CE Board Exam Randomizer

Question 1

A probability distribution assigns probabilities to possible outcomes. The expected value is the long-run weighted average.

$$E(X)=\sum xp(x), \qquad Var(X)=E(X^2)-[E(X)]^2$$

Answer

$$E(X)=0(0.20)+1(0.50)+2(0.30)=1.10$$

$$E(X^2)=0^2(0.20)+1^2(0.50)+2^2(0.30)=1.70$$

$$Var(X)=1.70-(1.10)^2=0.49$$

Final answer: $E(X)=1.10$ and $Var(X)=0.49$.

Question 2

Let $X=0,1,2$ with probabilities $0.20,0.50,0.30$. Find $E(X)$ and $Var(X)$.