The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of the data:
| No. of Mangoes | No. of Trees (f) |
|---|---|
| 50 – 100 | 33 |
| 100 – 150 | 30 |
| 150 – 200 | 90 |
| 200 – 250 | 80 |
| 250 – 300 | 17 |
Step 1: Find the cumulative frequency (c.f.).
| Class Interval | Frequency (f) | Cumulative Frequency (c.f.) |
|---|---|---|
| 50 - 100 | 33 | 33 |
| 100 - 150 | 30 | 63 |
| 150 - 200 | 90 | 153 |
| 200 - 250 | 80 | 233 |
| 250 - 300 | 17 | 250 |
Step 2: Find total frequency.
N = 250
Step 3: Find median class.
N/2 = 250 / 2 = 125
The c.f. just greater than 125 is 153. Hence, the median class is 150 - 200.
Step 4: Apply the median formula.
Median = L + ((N/2 - c.f.) / f) × h
Where:
L = 150, N = 250, c.f. = 63, f = 90, h = 50
Substitute the values:
Median = 150 + ((125 - 63) / 90) × 50
= 150 + (62 / 90 × 50)
= 150 + 34.44 = 184.44
Step 5: Conclusion.
The median number of mangoes produced per tree is approximately 184.4.
Final Answer: Median = 184.4
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.