# Probability Distribution Mean (Expectation), Variance :: Problems

 Problem Back to Problems Page
Prove that: (i) var(2) = 0, (ii) var(3x) = 9 (iii) var (x + 4x) = 16 var (x)

Net Answers :

 Solution

 We know, var (x) = E(x2) − (E (x))2 = Σ px2 − (Σ px)2

Therefore,
(i)
 var (2) = Σ p × (2)2 − (Σ p × (2))2 = Σ p × 4 − (2 Σ p)2 = 4 Σ p − (2 × 1)2 = 4 × 1 − (2)2 = 4 − 4 = 0
(ii)
 var (3x) = Σ p × (3x)2 − (Σ p × (3x))2 = Σ p × 9x2 − (3 Σ px)2 = 9 Σ p x2 − [ (3)2 × (Σ px)2] = 9 Σ p x2 − [9 (Σ px)2] = 9 {Σ p x2 − (Σ px)2} = 9 {E (x2) − (E (x))2} = 9 var (x)
(iii)
 var (3 + 4x) = Σ p (3 + 4x)2 − ( Σ p (3 + 4x))2 = Σ p {(3)2 + 2(3)(4)x + ((4)x)2)} − (Σ p × 3 + p × 4x)2 = Σ p (9 + 24x + 16x2) − {Σ (3p + 4px)2} = Σ (9p + 24px + 16px2) − (Σ 3p + Σ 4px)2 = {Σ 9p + Σ 24px + Σ 16px2} − {3 Σ p + 4 Σ px)2} = {9 Σ p + 24 Σ px + 16 Σ px2} − {3 × 1 + 4 Σ px)2} = {9 × 1 + 24 Σ px + 16 Σ px2} − {3 + 4 Σ px)2} = {9 + 24 Σ px + 16 Σ px2} − { (3)2 + 2 (3)(4 Σ px) + (4 Σ px)2} = {9 + 24 Σ px + 16 Σ px2} − { 9 + 24 Σ px + [(4)2 × (Σ px)2]} = {9 + 24 Σ px + 16 Σ px2} − { 9 + 24 Σ px + 16 (Σ px)2} = 9 + 24 Σ px + 16 Σ px2 − 9 − 24 Σ px − 16 (Σ px)2 = 16 Σ px2 − 16 (Σ px)2 = 16 [Σ px2 − (Σ px)2] = 16 [E(x2) − (E(x))2] = 16 var (x)

 Credit : Vijayalakshmi Desu

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