Consider the following probability distribution of X
X 1 2 3 4 5
P(X=x) 3k 5k 2k k k
i) Find the value of k
ii) P(x) is even
iii) Calculate distribution function of X

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

### Part (i) Find the value of \( k \)

Given the probability distribution of \( X \):

\[
\begin{array}{c|c|c|c|c|c}
X & 1 & 2 & 3 & 4 & 5 \\
\hline
P(X = x) & 3k & 5k & 2k & k & k \\
\end{array}
\]

The sum of all probabilities must equal 1 because they represent the complete probability distribution:

\[
3k + 5k + 2k + k + k = 1
\]

Simplifying:

\[
12k = 1
\]

So,

\[
k = \frac{1}{12}
\]

### Part (ii) \( P(X = x) \) when \( x \) is even

The even values of \( X \) are 2 and 4. The probability that \( X \) is even is:

\[
P(X \text{ is even}) = P(X = 2) + P(X = 4)
\]

Substituting the values:

\[
P(X \text{ is even}) = 5k + k = 6k
\]

Using the value of \( k \) found earlier:

\[
P(X \text{ is even}) = 6 \times \frac{1}{12} = \frac{6}{12} = \frac{1}{2}
\]

### Part (iii) Calculate the distribution function \( F(x) \) of \( X \)

The distribution function \( F(x) \) is defined as:

\[
F(x) = P(X \leq x)
\]

Let's compute \( F(x) \) for each possible value of \( x \):

\[
F(1) = P(X \leq 1) = P(X = 1) = 3k = 3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}
\]

\[
F(2) = P(X \leq 2) = P(X = 1) + P(X = 2) = 3k + 5k = 8k = 8 \times \frac{1}{12} = \frac{8}{12} = \frac{2}{3}
\]

\[
F(3) = P(X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3) = 3k + 5k + 2k = 10k = 10 \times \frac{1}{12} = \frac{10}{12} = \frac{5}{6}
\]

\[
F(4) = P(X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 3k + 5k + 2k + k = 11k = 11 \times \frac{1}{12} = \frac{11}{12}
\]

\[
F(5) = P(X \leq 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 12k = 12 \times \frac{1}{12} = 1
\]

So, the distribution function \( F(x) \) is:

\[
F(x) =
\begin{cases}
0 & \text{if } x < 1 \\
\frac{1}{4} & \text{if } 1 \leq x < 2 \\
\frac{2}{3} & \text{if } 2 \leq x < 3 \\
\frac{5}{6} & \text{if } 3 \leq x < 4 \\
\frac{11}{12} & \text{if } 4 \leq x < 5 \\
1 & \text{if } x \geq 5 \\
\end{cases}
\]

### Summary:
- \( k = \frac{1}{12} \)
- \( P(X \text{ is even}) = \frac{1}{2} \)
- The distribution function \( F(x) \) is piecewise defined as described above.

Would you like more details on any part of this solution? Here are 5 related questions to deepen your understanding:

1. How do you derive the cumulative distribution function (CDF) from a probability mass function (PMF)?
2. What is the significance of the normalization condition (sum of all probabilities equals 1)?
3. How would the distribution function \( F(x) \) change if the probability distribution were different?
4. Can you explain the relationship between the distribution function and the probability of an event?
5. How would you calculate \( P(X \) is odd)?

**Tip:** Remember, the cumulative distribution function (CDF) \( F(x) \) gives you the probability that a random variable \( X \) is less than or equal to \( x \).

Explore how to solve a probability distribution problem with variables X and P(X). Learn to find the value of k, calculate even probabilities, and derive the distribution function of X step-by-step. Suitable for advanced high school students and those studying probability theory.

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