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

Solution

"x" indicates the taxi drivers earnings [Since you are required to find the taxi drivers expected earnings, the variable would represent the taxi drivers earnings] The taxi drivers earnings on a day would be + Rs. 100 if it rains − Rs. 10 if it is fair ⇒ The values carried by the variable ("x") would be either − 10 or + 100 ⇒ "X" is a discrete random variable with range = {− 10, + 100} "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 for it to rain ⇒ P(Rain) = 0.4 it it to be fair ⇒ P(Fair) = 0.6 On the assumption that there are only two possibilities, i.e. for it to rain or for it to be fair, The two events "Rain" and "Fair" are exhaustive events. ⇒ P(Rain ∪ Fair) = 1       → (1) Since only one of the events can occur, the two events of "Rain" and "Fair" are mutually exclusive ⇒ P(Rain ∩ Fair) = 0 (Or) P(Rain ∪ Fair) = P(Rain) + P(Fair)     → (2) From (1) and (2) we can write P(Rain ∪ Fair) = P(Rain) + P(Fair) = 1 ⇒ P(Rain) + P(Fair) = 1 ⇒ 0.4 + P(Fair) = 1 ⇒ P(Fair) = 1 − 0.4 ⇒ P(Fair) = 0.6 Probability for the taxi drivers earnings to be + Rs. 100 ⇒ P(X = + 100) = P(+ 100) = 0.4 − Rs. 10 ⇒ P(X = − 10) = P(− 10) = 0.6 The probabilty distribution of "x" would be x − 10 + 100 P(X = x) or P 0.6 0.4 Calculations for Mean and Standard Deviations x P Px x2 Px2 − 10 0.6 − 6 100 60 + 100 0.4 40 10,000 4,000 Total 1 34 4,060 The taxi drivers expectation ⇒ Expectation of "x" ⇒ E (x) = Σ px = + Rs. 34 The taxi driver can expect an earnings of Rs. 34 Variance of the taxi drivers earnings ⇒ var (x) = E (x2) − (E(x))2 = Σ px2 − (Σ px)2 = 4,060 − (+ 34)2 = 4,060 − 1,156 = 2,904 Standard Deviation of the mans profit ⇒ SD (x) = + √ Var (x) = + √ 2,904 = + Rs. 53.89