Variance and standard deviation
Standard deviation: SD(X)=√Var(X). Measures spread around mean. Same units as X.
Worked example
RV X: values 1,2,3 with probs 0.2,0.5,0.3. Find Var(X).
Solution
- E(X)=1(0.2)+2(0.5)+3(0.3)=2.1
- E(X²)=1²(0.2)+4(0.5)+9(0.3)=4.7
- Var(X)=4.7-2.1²=4.7-4.41=0.29
- SD(X)=√0.29≈0.54
Final answer
Variance=0.29, SD≈0.54.
Variance properties
Independence: If X and Y independent: Var(X+Y)=Var(X)+Var(Y). Var(X-Y)=Var(X)+Var(Y) also!
Worked example
Var(X)=4, Var(Y)=9, independent. Find Var(X+Y) and Var(2X).
Solution
- Var(X+Y)=4+9=13
- Var(2X)=2²(4)=16
- SD(X+Y)=√13≈3.6, SD(2X)=4
Final answer
Var(X+Y)=13, Var(2X)=16.
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Interpreting spread
Larger SD means: More spread around mean. Values more likely to be far from mean. Less predictable outcome.
| SD value | Interpretation |
|---|---|
| Small (near 0) | Values cluster near mean |
| Large (>2) | Values spread out, high variability |
Comparison example
Distribution A: SD=0.1. Distribution B: SD=2. Which is more predictable?
Answer
- A has SD=0.1 (tiny spread)
- B has SD=2 (large spread)
- A is highly predictable: values stay near mean
- B is unpredictable: values vary widely
Final answer
A more predictable. Smaller SD = clustering.
Variance for grouped data
From frequency table: Use class midpoints as x values. Apply same variance formula.
Worked example
Classes [0-10) freq 5, [10-20) freq 8, [20-30) freq 7. Find variance.
Solution
- Midpoints: 5,15,25. Total n=20
- E(X)=(5×5+15×8+25×7)/20=2.5
- E(X²)=(25×5+225×8+625×7)/20
- Calculate Var(X) from formula
Final answer
Use midpoint method for grouped variance.