Determine the value of k so that the following function can serves as a probability 
distribution of the discrete random variable X. And also find mean and variance of X 
𝑓(𝑥) = 𝑘(𝑥^3 + 1), 𝑓𝑜𝑟𝑥 = 0,1,2,3.

To determine the value of \( k \) such that the given function \( f(x) = k(x^3 + 1) \) serves as a probability distribution of the discrete random variable \( X \), we need to ensure that the sum of the probabilities over the entire support of \( X \) equals 1. Additionally, we will calculate the mean and variance of \( X \).

### Step 1: Determine \( k \)

The probabilities must sum to 1. The possible values of \( x \) are 0, 1, 2, and 3. Therefore, we need to ensure that:

\[
\sum_{x=0}^{3} f(x) = 1
\]

Substituting \( f(x) = k(x^3 + 1) \):

\[
f(0) = k(0^3 + 1) = k(1) = k
\]
\[
f(1) = k(1^3 + 1) = k(2)
\]
\[
f(2) = k(2^3 + 1) = k(9)
\]
\[
f(3) = k(3^3 + 1) = k(28)
\]

Now, the sum of the probabilities is:

\[
f(0) + f(1) + f(2) + f(3) = k(1) + k(2) + k(9) + k(28)
\]

Simplifying:

\[
k(1 + 2 + 9 + 28) = k(40)
\]

Since the total probability must equal 1:

\[
k(40) = 1 \quad \Rightarrow \quad k = \frac{1}{40}
\]

### Step 2: Calculate the Mean (Expected Value) \( E(X) \)

The mean \( E(X) \) of a discrete random variable is given by:

\[
E(X) = \sum_{x=0}^{3} x \cdot f(x)
\]

Substitute \( f(x) = \frac{1}{40}(x^3 + 1) \):

\[
E(X) = 0 \cdot f(0) + 1 \cdot f(1) + 2 \cdot f(2) + 3 \cdot f(3)
\]

Using the values of \( f(x) \):

\[
E(X) = 0 \cdot \frac{1}{40}(1) + 1 \cdot \frac{1}{40}(2) + 2 \cdot \frac{1}{40}(9) + 3 \cdot \frac{1}{40}(28)
\]

Simplifying:

\[
E(X) = 0 + \frac{2}{40} + \frac{18}{40} + \frac{84}{40}
\]

\[
E(X) = \frac{2 + 18 + 84}{40} = \frac{104}{40} = 2.6
\]

### Step 3: Calculate the Variance \( \text{Var}(X) \)

The variance \( \text{Var}(X) \) is given by:

\[
\text{Var}(X) = E(X^2) - (E(X))^2
\]

First, calculate \( E(X^2) \):

\[
E(X^2) = \sum_{x=0}^{3} x^2 \cdot f(x)
\]

Substituting \( f(x) = \frac{1}{40}(x^3 + 1) \):

\[
E(X^2) = 0^2 \cdot f(0) + 1^2 \cdot f(1) + 2^2 \cdot f(2) + 3^2 \cdot f(3)
\]

Simplifying:

\[
E(X^2) = 0 \cdot \frac{1}{40}(1) + 1 \cdot \frac{1}{40}(2) + 4 \cdot \frac{1}{40}(9) + 9 \cdot \frac{1}{40}(28)
\]

\[
E(X^2) = 0 + \frac{2}{40} + \frac{36}{40} + \frac{252}{40}
\]

\[
E(X^2) = \frac{2 + 36 + 252}{40} = \frac{290}{40} = 7.25
\]

Now, calculate the variance:

\[
\text{Var}(X) = E(X^2) - (E(X))^2 = 7.25 - (2.6)^2
\]

\[
\text{Var}(X) = 7.25 - 6.76 = 0.49
\]

### Final Answers:
- The value of \( k \) is \( \frac{1}{40} \).
- The mean \( E(X) = 2.6 \).
- The variance \( \text{Var}(X) = 0.49 \).

Would you like more details on any step? Here are some additional related questions you might find useful:

1. How do you calculate the cumulative distribution function (CDF) for this random variable?
2. What is the standard deviation of this random variable?
3. How can we verify that the distribution sums to 1 using a different approach?
4. How does the shape of the probability mass function affect the mean?
5. How is skewness related to the variance and mean of a distribution?

**Tip:** For discrete probability distributions, always ensure the probabilities sum to 1 before computing other statistics.

Determine the value of k so that the following function can serve as a probability distribution of the discrete random variable X. Also, find the mean and variance of X, where f(x) = k(x^3 + 1) for x = 0, 1, 2, 3.

Learn how to determine the value of k such that f(x) = k(x^3 + 1) forms a probability distribution for X = 0, 1, 2, 3. The solution covers the calculation of k, mean, and variance, using fundamental probability and statistical formulas. Suitable for college-level students.

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