Question

Compute median of the following data :

Hint:

Class interval \(= [Mid-value - h/2, Mid-value + h/2]\). Correctly forming the table is 90\% of the work.

Answer 1

Mid–values: 115, 125, 135, 145, 155, 165, 175
Frequencies: 12, 15, 20, 16, 10, 16, 11

Total frequency:
12 + 15 + 20 + 16 + 10 + 16 + 11 = 100
So N = 100 → N/2 = 50

Step 1: Compute cumulative frequencies
115 → 12
125 → 12 + 15 = 27
135 → 27 + 20 = 47
145 → 47 + 16 = 63
155 → 63 + 10 = 73
165 → 73 + 16 = 89
175 → 89 + 11 = 100

N/2 = 50 lies in the class whose cumulative frequency first exceeds 50.
CF just before 50 = 47 (for mid–value 135).
Next class (median class) has mid–value = 145 with frequency f = 16.

Step 2: Determine class boundaries
Mid-values differ by 10, so class width h = 10.
Class corresponding to mid = 145 is 140 – 150.
Therefore L = 140.

Step 3: Apply median formula
\[ \text{Median} = L + \left(\frac{\frac{N}{2} - CF}{f}\right)h \] Substitute values:
\[ \text{Median} = 140 + \left(\frac{50 - 47}{16}\right)10 \] \[ = 140 + \frac{3}{16} \times 10 \] \[ = 140 + 1.875 \] \[ = 141.875 \]

Final Answer:
The median of the given data is \[ \boxed{141.875} \]