The abdomen length (in millimeters) was measured in 15 male fruit flies, and the following data were obtained:
1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3 and 2.2.
Q.53 The value of standard deviation (SD) will be
Raw Data (15 Male Fruit Flies)
Abdomen lengths (mm): 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3, 2.2
Range: 1.6 – 2.4 mm | n = 15 samples
Step-by-Step Standard Deviation Calculation
Step 1: Calculate Mean (¯x)
¯x = (1.9 + 2.4 + 2.1 + 2.0 + 2.2 + 2.4 + 1.7 + 1.8 + 2.0 + 2.0 + 2.3 + 2.1 + 1.6 + 2.3 + 2.2) / 15
¯x = 30.7 / 15 = 2.047 mm
Step 2: Sample Standard Deviation Formula
s = √[Σ(xᵢ – ¯x)² / (n-1)]
Where n = 15, so denominator = 14 (sample data)
Step 3: Squared Deviations Summary
Σ(xᵢ – 2.047)² = 0.90
Variance (s²) = 0.90 / 14 = 0.0643
Standard Deviation (s) = √0.0643 = 0.254 mm ≈ 0.25 mm
Option Analysis Table
| Option | SD Value | Calculation Method | Result | Status |
|---|---|---|---|---|
| (A) | 0.061 | Population SD (÷15) | √(0.90/15) = 0.061 | ❌ Wrong formula for sample |
| (B) | 0.25 | Sample SD (÷14) | √(0.90/14) = 0.25 | ✅ CORRECT |
| (C) | 0.61 | ~2.4× actual SD | Overestimate | ❌ Too High |
| (D) | 0.85 | ~3.4× actual SD | Major overestimate | ❌ Too High |
Why Sample SD (Not Population)?
- Biological samples always use
n-1denominator - Unbiased estimation of population variance from sample
- Option A error: Uses population formula (÷15) instead of sample (÷14)
- Options C/D: No mathematical basis, pure distractors
Final Answer: Option (B) 0.25 mm
Master fruit fly abdomen length standard deviation calculations for biology exams!
Key Formula: s = √[Σ(xᵢ – ¯x)² / (n-1)] = 0.25 mm
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