To solve the given problem, we need to determine the value of $k$ and evaluate various probabilities. Let’s go through the steps:

(i) Find $k$

The sum of all probabilities in a probability distribution must equal 1:

$k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$

Simplify the expression:

$8k + 10k^2 = 1$

This is a quadratic equation:

$10k^2 + 8k - 1 = 0$

To find $k$, use the quadratic formula:

$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 10$, $b = 8$, and $c = -1$.

$k = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 10 \cdot (-1)}}{2 \cdot 10}$ $k = \frac{-8 \pm \sqrt{64 + 40}}{20}$ $k = \frac{-8 \pm \sqrt{104}}{20}$ $k = \frac{-8 \pm 2\sqrt{26}}{20}$ $k = \frac{-4 \pm \sqrt{26}}{10}$

Choose the positive value for $k$:

$k = \frac{-4 + \sqrt{26}}{10}$

(ii) Evaluate $P(X < 6)$, $P(X \geq 6)$, and $P(0 < X < 5)$

• $P(X < 6)$:

$P(X < 6) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$ $= k + 2k + 2k + 3k + k^2$ $= 8k + k^2$

• $P(X \geq 6)$:

$P(X \geq 6) = P(X=6) + P(X=7)$ $= 2k^2 + (7k^2 + k)$ $= 9k^2 + k$

• $P(0 < X < 5)$:

$P(0 < X < 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4)$ $= k + 2k + 2k + 3k$ $= 8k$

(iii) If $P(X \leq k) > \frac{1}{2}$, find the minimum value of $k$ and determine the distribution function of $X$.

From part (i), we already solved for $k$. Because the probabilities must sum to 1 and follow the general probability conditions, this scenario naturally calculates to a correct probability distribution already.

For the distribution function of $X$, calculate cumulative probabilities:

• $F(0) = 0$
• $F(1) = k$
• $F(2) = k + 2k = 3k$
• $F(3) = 3k + 2k = 5k$
• $F(4) = 5k + 3k = 8k$
• $F(5) = 8k + k^2$
• $F(6) = 8k + k^2 + 2k^2 = 8k + 3k^2$
• $F(7) = 1$

This process completes the determination for $k$ and valid probabilities based on given conditions.