Find k and the probability that on a given day the electricity consumption is more than expected electricity consumption.

For any probability density function $\int\limits_{-\infty}^{\infty}f(x)dx=1\\ \therefore \int\limits_0^{\infty}kxe^{-x/3}.dx=1\\ \therefore k\Bigg[\dfrac {xe^{-x/3}}{-1/3}-1.\dfrac {e^{-x/3}}{(1/3)^2}\Bigg]_0^{\infty}=1\\ \therefore k\Big[(0-0)-\Big(0-\dfrac {e^0}{1/9}\Big)\Big]=1\\ \therefore 9k=1\\ \therefore k=\dfrac 19\\ \therefore f(x)=\dfrac 19xe^{-x/3}$

Expected electric consumption =f(x)

$$=\int\limits_{-\infty}^{\infty}x.f(x)dx$$

$=\int\limits_{0}^{\infty}x\dfrac 19xe^{-x/3}dx\\ =\dfrac 19\int\limits_0^{\infty}x^2.e^{-x/3}dx\\ =\dfrac 19\Bigg[x^2.\dfrac {e^{-x/3}}{-1/3}-2x.\dfrac {e^{-x/3}}{(-1/3)^2}+2.\dfrac {e^{-x/3}}{(-1/3)^3}\Bigg]_0^{\infty}\\ =\dfrac 19\Bigg[(0-0+0)-\Bigg(0-0+2.\dfrac {e^0}{-1/27}\Bigg)\Bigg]\\ =\dfrac 19(2\times 27)\\ =6$

P (consumption is more than the expected value)

$$= P (x\gt6)$$

$\int\limits_6^{\infty}\dfrac 19x.e^{-x/3}.dx\\ =\dfrac 19\Big[x.\dfrac {e^{-x/3}}{-1/3}-1.\dfrac {e^{-x/3}}{(-1/3)^2}\Big]_6^{\infty}\\ =\dfrac 19\Bigg[(0-0)-\Bigg(6.\dfrac {e^{-2}}{-1/3}-\dfrac {e^{-2}}{1/9}\Bigg)\Bigg]\\ =\dfrac 19\times (6\times 3e^{-2}+9^{e^{-2}})\\ =3e^{-2}\\ =0.406$

$K = 1/9;$ P (Consumption is more than the expected value) $= 0.406$