Answer to Question #321122 in Statistics and Probability for Konjo

Question

The probability density function of a continuous random variable X is given below:

𝑓(𝑥) = {

1.25(1 − 𝑥

4

) , 0 < 𝑥 < 1

0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

a. Find P(0.4< X<0.9)

b. Compute the expected value and variance of X.

Answer 1

$a:\\P\left( 0.4<X<0.9 \right) =\int_{0.4}^{0.9}{1.25\left( 1-x^4 \right) dx}=1.25\left( 0.5-\frac{0.9^5-0.4^5}{5} \right) =0.479938\\b:\\EX=\int_0^1{x\cdot 1.25\left( 1-x^4 \right) dx}=1.25\left( \frac{1}{2}-\frac{1}{6} \right) =0.416667\\EX^2=\int_0^1{x^2\cdot 1.25\left( 1-x^4 \right) dx}=1.25\left( \frac{1}{3}-\frac{1}{7} \right) =0.238095\\Var\left( X \right) =0.238095-0.416667^2=0.0644836$