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Question
Consider the probability distribution of a random variable X:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.25 | 0.3 | 0.2 | 0.15 |
Variance of X.
Solution
Here, we have
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | 0.25 | 0.3 | 0.2 | 0.15 |
We know that: Var(X) = E(X2) – [E(X)]2
Where E(X) = `sum_("i" = 1)^"n" x_"i""p"_"i"` and E(X2) = `sum_("i" = 1)^"n" "p"_"i"x"i"^2`
∴ E(X) = 0 × 0.1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15
= 0 + 0.25 + 0.6 + 0.6 + 0.6
= 2.05
E(X2) = 0 × 0.1 + 1 × 0.25 + 4 × 0.3 + 9 × 0.2 + 16 × 0.15
= 0 + 0.25 + 1.2 + 1.8 + 2.40
= 5.65
Var(X) = 1.4475
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