Answer to Question #210272 in Statistics and Probability for Prash

Question

The p.d.f of a continuous random variable is give

P(x) ={ kx (1-x) e*, 0 ≤ x ≤ 1

{ 0, otherwise

Find k and hence find mean and standard deviation

Answer 1

$$1=\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1}kx(1-x)dx$$

$$=k\big[\dfrac{x^2}{2}-\dfrac{x^3}{3}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{k}{6}$$

$$k=6$$

$$f(x)=\begin{cases} 6x(1-x), & 0\leq x\leq 1 \\ 0, &\text{otherwise} \end{cases}$$

$$\mu=E[X]=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{1}6xx(1-x)dx$$

$$=6\big[\dfrac{x^3}{3}-\dfrac{x^4}{4}\big]\begin{matrix} 1 \\ 0 \end{matrix}=2-\dfrac{3}{2}=\dfrac{1}{2}=0.5$$

$$E[X^2]=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{1}6x^2x(1-x)dx$$

$$=6\big[\dfrac{x^4}{4}-\dfrac{x^5}{5}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{2}-\dfrac{6}{5}=\dfrac{3}{10}$$

$$V(X)=\sigma^2=E[X^2]-(E[X])^2$$

$$=\dfrac{3}{10}-(\dfrac{1}{2})^2=\dfrac{1}{20}=0.05$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{0.05}=0.1\sqrt{5}\approx0.2236$$