Subject: Applied Mathematics 2

Topic: Probability Distribution

Difficulty: Medium

The daily consumption of electric power is a random variable X with p.d.f (probability density function) Find the value of k ,the expectation of X and the probability that on a given day the electric consumption is more than expected value.

Since the total probability is 1

$∫_{-∞}^∞f(x) $ = 1

$∫_0^∞kxe^{-x/3} \ dx = k∫_0^∞xe^{-x/3}= k[x \frac{e^{-x/3}}{(-1/3)}- 1.\frac{e^{-x/3}}{(1/9)}]_0^∞ = k [0+9] = 1$

∴ k = $\frac{1}{9}$

∴ f(x) = $ \begin{cases} \frac{1}{9}xe^{-x/3} \,\,\,\,\, for \ x\gt0 \\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, for \ x \leq 0 \end{cases} $

By definition, $E(X) =∫_0^∞xf(x) \ dx = ∫_0^∞x. \frac{1}{9} xe^{-x/3} \ dx = ∫_0^∞ \frac{1}{9} x^2 e^{-x/3} \ dx$

= $ \frac{1}{9} [x^2. \frac{e^{-x/3}}{(-1/3)}- 2x \frac{e^{-x/3}}{(1/9)}+2.\frac{e^{-x/3}}{(-1/27)}]_0^∞ = \frac{1}{9} [54] $= 6

∴ the probability that on a given day the electric consumption is more than expected value i.e.
p(X>6) = $∫_0^6\frac{1}{9} xe^{-x/3} \ dx = \frac{1}{9} ∫_0^6xe^{-x/3}= \frac{1}{9} [x \frac{e^{-x/3}}{(-1/3)}-(1)\frac{e^{-x/3}}{(1/9)}]_0^6 = 3e^{-2}$