Question Type 6: Finding the expected value for a given continuous distribution and PDF Bootcamps

A random variable $X$ has pdf $f(x)=ke^{-x/2}$ for $x>0$. Find $k$ and $E(X)$.

If $X$ has pdf $f(x)=6x(1-x)$ for $0<x<1$, find the expected value $E(X)$.

Find the expected value of a random variable $X$ with pdf $f(x)=4e^{-4x}$ for $x>0$.

Given a random variable $X$ with pdf $f(x)=4e^{-4x}$ for $x>0$, calculate $P(X>12)$.

The pdf of $X$ is defined by $f(x)=\begin{cases} ax,&0<x<1,\\ a(2-x),&1<x<2,\\ 0,&\text{otherwise}. \end{cases}$Determine $a$ and compute $E(X)$.

Let $X$ have pdf $\displaystyle f(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}\quad\text{for }x>0.$Express $E(X)$ in terms of $\alpha$ and $\beta$.

Consider the pdf $f(x)=\tfrac{c}{x^4}$ for $x\ge1$. Find $c$ and evaluate whether $E(X)$ converges or diverges.

The pdf of $X$ is $f(x)=kx^{-3}$ for $x\ge1$ and 0 otherwise. Determine $k$ and state whether $E(X)$ exists.