Theory of Expectation :: Problems on Profits, Business : Probability Distribution

Solution

"x" indicates the amount of profit made by the man [Since you are required to find the man's expected profit, the variable would represent the man's expected profit] The profit earned by the man would be + Rs. 2,800 (when he makes a profit of Rs. 2000) − 5,500 (when he makes a loss of Rs. 5,500) 0 (when he neither makes a profit nor a loss) ⇒ The values carried by the variable ("x") would be either − 5,500 or 0 or + 2,800 ⇒ "X" is a discrete random variable with range = {− 5,500, 0, + 2,800} "X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x" Probabilty that the man Makes/Earns a profit of Rs. 2,800 ⇒ P(+2,800) = 0.5 Makes a loss of Rs. 5,500 ⇒ P(− 5,500) = 0.3 Makes neither a profit nor a loss ⇒ P(0) = 0.2 Probability for the mans earnings to be + Rs. 2,800 ⇒ P(X = + 2,800) = P(+ 2,800) = 0.5 − Rs. 5,500 ⇒ P(X = − 5,500) = P(− 5,500) = 0.3 0 ⇒ P(X = 0) = P(0) = 0.2 The probabilty distribution of "x" would be x − 5,500 0 + 2,800 P(X = x) or P 0.3 0.2 0.5 Calculations for Mean and Standard Deviations x P Px x2 Px2 − 5,500 0.3 − 1,650 3,02,50,000 90,75,000 0 0.2 0 0 0 + 2,800 0.5 + 1,400 78,40,000 39,20,000 Total 1 − 250 1,29,95,000 The mans expected profit ⇒ Expectation of "x" ⇒ E (x) = Σ px = − 250 The man can expect to make a loss of Rs. 250 Variance of the mans profit ⇒ var (x) = E (x2) − (E(x))2 = Σ px2 − (Σ px)2 = 1,29,95,000 − (− 250)2 = 1,29,95,000 − 62,500 = 1,29,32,500 Standard Deviation of the mans profit ⇒ SD (x) = + √ Var (x) = + √ 1,29,32,500 = + Rs. 3,596.18