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) Obtain cumulative distribution function (c.d.f) 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 \( X \):

\[
P(X = x) = \begin{cases} 
3k & \text{for } X = 1 \\
5k & \text{for } X = 2 \\
2k & \text{for } X = 3 \\
k & \text{for } X = 4 \\
k & \text{for } X = 5 
\end{cases}
\]

Since the sum of all probabilities must equal 1 (the total probability must be 1):

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

\[
12k = 1
\]

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

So, the value of \( k \) is \( \frac{1}{12} \).

### Part ii) Find \( P(X) \) when \( X \) is even

The even values of \( X \) are 2 and 4. We need to find the probability that \( X \) is even:

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

Substitute the values from the probability distribution:

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

Using the value of \( k = \frac{1}{12} \):

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

So, \( P(X \text{ is even}) = \frac{1}{2} \).

### Part iii) Obtain the cumulative distribution function (c.d.f) of \( X \)

The cumulative distribution function (c.d.f) \( F(x) \) is defined as:

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

To obtain \( F(x) \), we calculate the cumulative probabilities up to each value of \( X \):

- For \( X = 1 \):

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

- For \( X = 2 \):

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

- For \( X = 3 \):

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

- For \( X = 4 \):

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

\[
F(4) = \frac{11}{12}
\]

- For \( X = 5 \):

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

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

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

Would you like any additional details or have any questions?

Here are some related questions to consider:

1. What is the expected value (mean) of \( X \) for this distribution?
2. How would the variance of \( X \) be calculated for this distribution?
3. How does changing the value of \( k \) affect the distribution?
4. How do you interpret the cumulative distribution function in practical terms?
5. What is the probability that \( X \) is greater than 3?

**Tip:** Understanding the cumulative distribution function (c.d.f) helps in determining the probability that a random variable takes on a value less than or equal to a specific value.

Explore the probability distribution of X with detailed solutions: finding the value of k, calculating even probabilities, and deriving the cumulative distribution function (CDF). Learn key concepts in probability theory applied to discrete random variables. Suitable for college-level students.

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