Question

A random variable \(X\) has the probability density function \[ f(x)=Kx,\qquad 0\lt x\lt 2, \] then the ratio between \(K\) and the mean of \(X\) is

• A. \(3:2\)
• B. \(2:3\)
• C. \(5:3\)
• D. \(3:8\)

Hint:

For a pdf, \[ \int f(x)\,dx=1, \qquad E(X)=\int xf(x)\,dx. \]

Answer 1

The Correct Option is D

Since total probability is \(1\), \[ \int_0^2Kx\,dx=1. \] Therefore, \[ K\left[\frac{x^2}{2}\right]_0^2=1 \] \[ 2K=1 \] \[ K=\frac12. \] Now, \[ E(X)=\int_0^2x(Kx)\,dx =\frac12\int_0^2x^2\,dx \] \[ =\frac12\left[\frac{x^3}{3}\right]_0^2 =\frac12\cdot\frac83 =\frac43. \] Hence, \[ K:E(X) = \frac12:\frac43 = 3:8. \] Therefore, \[ \boxed{(D)} \]