Q.1 (b) Solution
Given: Median = 86
Class | Frequency |
|---|---|
45–50 | 2 |
50–60 | 1 |
60–70 | 6 |
70–80 | 6 |
80–90 | x |
90–100 | 12 |
100–110 | 5 |
Step 1: Total Frequency
Total frequency
N = 2 + 1 + 6 + 6 + x + 12 + 5
N = 32 + x
N/2 = (32 + x) / 2
Step 2: Cumulative Frequency
C.F. up to:
- 45–50 = 2
- 50–60 = 3
- 60–70 = 9
- 70–80 = 15
- 80–90 = 15 + x ← Median class
- 90–100 = 27 + x
- 100–110 = 32 + x
Median class = 80–90 (because median value 86 lies here)
Step 3: Median Formula
Median = l + [(N/2 – c.f.) / f] × h
Here,
l = 80
c.f. = 15
f = x
h = 10
So,
86 = 80 + [( (32 + x)/2 – 15 ) / x ] × 10
Step 4: Solving
Subtract 80 from both sides:
6 = [ ( (32 + x)/2 – 15 ) / x ] × 10
Divide both sides by 10:
0.6 = [ (32 + x)/2 – 15 ] / x
Simplify numerator:
(32 + x)/2 – 15
= (32 + x – 30) / 2
= (x + 2) / 2
So,
0.6 = (x + 2) / (2x)
Cross-multiply:
0.6 × 2x = x + 2
1.2x = x + 2
1.2x – x = 2
0.2x = 2
x = 2 / 0.2
x = 10
Final Answer
Hence, the missing frequency = 10
Missing frequency
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