Given, X = 0, 1, 2, 3
P(X = r) = nCr pr qn-r
Where n = 3, p = ½, q = ½ and r = 0, 1, 2, 3
P(X = 0) = ½ x ½ x ½ = 1/8
P(X = 1) = 3 x ½ x ½ x ½ = 3/8
P(X = 2) = 3 x ½ x ½ x ½ = 3/8
P(X = 3) = ½ x ½ x ½ = 3/8
The probability distribution table is:
Now,
E(X) = 0 + 1 x 3/8 + 2 x 3/8 + 3 x 1/8 = 3/8 + 6/8 + 3/8 = 12/8 = 3/2
E(X2) = 0 + 1 x 3/8 + 4 x 3/8 + 9 x 1/8 = 3/8 + 12/8 + 9/8 = 24/8 = 3
W.k.t, Var(X) = E(X2) – [E(X)]2 = 3 – (3/2)2 = 3 – (3/2)2 = 3 – 9/4 = ¾
Thus, the standard deviation = Var(X) = √(¾) = √3/2.
