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. 2,000 (when he makes a profit of Rs. 2000) − 1,000 (when he makes a loss of Rs. 1,000) ⇒ The values carried by the variable ("x") would be either − 1,000 or + 2,000 ⇒ "X" is a discrete random variable with range = {− 1,000, + 2,000} "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. 2,000 ⇒ P(+2,000) = 0.4 Makes a loss of Rs. 1,000 ⇒ P(− 1,000) = 0.6 Probability for the mans earnings to be + Rs. 2,000 ⇒ P(X = +2,000) = P(+ 2,000) = 0.4 − Rs. 1,000 ⇒ P(X = − 1,000) = P(− 1,000) = 0.6 The probabilty distribution of "x" would be x − 1,000 + 2,000 P(X = x) or p 0.6 0.4 Calculations for Mean and Standard Deviations x p px x2 px2 − 1,000 0.6 − 600 10,00,000 6,00,000 + 2,000 0.4 + 800 40,00,000 16,00,000 Total 1 + 200 22,00,000 The mans expected profit ⇒ Expectation of "x" ⇒ E (x) = Σ px = + Rs. 200 The man can expect to make a profit of Rs. 200 Variance of the mans profit ⇒ var (x) = E (x2) − (E(x))2 = Σ px2 − (Σ px)2 = 22,00,000 − (+ 200)2 = 22,00,000 − 40,000 = Rs. 21,60,000 Standard Deviation of the number of red balls drawn ⇒ SD (x) = + √ Var (x) = + √ 21,60,000 = Rs. 1,469.69