Answered

Get detailed and reliable answers to your questions on IDNLearn.com. Whether your question is simple or complex, our community is here to provide detailed and trustworthy answers quickly and effectively.

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]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.