Alright, let’s solve the problem step-by-step.
### Step 1: Correct the Table of the Probability Distribution
First, let’s rewrite the properly formatted table for the given discrete random variable [tex]\( X \)[/tex].
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
X & -1 & 0 & 2 & 5 & 6 \\
\hline
P(X) & 0.1 & 0.05 & 0.15 & 0.4 & 0.3 \\
\hline
\end{array}
\][/tex]
This table shows the values that [tex]\( X \)[/tex] can take and their respective probabilities.
### Step 2: Verify the Probability Distribution
The probabilities should sum to 1 to confirm that this is a valid probability distribution. Adding these probabilities:
[tex]\[
0.1 + 0.05 + 0.15 + 0.4 + 0.3 = 1
\][/tex]
This confirms the table represents a valid probability distribution.
### Step 3: Calculate the Mean ([tex]\(\mu\)[/tex])
The mean (or expected value) is given by:
[tex]\[
\mu = E[X] = \sum (x \cdot P(x))
\][/tex]
Substituting the given values:
[tex]\[
\mu = (-1 \cdot 0.1) + (0 \cdot 0.05) + (2 \cdot 0.15) + (5 \cdot 0.4) + (6 \cdot 0.3)
\][/tex]
[tex]\[
= -0.1 + 0 + 0.3 + 2 + 1.8
\][/tex]
[tex]\[
= 4.0
\][/tex]
So, the mean is [tex]\( \mu = 4.0 \)[/tex].
### Step 4: Calculate the Second Moment
The second moment about the origin is:
[tex]\[
E[X^2] = \sum (x^2 \cdot P(x))
\][/tex]
Substituting the given values:
[tex]\[
E[X^2] = ((-1)^2 \cdot 0.1) + (0^2 \cdot 0.05) + (2^2 \cdot 0.15) + (5^2 \cdot 0.4) + (6^2 \cdot 0.3)
\][/tex]
[tex]\[
= (1 \cdot 0.1) + (0 \cdot 0.05) + (4 \cdot 0.15) + (25 \cdot 0.4) + (36 \cdot 0.3)
\][/tex]
[tex]\[
= 0.1 + 0 + 0.6 + 10 + 10.8
\][/tex]
[tex]\[
= 21.5
\][/tex]
### Step 5: Calculate the Variance ([tex]\(\sigma^2\)[/tex])
The variance is given by:
[tex]\[
\sigma^2 = E[X^2] - (E[X])^2
\][/tex]
Substituting the values:
[tex]\[
\sigma^2 = 21.5 - (4.0)^2
\][/tex]
[tex]\[
= 21.5 - 16
\][/tex]
[tex]\[
= 5.5
\][/tex]
So, the variance is [tex]\( \sigma^2 = 5.5 \)[/tex].
### Step 6: Find the Moment Generating Function (MGF)
The moment generating function [tex]\( M_X(t) \)[/tex] for a discrete random variable [tex]\( X \)[/tex] is given by:
[tex]\[
M_X(t) = E[e^{tX}] = \sum e^{tx}P(x)
\][/tex]
For this random variable [tex]\( X \)[/tex]:
[tex]\[
M_X(t) = (e^{-t} \cdot 0.1) + (e^{0 \cdot t} \cdot 0.05) + (e^{2t} \cdot 0.15) + (e^{5t} \cdot 0.4) + (e^{6t} \cdot 0.3)
\][/tex]
### Step 7: Evaluate the MGF at [tex]\( t = 0 \)[/tex]
By definition, [tex]\( M_X(0) \)[/tex] should be 1, as:
[tex]\[
M_X(0) = \sum e^{0 \cdot x} P(x) = \sum 1 \cdot P(x) = \sum P(x) = 1
\][/tex]
So,
[tex]\[
M_X(0) = 1.0
\][/tex]
### Summary:
- Mean ([tex]\(\mu\)[/tex]): [tex]\( 4.0 \)[/tex]
- Variance ([tex]\(\sigma^2\)[/tex]): [tex]\( 5.5 \)[/tex]
- MGF evaluated at [tex]\( t=0 \)[/tex]: [tex]\( 1.0 \)[/tex]
Thus, we have successfully determined the mean, variance, and moment generating function [tex]\( M_X(t) \)[/tex] of the discrete random variable [tex]\( X \)[/tex].
