Question

A continuous random variable M has a pdf given by f(m)={k(1-m/10)for m<or equal to 10

{0,elsewhere

find the value of the constant k,the mean and the variance of x

Answer 1

$$\displaystyle\int_{-\infin}^{\infin}f(m)dm=\displaystyle\int_{0}^{10}k(1-\dfrac{m}{10})dm$$

$$=k[m-\dfrac{m^2}{20}]\begin{matrix} 10 \\ 0 \end{matrix}=5k=1$$

$$k=\dfrac{1}{5}$$

$$mean=\mu=E[M]=\displaystyle\int_{-\infin}^{\infin}mf(m)dm$$

$$=\displaystyle\int_{0}^{10}\dfrac{1}{5}m(1-\dfrac{m}{10})dm=[\dfrac{m^2}{10}-\dfrac{m^3}{150}]\begin{matrix} 10 \\ 0 \end{matrix}=\dfrac{10}{3}$$

$$E[M^2]=\displaystyle\int_{-\infin}^{\infin}m^2f(m)dm=\displaystyle\int_{0}^{10}\dfrac{1}{5}m^2(1-\dfrac{m}{10})dm$$

$$=[\dfrac{m^3}{15}-\dfrac{m^4}{200}]\begin{matrix} 10 \\ 0 \end{matrix}=\dfrac{50}{3}$$

$$Var(M)=E[M^2]-(E[M])^2$$

$$=\dfrac{50}{3}-(\dfrac{10}{3})^2=\dfrac{50}{9}$$