Random Variables and Distributions Expectation and variance of a random variable Let $X$ be a random variable with the following probability distribution. \begin{tabular}{|c|c|} \hline Value $x$ of $X$ & $P(X=x)$ \\ \hline-3 & 0.05 \\ \hline-2 & 0.10 \\ \hline-1 & 0.10 \\ \hline 0 & 0.10 \\ \hline 1 & 0.65 \\ \hline \end{tabular} Complete the following. (If necessary, consult a list of formulas.) (a) Find the expectation $E(X)$ of $X$. \[ E(X)=\square \] (b) Find the variance $\operatorname{Var}(X)$ of $X$. \[ \operatorname{Var}(X)= \]

Step 1 ：Let $X$ be a random variable with the following probability distribution.

Step 2 ：\begin{tabular}{|c|c|}\n\hline Value $x$ of $X$ & $P(X=x)$ \\ \hline-3 & 0.05 \\ \hline-2 & 0.10 \\ \hline-1 & 0.10 \\ \hline 0 & 0.10 \\ \hline 1 & 0.65 \\ \hline\n\end{tabular}

Step 3 ：Find the expectation $E(X)$ of $X$.

Step 4 ：The expectation of a random variable is calculated by multiplying each possible value of the variable by its probability, and then summing these products.

Step 5 ：$E(X) = \sum xP(X=x)$

Step 6 ：Substitute the given values into the formula to get $E(X) = (-3)(0.05) + (-2)(0.10) + (-1)(0.10) + (0)(0.10) + (1)(0.65) = 0.2$

Step 7 ：Find the variance $\operatorname{Var}(X)$ of $X$.

Step 8 ：The variance of a random variable is calculated by subtracting the square of the expectation from the expectation of the square of the variable.

Step 9 ：$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$

Step 10 ：Substitute the given values into the formula to get $\operatorname{Var}(X) = [(-3)^2(0.05) + (-2)^2(0.10) + (-1)^2(0.10) + (0)^2(0.10) + (1)^2(0.65)] - (0.2)^2 = 1.56$

Step 11 ：The expectation of X is \(\boxed{0.2}\) and the variance of X is \(\boxed{1.56}\).