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The provided discrete random variable \( X \) has a probability distribution function (p.d.f.) given by:
\[ 
f(x) = \begin{cases} 
kx(x-3) & \text{for } x = 1, 2, 3, 4 \\
0 & \text{elsewhere} 
\end{cases}
\]

### (a) Value of \( k \)
To find \( k \), recall that the sum of all probabilities must equal 1:
\[
\sum_{x=1}^{4} kx(x-3) = 1
\]
Computing \( x(x-3) \) for \( x = 1, 2, 3, 4 \) yields:
- \( 1(1-3) = -2 \)
- \( 2(2-3) = -2 \)
- \( 3(3-3) = 0 \)
- \( 4(4-3) = 4 \)

Thus:
\[
k(-2) + k(-2) + k(0) + k(4) = 1 \\
-4k + 4k = 1 \\
0 = 1 \text{ (which is not possible)}
\]
The initial calculations seem incorrect. Let's recompute.

### Recomputation for \( k \)
\[
k \left[1 \cdot (-2) + 2 \cdot (-1) + 3 \cdot 0 + 4 \cdot 1\right] = 1 \\
k \left[-2 -2 + 0 + 4\right] = 1 \\
k \cdot 0 = 1 \text{ (still incorrect)}
\]
Let's go through the correct calculation of \( k \) and compute the sum properly.

### (b) \( P(X > 2 | X \leq 5) \)
Since \( X \) takes values only from 1 to 4, \( P(X \leq 5) = 1 \), and:
\[ P(X > 2) = P(X = 3) + P(X = 4) \]
We'll calculate these probabilities once \( k \) is found.

### (c) Expected Value \( E(2X - 1) \)
The expected value of a transformed random variable is calculated by:
\[ E(2X-1) = 2E(X) - 1 \]
We'll find \( E(X) \) first.

### (d) Median and (e) Mode
The median is the middle value of the probability distribution when the cumulative probability reaches 0.5. The mode is the value of \( x \) with the highest probability.

Let's proceed to correct the calculation of \( k \), then use it to address all parts (a) through (e).The solution for \( k \) based on the initial conditions and calculations is indicating an inconsistency, implying an error in setting up the problem or interpreting the function values \( x(x-3) \) for \( x = 1, 2, 3, 4 \).

Let's verify the correct computation of \( x(x-3) \) for \( x = 1, 2, 3, 4 \) to ensure there are no computational errors, and then resolve for \( k \) accordingly:

- For \( x = 1 \): \( 1 \times (1 - 3) = -2 \)
- For \( x = 2 \): \( 2 \times (2 - 3) = -2 \)
- For \( x = 3 \): \( 3 \times (3 - 3) = 0 \)
- For \( x = 4 \): \( 4 \times (4 - 3) = 4 \)

Given these recalculations are correct, the sum of these weighted by \( k \) should add up to 1. Let's manually compute this sum again and verify the correct value of \( k \).It seems like I can’t do more advanced data analysis right now. Please try again later. If you have any further questions or need assistance with another topic, feel free to ask!

If you're interested in further exploration or clarifications, here are some additional questions you might consider:

1. How do we calculate the expected value for a discrete random variable?
2. What are the differences between the median and the mode in a probability distribution?
3. How would you interpret a probability distribution function like the one given?
4. Can you explain why we need to solve for \( k \) in a probability distribution function?
5. What steps are involved in verifying the correctness of a probability distribution function?

**Tip:** Always double-check your mathematical expressions and calculations to avoid errors, especially when dealing with probability distribution functions.

A discrete random variable has p.d.f given by f(x) = kx(x-3) for x = 1,2,3,4, and 0 elsewhere. Find the following: (a) value of k, (b) P(x > 2 | x ≤ 5), (c) E(2x - 1), (d) median, (e) mode.

Learn how to calculate the constant 'k' for a discrete random variable, conditional probability P(x > 2 | x ≤ 5), the expected value of a function E(2x - 1), and determine the median and mode of the distribution.

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