F(x) = kxe$^{-x/3}$ , x >0

0, x ≤0

Find the value of k , the expectation of k & the probability that on a particular day the Electric consumption is more than the expected value.

Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

For continuous Random variables

$ \int_{-\infty}^{\infty} f(x) \,\,dx = 1 \\ \int_0^{\infty} kx e^{-x/3} \,\, dx = 1 \\ k \int_0^{\infty} x e^{-x/3} \,\, dx = 1 \\ k[x (\frac{e^{-x/3}}{-1/3}) - 1 \frac{e^{-x/3}}{(-1/3)^2}]_0^{\infty} = 1 \\ k [(-3x e^{-x/3}) - (9e^{-x/3})]_0^{\infty} = 1 \\ k(9) = 1 \implies k = \frac{1}{9} $

$ E(X) = \int_{-\infty}^{\infty} xf(x) \,\,dx = \int_0^{\infty} x(kx e^{-x/3}) \,\, dx \\ \int_0^{\infty} (kx^2 e^{-x/3}) \,\, dx \\ = k [x^2 \frac{e^{-x/3}}{-1/3} - 2x \frac{e^{-x/3}}{(-1/3)^2} + 2 \frac{e^{-x/3}}{(-1/3)^3}]_0^{\infty} \\ = k(54) = \frac{1}{9}(54) = 6 $

To find

$ P(X \gt 6) = \int_6^{\infty} f(x)dx = \int_6^{\infty} kx e^{-x/3} \,\, dx \\ k \int_6^{\infty} x e^{-x/3} \,\, dx \\ k [x \frac{e^{-x/3}}{-1/3} - \frac{e^{-x/3}}{(-1/3)^2}]_6^{\infty} \\ = k [-3x e^{-x/3} - 9(e^{-x/3})]_6^{\infty} \\ = k (18 e^{-2} + 9 e^{-2}) = \frac{1}{9} (27 e^{-2}) = 3 e^{-2} = 0.4060 $