To solve the problem, let's determine the following:

1. The value of $k$
2. The mean of $X$ using the PMF.

1. Finding the value of $k$

The given PMF (Probability Mass Function) table is:

The property of a PMF is that the total probability must sum to $1$:

$f(0) + f(1) + f(2) + f(3) = 1$

Substitute the values of $f(x)$:

$\frac{k}{2} + \frac{k}{3} + \frac{k+1}{3} + \frac{2k-1}{6} = 1$

To simplify, find the LCM of 2, 3, and 6, which is $6$. Rewrite each term with denominator $6$:

$\frac{3k}{6} + \frac{2k}{6} + \frac{2(k+1)}{6} + \frac{2k-1}{6} = 1$

Combine all terms over $6$:

$\frac{3k + 2k + 2(k+1) + (2k-1)}{6} = 1$

Simplify the numerator:

$3k + 2k + 2k + 2 + 2k - 1 = 9k + 1$

Thus, the equation becomes:

$\frac{9k + 1}{6} = 1$

Multiply through by $6$ to eliminate the denominator:

$9k + 1 = 6$

Solve for $k$:

$9k = 5 \quad \implies \quad k = \frac{5}{9}$

2. Finding the mean of $X$

The mean (expected value) $E(X)$ is given by:

$E(X) = \sum x \cdot f(x)$

Using the values of $X$ and $f(x)$:

• For $X = 0$, $f(0) = \frac{k}{2}$
• For $X = 1$, $f(1) = \frac{k}{3}$
• For $X = 2$, $f(2) = \frac{k+1}{3}$
• For $X = 3$, $f(3) = \frac{2k-1}{6}$

Substitute $k = \frac{5}{9}$ into these values:

1. $f(0) = \frac{\frac{5}{9}}{2} = \frac{5}{18}$
2. $f(1) = \frac{\frac{5}{9}}{3} = \frac{5}{27}$
3. $f(2) = \frac{\frac{5}{9} + 1}{3} = \frac{\frac{5+9}{9}}{3} = \frac{\frac{14}{9}}{3} = \frac{14}{27}$
4. $f(3) = \frac{2 \cdot \frac{5}{9} - 1}{6} = \frac{\frac{10}{9} - 1}{6} = \frac{\frac{10-9}{9}}{6} = \frac{1}{54}$

Now calculate $E(X)$:

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

Substitute the values:

$E(X) = 0 \cdot \frac{5}{18} + 1 \cdot \frac{5}{27} + 2 \cdot \frac{14}{27} + 3 \cdot \frac{1}{54}$

Simplify each term:

• $0 \cdot \frac{5}{18} = 0$
• $1 \cdot \frac{5}{27} = \frac{5}{27}$
• $2 \cdot \frac{14}{27} = \frac{28}{27}$
• $3 \cdot \frac{1}{54} = \frac{3}{54} = \frac{1}{18}$

Convert $\frac{1}{18}$ to denominator $27$ (since $\frac{1}{18} = \frac{3}{54} = \frac{2}{27}$):

$E(X) = \frac{5}{27} + \frac{28}{27} + \frac{2}{27}$

Combine terms:

$E(X) = \frac{5 + 28 + 2}{27} = \frac{35}{27}$

Final Answers:

1. The value of $k$ is $\frac{5}{9}$.
2. The mean of $X$ is $\frac{35}{27}$.

Do you need further explanation or simplification? Let me know!

Related Questions:

1. How do you calculate the variance of $X$ using the PMF?
2. What is the cumulative distribution function (CDF) of $X$?
3. How can you check if the PMF satisfies all properties of a probability distribution?
4. How does the mean change if we adjust the PMF values slightly?
5. How do you calculate $P(X \leq 2)$ using this PMF?

Tip: Always verify that the PMF sums to 1 before proceeding with further calculations.