Probability Distribution Mean (Expectation), Variance :: Problems

Solution

Let "x" represent the number of cakes demanded (in hundreds) per day. The values carried by "x" are 0, 1, 2, ... ,9 ==> "x" is finite. Sum of the probabilities of the various values of "x" = 0.02 + 0.07 + 0.09 + 0.12 + 0.20 + 0.20 + 0.18 + 0.10 + 0.01 + 0.01 = 1 The values of "x" represent the values in the range of a discrete random variable ⇒ "X" is a discrete random variable with range = {0, 1, 2, ... .9} The given distribution therefore, represents a discrete probability distribution of "X" Calculations for Mean and Standard Deviations x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 0 0.02 0 0 0 1 0.07 0.07 1 0.07 2 0.09 0.18 4 0.36 3 0.12 0.36 9 1.08 4 0.20 0.80 16 3.20 5 0.20 1.00 25 5.00 6 0.18 1.08 36 6.48 7 0.10 0.70 49 4.90 8 0.01 0.08 64 0.64 9 0.01 0.09 81 0.81 Total 1.00 4.36 22.54 Since the given distribution is the probability distribution of a discrete random variable "X", Σ p = 1. From the probability distribution: Expected (Mean) Number of cakes demanded per day ⇒ Expectation/Mean of the distribution ⇒ E (x) (Or) x = Σ px = 4.36 (in hundreds) = 436 Variance of the number of caked demanded per day ⇒ Variance of the distribution ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 22.54 − (4.36)2 = 22.54 − 19.0096 = 3.5304 Standard Deviation of the number of cakes demanded per day ⇒ standard deviation of the distribution ⇒ SD (x) = + √ Var (x) ⇒ SD (x) = + √ 3.5304 ⇒ SD (x) = + 1.879