Question

Express the given probability distribution in tabular form.

Answer 1

Step 1: Clarify the provided probability distribution.
The probability mass function \( P(X = x) \) is defined as follows: \[ P(X = x) = \begin{cases} kx^2 & \text{for } x = 1, 2, 3, \\ 2kx & \text{for } x = 4, 5, 6, \\ 0 & \text{otherwise}. \end{cases} \] Step 2: Present the distribution in a table.
By substituting the values of \( x \), the distribution can be represented as: \[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 1 & k \cdot 1^2 = k \\ 2 & k \cdot 2^2 = 4k \\ 3 & k \cdot 3^2 = 9k \\ 4 & 2k \cdot 4 = 8k \\ 5 & 2k \cdot 5 = 10k \\ 6 & 2k \cdot 6 = 12k \\ \hline \end{array} \]

Answer 2

Step 1: Apply the total probability rule. The sum of all probabilities must be 1: \[ k + 4k + 9k + 8k + 10k + 12k = 1. \]
Step 2: Determine the value of \( k \). Simplify the equation: \[ 44k = 1 \implies k = \frac{1}{44}. \]

Answer 3

Step 1: Apply the mean formula.
The mean (μ) is computed as the sum of each value (X) multiplied by its probability P(X): \[ \mu = \sum X \cdot P(X). \] 
Step 2: Insert values from the distribution table.
Using the table data: \[ \mu = (1 \cdot k) + (2 \cdot 4k) + (3 \cdot 9k) + (4 \cdot 8k) + (5 \cdot 10k) + (6 \cdot 12k). \] Factor out k: \[ \mu = k(1 + 8 + 27 + 32 + 50 + 72). \] 
Step 3: Simplify and determine μ.
\[ \mu = k \cdot 190. \] With \( k = \frac{1}{44} \), the mean is: \[ \mu = \frac{190}{44} = \frac{95}{22}. \]

Answer 4

Step 1: Determine the relevant range. We need the probability for \( X = 2, 3, 4, 5 \). This can be written as: \[ P(1 < X < 6) = P(X=2) + P(X=3) + P(X=4) + P(X=5) \]

Step 2: Substitute probabilities from the distribution table: \[ P(1 < X < 6) = 4k + 9k + 8k + 10k \]

Step 3: Simplify and calculate: \[ P(1 < X < 6) = 31k \] Substituting \( k = \frac{1}{44} \), \[ P(1 < X < 6) = \frac{31}{44} \]

Final Answer: \[ \boxed{\frac{31}{44}} \]