Q.32 Runs scored by a batsman in five one-day matches are
55, 75, 67, 88 and 15. The standard deviation is ____________.
✅ Correct Answer: 27.87
The standard deviation measures variability in a batsman’s scores across five one-day matches: 55, 75, 67, 88, and 15. For this sample data, the correct value is approximately 27.87
Step-by-Step Calculation
- Compute the mean: (55 + 75 + 67 + 88 + 15) / 5 = 300 / 5 = 60
- Squared differences from mean:
- (55-60)² = 25
- (75-60)² = 225
- (67-60)² = 49
- (88-60)² = 784
- (15-60)² = 2025
Sum = 3108
- Sample variance: 3108 / (5-1) = 3108 / 4 = 777
- Standard deviation: √777 ≈ 27.87
Calculation Table
| Runs (xᵢ) | (xᵢ – mean) | (xᵢ – mean)² |
|---|---|---|
| 55 | -5 | 25 |
| 75 | 15 | 225 |
| 67 | 7 | 49 |
| 88 | 28 | 784 |
| 15 | -45 | 2025 |
| Total | 0 | 3108 |
Sample vs Population Formula
Use the sample standard deviation formula for exam data like this: s = √[Σ(xᵢ - x̄)² / (n-1)], where n=5
Population formula divides by n instead of n-1, yielding √(3108/5) = √621.6 ≈ 24.93—incorrect here as problems treat such sets as samples
Common Mistakes Explained
- Dividing by n: ~24.93, underestimates variability—wrong for sample data
- Forgetting squares: √(108/4)=5.2—completely invalid
- Wrong mean (59): ~28.1—close but imprecise
Cricket Stats Application
Standard deviation of 27.87 shows high inconsistency in these runs, useful for analyzing batsman performance variability
Sports analysts apply this to predict reliability over averages alone
