Question

let (x) be a random variable with the following probability distribution.
value (x) of (x) (p(x = x)
-3 0.25
-2 0.05
-1 0.15
0 0.15
1 0.40
complete the following. (if necessary, consult a list of formulas.)
(a) find the expectation (e(x)) of (x).
(e(x)=)
(b) find the variance (var(x)) of (x).
(var(x)=)

Explanation:

Step1: Recall expectation formula

The formula for the expectation $E(X)$ of a discrete - random variable is $E(X)=\sum_{i}x_ip_i$, where $x_i$ are the values of the random variable and $p_i$ are their corresponding probabilities.
\[E(X)=(-3)\times0.25+(-2)\times0.05+(-1)\times0.15 + 0\times0.15+1\times0.40\]

Step2: Calculate expectation

\[

$$\begin{align*} E(X)&=(-3)\times0.25+(-2)\times0.05+(-1)\times0.15 + 0\times0.15+1\times0.40\\ &=- 0.75-0.1 - 0.15+0 + 0.4\\ &=-0.6 \end{align*}$$

\]

Step3: Recall variance formula

The formula for the variance $Var(X)=E((X - E(X))^{2})=\sum_{i}(x_i - E(X))^{2}p_i$. First, we calculate $(x_i - E(X))^{2}$ for each $i$.
For $x_1=-3$: $(-3-(-0.6))^{2}=(-2.4)^{2}=5.76$
For $x_2=-2$: $(-2 - (-0.6))^{2}=(-1.4)^{2}=1.96$
For $x_3=-1$: $(-1-(-0.6))^{2}=(-0.4)^{2}=0.16$
For $x_4 = 0$: $(0-(-0.6))^{2}=0.36$
For $x_5 = 1$: $(1-(-0.6))^{2}=2.56$
Then $Var(X)=5.76\times0.25+1.96\times0.05+0.16\times0.15+0.36\times0.15+2.56\times0.40$

Step4: Calculate variance

\[

$$\begin{align*} Var(X)&=5.76\times0.25+1.96\times0.05+0.16\times0.15+0.36\times0.15+2.56\times0.40\\ &=1.44 + 0.098+0.024+0.054+1.024\\ &=2.64 \end{align*}$$

\]

Answer:

(a) $E(X)=-0.6$
(b) $Var(X)=2.64$