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

Step 1: Find K from the total probability rule.
Since $f(x)=Kx$ must integrate to 1 over $(0,2)$, $K\int_0^2 x\,dx=K\times2=1$, so $K=\dfrac{1}{2}$.
Step 2: Use the shortcut mean formula for this shape of pdf.
For a straight-line pdf $f(x)=kx$ on $(0,b)$, the mean always works out to $E(X)=\dfrac{2b}{3}$. Here $b=2$, so $E(X)=\dfrac{2\times2}{3}=\dfrac{4}{3}$.
Step 3: Form the ratio.
$K:E(X)=\dfrac12:\dfrac43$. Multiplying both sides by 6 to clear denominators gives $3:8$.
\[ \boxed{3:8} \]