\"A continuous random variable X has the probability density function f(x) = 1/5

Question

"A continuous random variable X has the probability density function f(x) = 1/5, 1 <= X <= 5. What is P(2<= X <= 4)?" Answer 0.600 0.400 0.200 0.800

Explanation / Answer

a) We need ?(x = 0 to 1) (e - ke^(kx)) dx = 1 ==> (ex - e^(kx)) {for x = 0 to 1} = 1 ==> (e - e^k) - (0 - 1) = 1 ==> e^k = e ==> k = 1. b) Simply compute ?(x = 1/4 to 1/2) (e - e^x) dx = (ex - e^x) {for x = 1/4 to 1/2} = (e/2 - e^(1/2)) - (e/4 - e^(1/4)) = e/4 - e^(1/2) + e^(1/4). c) µ = E(X) ......= ?(x = 0 to 1) x(e - e^x) dx ......= ?(x = 0 to 1) (ex - xe^x) dx ......= [ex^2/2 - (xe^x - e^x)] {for x = 0 to 1} ......= e/2 - 1. E(X^2) = ?(x = 0 to 1) x^2(e - e^x) dx ......= ?(x = 0 to 1) (ex^2 - x^2 e^x) dx ......= [ex^3/3 - (x^2 - 2xe^x + 2e^x)] {for x = 0 to 1} ......= -2e/3 + 2. Finally, Var X = E(X^2) - (E(X))^2 ........= (-2e/3 + 2) - (e/2 - 1)^2 ........= 2 + e/3 - e^2/4.

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