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Let's address each part of the problem step by step.

### Given:
The probability distribution function is:
\[
P(X = x) = 
\begin{cases}
k(1 - x)^2, & x = -1, 0, 1, 2 \\
0, & \text{otherwise}
\end{cases}
\]

We need to solve for:
1. \( k \)
2. \( E(X) \)
3. \( E(X^2) \)
4. \( \text{Var}(X) \)

### Part (i): Show that \( k = \frac{1}{6} \)

To find \( k \), we use the fact that the total probability must be equal to 1. That is:
\[
P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) = 1
\]
Substituting the values:
\[
k(1 - (-1))^2 + k(1 - 0)^2 + k(1 - 1)^2 + k(1 - 2)^2 = 1
\]
\[
k(2^2) + k(1^2) + k(0^2) + k(-1^2) = 1
\]
\[
k(4 + 1 + 0 + 1) = 1
\]
\[
k(6) = 1 \implies k = \frac{1}{6}
\]

So, we have shown that \( k = \frac{1}{6} \).

### Part (ii): Find \( E(X) \)

The expected value \( E(X) \) is calculated as:
\[
E(X) = \sum_{x} x \cdot P(X = x)
\]
\[
E(X) = -1 \cdot P(X = -1) + 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2)
\]

Using \( k = \frac{1}{6} \), we have:
\[
P(X = -1) = \frac{1}{6}(1 - (-1))^2 = \frac{1}{6}(2^2) = \frac{4}{6} = \frac{2}{3}
\]
\[
P(X = 0) = \frac{1}{6}(1 - 0)^2 = \frac{1}{6}(1) = \frac{1}{6}
\]
\[
P(X = 1) = \frac{1}{6}(1 - 1)^2 = \frac{1}{6}(0) = 0
\]
\[
P(X = 2) = \frac{1}{6}(1 - 2)^2 = \frac{1}{6}(-1)^2 = \frac{1}{6}
\]

Now, substituting these values:
\[
E(X) = -1 \cdot \frac{2}{3} + 0 \cdot \frac{1}{6} + 1 \cdot 0 + 2 \cdot \frac{1}{6}
\]
\[
E(X) = -\frac{2}{3} + 0 + 0 + \frac{2}{6}
\]
\[
E(X) = -\frac{2}{3} + \frac{1}{3} = -\frac{1}{3}
\]

### Part (iii): Find \( E(X^2) \)

The expected value \( E(X^2) \) is calculated as:
\[
E(X^2) = \sum_{x} x^2 \cdot P(X = x)
\]
\[
E(X^2) = (-1)^2 \cdot P(X = -1) + 0^2 \cdot P(X = 0) + 1^2 \cdot P(X = 1) + 2^2 \cdot P(X = 2)
\]
\[
E(X^2) = 1 \cdot \frac{2}{3} + 0 \cdot \frac{1}{6} + 1 \cdot 0 + 4 \cdot \frac{1}{6}
\]
\[
E(X^2) = \frac{2}{3} + 0 + 0 + \frac{4}{6}
\]
\[
E(X^2) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}
\]

### Part (iv): Find \( \text{Var}(X) \)

The variance \( \text{Var}(X) \) is:
\[
\text{Var}(X) = E(X^2) - (E(X))^2
\]
\[
\text{Var}(X) = \frac{4}{3} - \left(-\frac{1}{3}\right)^2
\]
\[
\text{Var}(X) = \frac{4}{3} - \frac{1}{9}
\]

Finding a common denominator:
\[
\text{Var}(X) = \frac{12}{9} - \frac{1}{9} = \frac{11}{9}
\]

### Final Answers:
1. \( k = \frac{1}{6} \)
2. \( E(X) = -\frac{1}{3} \)
3. \( E(X^2) = \frac{4}{3} \)
4. \( \text{Var}(X) = \frac{11}{9} \)

Would you like more details or have any questions about these solutions?

### Further Related Questions:
1. How would the probabilities change if \( k \) were different?
2. Can you explain the significance of variance in a probability distribution?
3. What if \( X \) took different values, like from -2 to 3; how would \( k \) be affected?
4. How can you verify whether a function is a valid probability distribution?
5. What is the role of \( k \) in ensuring this is a proper distribution?

### Tip:
Always check that the total probability sums up to 1 when dealing with discrete probability distributions. This confirms that you've correctly determined any constants like \( k \).

A discrete random variable X has the distribution function P(X = x) = k(1 - x)^2 for x = -1, 0, 1, 2, and 0 otherwise. Find k, E(X), E(X²), and Var(X).

Learn how to calculate the normalization constant k, expected value E(X), second moment E(X²), and variance Var(X) for a discrete random variable X with a given probability distribution. Includes detailed steps and formulas.

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