Question

Express the given probability distribution in tabular form.

Answer 1

Step 1: Understand the given probability distribution
The probability distribution of \( P(X = x) \) is given as: \[ P(X = x) = \begin{cases} kx^2 & \text{for } x = 1, 2, 3, \\ 2kx & \text{for } x = 4, 5, 6, \\ 0 & \text{otherwise}. \end{cases} \] Step 2: Write the distribution in tabular form
Substituting the respective values of \( x \), we get: \[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 1 & k \cdot 1^2 = k \\ 2 & k \cdot 2^2 = 4k \\ 3 & k \cdot 3^2 = 9k \\ 4 & 2k \cdot 4 = 8k \\ 5 & 2k \cdot 5 = 10k \\ 6 & 2k \cdot 6 = 12k \\ \hline \end{array} \]

Answer 2

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

Answer 3

Step 1: Use the formula for the mean
 The mean is calculated as: \[ \mu = \sum X \cdot P(X). \] 
Step 2: Substitute the values from the distribution table
From the table: \[ \mu = (1 \cdot k) + (2 \cdot 4k) + (3 \cdot 9k) + (4 \cdot 8k) + (5 \cdot 10k) + (6 \cdot 12k). \] \[ \mu = k(1 + 8 + 27 + 32 + 50 + 72). \] 
Step 3: Simplify and solve for \( \mu \)
\[ \mu = k \cdot 190. \] Substitute \( k = \frac{1}{44} \): \[ \mu = \frac{190}{44} = \frac{95}{22}. \]

Answer 4

Step 1: Identify the required range
We need to find the probabilities for \( X = 2, 3, 4, 5 \). \[ P(1<X<6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5). \] 
Step 2: Substitute the values from the distribution table
 \[ P(1<X<6) = 4k + 9k + 8k + 10k. \] 
Step 3: Simplify and solve
\[ P(1<X<6) = 31k. \] 
Substitute 
\( k = \frac{1}{44} \): \[ P(1<X<6) = \frac{31}{44}. \]