Question

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	10
800–900	y
900–1000	2
Total	100

If the median of the given data is 525, find the values of x and y.

Hint:

Use the formula for median in grouped data: \[ \text{Median} = l + \frac{\frac{N}{2} - CF}{f} \cdot h \] Fill missing values using total frequency condition.

Answer 1

Given Data:

Number of lions	Number of regions (Frequency \(f\))
0–100	2
100–200	5
200–300	9
300–400	12
400–500	x
500–600	20
600–700	15
700–800	10
800–900	y
900–1000	2
Total	100

Median = 525
Total frequency \(n = 100\)
Median class is the class where cumulative frequency just exceeds \(\frac{n}{2} = 50\).

Step 1: Express \(y\) in terms of \(x\) using total frequency
Sum of frequencies = 100, so:
\[ 2 + 5 + 9 + 12 + x + 20 + 15 + 10 + y + 2 = 100 \] \[ 75 + x + y = 100 \Rightarrow x + y = 25 \quad ...(1) \]

Step 2: Find cumulative frequency up to each class
 

Class	Frequency (f)	Cumulative Frequency (CF)
0–100	2	2
100–200	5	7
200–300	9	16
300–400	12	28
400–500	\(x\)	\(28 + x\)
500–600	20	\(48 + x\)
600–700	15	\(63 + x\)
700–800	10	\(73 + x\)
800–900	\(y\)	\(73 + x + y\)
900–1000	2	\(75 + x + y = 100\)

Step 3: Identify median class
Median lies in the class where cumulative frequency \(\geq 50\). From above:
- CF up to 500–600 class = \(48 + x\)
- CF up to 600–700 class = \(63 + x\)
Since \(48 + x < 50 \leq 63 + x\), median class is 600–700.

Step 4: Use median formula
\[ \text{Median} = l + \left( \frac{\frac{n}{2} - F}{f_m} \right) \times h \] Where:
- \(l = 600\) (lower limit of median class)
- \(n = 100\)
- \(F = 48 + x\) (CF before median class)
- \(f_m = 15\) (frequency of median class)
- \(h = 100\) (class width)
- Median = 525

Substitute values:
\[ 525 = 600 + \left( \frac{50 - (48 + x)}{15} \right) \times 100 \] \[ 525 - 600 = \frac{50 - 48 - x}{15} \times 100 \] \[ -75 = \frac{2 - x}{15} \times 100 \] \[ -75 = \frac{100(2 - x)}{15} \] Multiply both sides by 15: \[ -75 \times 15 = 100(2 - x) \] \[ -1125 = 200 - 100x \] \[ -1125 - 200 = -100x \] \[ -1325 = -100x \] \[ x = \frac{1325}{100} = 13.25 \] Since this is not an integer, check median class again:

Re-evaluate median class with median = 525, which lies in 500–600 interval
Use median class 500–600:
- \(l = 500\)
- \(f_m = 20\)
- \(F = 28 + x\)
Using median formula:
\[ 525 = 500 + \left( \frac{50 - (28 + x)}{20} \right) \times 100 \] \[ 25 = \frac{22 - x}{20} \times 100 \] \[ 25 = 5 (22 - x) \] \[ 25 = 110 - 5x \] \[ 5x = 110 - 25 = 85 \] \[ x = 17 \]

Step 5: Find \(y\) using \(x + y = 25\)
\[ 17 + y = 25 \Rightarrow y = 8 \]

Final Answer:
\[ x = 17, \quad y = 8 \]