The population variance for abdomen lengths of 15 male fruit flies measures data dispersion from the mean, calculated using σ² = Σ(xi – μ)²/N formula. Correct answer is 0.061 matching option (C).
Data Summary
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 (n=15). First calculate population mean μ = Σxi/n = 31.0/15 = 2.067 mm.
Step-by-Step Calculation
Step 1: Mean (μ) = (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 = 2.0667 mm
Step 2: Deviations squared Σ(xi – μ)² = 0.92 total (detailed table below):
| Value |
Freq |
(x-μ) |
(x-μ)² |
| 1.6 |
1 |
-0.467 |
0.218 |
| 1.7 |
1 |
-0.367 |
0.134 |
| 1.8 |
1 |
-0.267 |
0.071 |
| 1.9 |
1 |
-0.167 |
0.028 |
| 2.0 |
3 |
-0.067 |
0.057 |
| 2.1 |
2 |
0.033 |
0.004 |
| 2.2 |
2 |
0.133 |
0.035 |
| 2.3 |
2 |
0.233 |
0.109 |
| 2.4 |
2 |
0.333 |
0.222 |
| Total |
15 |
|
0.918 |
Step 3: Variance = 0.918/15 = 0.0612 ≈ 0.061
Options Analysis
-
(A) 0.85: Wrong; possibly confused with range (2.4-1.6=0.8) or sample SD estimate
-
(B) 0.25: Wrong; might be (SD)² where SD≈0.5, but actual SD=0.247
-
(C) 0.061: Correct; matches σ²=0.0612 from population formula N=15 denominator
-
(D) 0.08: Wrong; sample variance using n-1=14: 0.918/14=0.0656≠0.08
Common Errors Explained
Population vs Sample Variance: Population uses N=15 denominator (0.061), sample uses n-1=14 (0.066). Question specifies “population variance Vx”.
Rounding: Exact calculation gives 0.0612, closest to option (C) 0.061.
Units: Variance in mm², numerical value only asked.
Key Formula Reference
σ² = Σ(xi - μ)² / N [Population Variance]
s² = Σ(xi - x̄)² / (n-1) [Sample Variance]
For GATE BT, always use N for population data as specified.
This calculation confirms option (C) 0.061 as the population variance for fruit fly abdomen length measurements.
