Answer to Question #228037 in Statistics and Probability for cat

Question

A random variable X has the following density function fx(x) = 2 − kx, 0 < x < 1/k 

a) Find the constant k.

b) Find the mean.

c) Find the standard deviation.

Answer 1

(a)

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1/k}(2-kx)dx=[2x-\dfrac{kx^2}{2}]\begin{matrix} 1/k \\ 0 \end{matrix}$$

$$=2/k-1/(2k)=\dfrac{3}{2k}=1=>k=\dfrac{3}{2}$$

(b)

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{2/3}(2x-\dfrac{3}{2}x^2)dx$$

$$=[x^2-\dfrac{x^3}{2}]\begin{matrix} 2/3 \\ 0 \end{matrix}=\dfrac{8}{27}$$

(c)

$$E(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{2/3}(2x^2-\dfrac{3}{2}x^3)dx$$

$$=[\dfrac{2}{3}x^3-\dfrac{3}{8}x^4]\begin{matrix} 2/3 \\ 0 \end{matrix}=\dfrac{10}{81}$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2$$

$$=\dfrac{10}{81}-(\dfrac{8}{27})^2=\dfrac{26}{729}$$

$$\sigma=\sqrt{\sigma^2}=\dfrac{\sqrt{26}}{27}$$