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Let [tex]\( X \)[/tex] be a random variable with the following probability distribution.

[tex]\[
\begin{tabular}{|c|c|}
\hline
Value \( x \) of \( X \) & \( P(X = x) \) \\
\hline
-1 & 0.05 \\
\hline
0 & 0.10 \\
\hline
1 & 0.10 \\
\hline
2 & 0.30 \\
\hline
3 & 0.05 \\
\hline
4 & 0.40 \\
\hline
\end{tabular}
\][/tex]

Complete the following. (If necessary, consult a list of formulas.)

(a) Find the expectation [tex]\( E(X) \)[/tex] of [tex]\( X \)[/tex].
[tex]\[
E(X) = \square
\][/tex]

(b) Find the variance [tex]\( \operatorname{Var}(X) \)[/tex] of [tex]\( X \)[/tex].
[tex]\[
\operatorname{Var}(X) = \square
\][/tex]

Sagot :

GinnyAnsweridn
Let's solve the problem step by step.

### Part (a): Finding the expectation [tex]\( E(X) \)[/tex] of [tex]\( X \)[/tex].
The expectation (or mean) [tex]\( E(X) \)[/tex] of a discrete random variable [tex]\( X \)[/tex] with values [tex]\( x_i \)[/tex] and corresponding probabilities [tex]\( P(X = x_i) \)[/tex] is given by the formula:
[tex]\[ E(X) = \sum_{i} x_i \cdot P(X = x_i) \][/tex]

Given values and their probabilities:
[tex]\[ \begin{array}{|c|c|} \hline \text{Value } x \text{ of } X & P(X = x) \\ \hline -1 & 0.05 \\ \hline 0 & 0.10 \\ \hline 1 & 0.10 \\ \hline 2 & 0.30 \\ \hline 3 & 0.05 \\ \hline 4 & 0.40 \\ \hline \end{array} \][/tex]

Plugging in these values, we get:
[tex]\[ E(X) = (-1 \cdot 0.05) + (0 \cdot 0.10) + (1 \cdot 0.10) + (2 \cdot 0.30) + (3 \cdot 0.05) + (4 \cdot 0.40) \][/tex]
[tex]\[ E(X) = -0.05 + 0 + 0.10 + 0.60 + 0.15 + 1.60 \][/tex]
[tex]\[ E(X) = 2.4 \][/tex]

Thus, the expectation [tex]\( E(X) \)[/tex] is:
[tex]\[ E(X) = 2.4 \][/tex]

### Part (b): Finding the variance [tex]\( \operatorname{Var}(X) \)[/tex] of [tex]\( X \)[/tex].
The variance [tex]\( \operatorname{Var}(X) \)[/tex] of a discrete random variable [tex]\( X \)[/tex] is given by:
[tex]\[ \operatorname{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]

First, we need to find [tex]\( E(X^2) \)[/tex], which is the expectation of [tex]\( X^2 \)[/tex]:
[tex]\[ E(X^2) = \sum_{i} (x_i^2) \cdot P(X = x_i) \][/tex]

Again, using the given values and probabilities:
[tex]\[ \begin{array}{|c|c|} \hline \text{Value } x & P(X = x) \\ \hline -1 & 0.05 \\ \hline 0 & 0.10 \\ \hline 1 & 0.10 \\ \hline 2 & 0.30 \\ \hline 3 & 0.05 \\ \hline 4 & 0.40 \\ \hline \end{array} \][/tex]

We calculate [tex]\( E(X^2) \)[/tex] as follows:
[tex]\[ E(X^2) = (-1^2 \cdot 0.05) + (0^2 \cdot 0.10) + (1^2 \cdot 0.10) + (2^2 \cdot 0.30) + (3^2 \cdot 0.05) + (4^2 \cdot 0.40) \][/tex]
[tex]\[ E(X^2) = (1 \cdot 0.05) + (0 \cdot 0.10) + (1 \cdot 0.10) + (4 \cdot 0.30) + (9 \cdot 0.05) + (16 \cdot 0.40) \][/tex]
[tex]\[ E(X^2) = 0.05 + 0 + 0.10 + 1.20 + 0.45 + 6.40 \][/tex]
[tex]\[ E(X^2) = 8.2 \][/tex]

Now, we use [tex]\( E(X^2) \)[/tex] and [tex]\( E(X) \)[/tex] to find the variance:
[tex]\[ \operatorname{Var}(X) = E(X^2) - [E(X)]^2 \][/tex]
[tex]\[ \operatorname{Var}(X) = 8.2 - (2.4)^2 \][/tex]
[tex]\[ \operatorname{Var}(X) = 8.2 - 5.76 \][/tex]
[tex]\[ \operatorname{Var}(X) = 2.44 \][/tex]

Finally, we need to raise the variance to the power of 3:
[tex]\[ \operatorname{Var}(X)^3 = (2.44)^3 \][/tex]
[tex]\[ \operatorname{Var}(X)^3 = 14.61184 \][/tex]

Therefore, the variance [tex]\( \operatorname{Var}(X)^3 \)[/tex] is:
[tex]\[ \operatorname{Var}(X)^3 = 14.61184 \][/tex]
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