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Question
The following table is describing about the probability mass function of the random variable X
| x | 3 | 4 | 5 |
| P(x) | 0.2 | 0.3 | 0.5 |
Find the standard deviation of x.
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Solution
E(X) = ∑xP(x)
= (3 × 0.2) + (4 × 0.3) + (5 × 05)
= 0.6 + 1.2 + 2.5
= 4.3
E(X2) = ∑x2P(x)
= (32 × 0.2) + (42 × 0.3) + (52 × 05)
= (9 × 0.2) + (16 × 0.3) + (25 × 0.5)
= 1.8 + 4.8 + 12.5
= 19.1
Variance of the Random Variable
σ2 = Var(X) = E(X2) − (E(X))2
Var(X) = 19.1 − (4.3)2
= 19.1 − 18.49
= 0.61
Standard Deviation (σ)
σ = `sqrt(Var(X))`
= `sqrt0.61`
= 0.781
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