{N}{2} - CF}{f}\right) \times h\)
\(525 = 500 + \left(\frac{50 - (28+x)}{20}\right) \times 100\)
\(25 = (22-x) \times 5\)
\(5 = 22 - x \Rightarrow x = 17\).
Sum of frequencies = 100.
\(74 + x + y = 100 \Rightarrow 74 + 17 + y = 100 \Rightarrow y = 9\).
Answer: \(x=17, y=9\).