Answer to Question #305499 in Statistics and Probability for Cons

Question

Let X be a random variable with the following probability distribution:

x 1 2 3 4

P(X) 0.4 0.3 0.2 0.1

Find the expected value and variance.

Answer 1

$$E(X)=1(0.4)+2(0.3)+3(0.2)+4(0.1)=2$$

$$E(X^2)=1^2(0.4)+2^2(0.3)+3^2(0.2)+4^2(0.1)=5$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2=5-(2)^2=1$$