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Question
Find the expected value and variance X using the following p.m.f.
| x | – 2 | – 1 | 0 | 1 | 2 |
| P(x) | 0.2 | 0.3 | 0.1 | 0.15 | 0.25 |
Solution
Expected value of X
E(X) = `sum_("i" = 1)^5 x_"i"*"P"(x_"i")`
= (– 2) × (0.2) + (–1) × (0.3) + 0 × (0.1) + 1 × (0.15) + 2 × (0.25)
= – 0.4 – 0.3 + 0 + 0.15 + 0.5
= – 0.05
E(X2) = `sum_("i" = 1)^5 x_"i"^2*"P"(x_"i")`
= (– 2)2 × (0.2) + (–1)2 × (0.3) + 02 × (0.1) + 12 × (0.15) + 22 × (0.25)
= 0.8 + 0.3 + 0 + 0.15 + 1
= 2.25
∴ Variance of X
= Var(X)
= E(X2) – [E(X)]2
= 2.25 – (– 0.05)2
= 2.2475
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