Question

A continuous r.v. X has the p.d.f. as follows: f(x) = Zx(9 – x2 ),​0 ≤ x ≤ 3
= 0,​otherwise

Calculate the value of Z?

Answer 1

solution

The PDF of X is

$f(x) = Zx(9-x^2)$ Where $0 \leq x \leq3$

The sum of all probabilities i.e the integral of the PDF for all values of X $=1$

$$1=\textstyle\int_0^3 Zx(9-x^2)$$

$$\frac{1}{Z}=\textstyle\int_0^3 (9x-x^3)$$

$$\frac{1}{Z}= (4.5x^2-0.25x^4) \textstyle\mid_0^3$$

$$\frac{1}{Z}= 4.5(3)^2-0.25(3)^4-0$$

$$\frac{1}{Z} = 20.25$$

$$Z= \frac{4}{81} = 0.0494$$

Answer: Z is 0.0494