Question

What is the mean of $f\left(x\right)=3x+2$ where $x$ is a random variable with probability distribution
$\begin{array}{ccccc}X=x& 1& 2& 3& 4\\ P(X=x)& \frac{1}{6}& \frac{1}{3}& \frac{1}{3}& \frac{1}{6}\end{array}$

• A. $\frac{19}{2}$
• B. $\frac{15}{2}$
• C. $\frac{5}{2}$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{19}{2}$

Mean of the given probability distribution, $E\left(X\right)=\sum p_i{x}_i=\frac{1}{6}+\frac{2}{3}+\frac{3}{3}+\frac{4}{6}=\frac{1+4+6+4}{6}=\frac{5}{2}$

Therefore, mean of $f\left(x\right)=3x+2$ is $E(3X+2)=3E\left(X\right)+2=\frac{15}{2}+2=\frac{19}{2}$