Question

Engineering Economic

10% interest rate

The probability mass functions of the initial investment and useful life of a system are given below. The system earns an annual revenue of $2000. Find the mean, variance and standard deviation of the probability distribution of the present worth of the system. Initial investment Value Probability $5000 0.2 $6000 0.3 $7000 0.4 $8000 0.1 Useful life (in years) Value Probability 6 0.3 7 0.6 9 0.1

Answer 1

solution given data from Problem Nord 1 $5000 Initial investment Valul Probability 0.2 $6000 3 $7000 0.4 나 $6000 0.1 0.3 calculation of mean by using the given formala 4 E(X)= X P(x=x) E(X) = ($5000X0-2) + ($6000xo. 3)+($7000 x0:4)+ ($8000 XO:1) E(x) = $6400 : Value of the meam is $6400 Calculation of varience by using the given formula bela Varizurefx) = EXPEIX] there E(X) ² = 1 4 4 x ? p(x=2) = (($5000)?6.2)] + [$6000 x 0.3] + [67000 ²x0.44] + [$60ooxoi] EX)= 218200000 E(X) = $6400 is already finded. then EX)?-($ 6400) 40960000 substituting E(X)" and [Ex] ² in the equation we get the valience(x),

Wh we know the value of Ex)= 6.9, then [Ex] ² - 6.92 [Ex] 2= 47.61 substituting the Ex)2 and [Ex] Vales in earnationa iwe get the varis), Varience(x)= 48.3 - 47.61 - 0.69 (iii) the calculation of the standard deviation by the squarrooting of the varience(x), then -0.69 - 0.8306