Question 1193544
**i. Find two equations involving p and q:**

1. **Sum of Probabilities:**
   The sum of all probabilities in a probability distribution must equal 1. 
   Therefore: 0.1 + p + q + 0.3 + 0.1 = 1
   Simplifying: p + q = 0.5 

2. **Expected Value (E(X)):**
   E(X) = Σ [x * P(X = x)] 
   Given E(X) = 3.1, we have: 
   3.1 = (1 * 0.1) + (2 * p) + (3 * q) + (4 * 0.3) + (5 * 0.1) 
   3.1 = 0.1 + 2p + 3q + 1.2 + 0.5 
   3.1 = 1.8 + 2p + 3q
   Simplifying: 2p + 3q = 1.3

**ii. Find the values of p and q:**

* We have two equations:
    * p + q = 0.5 
    * 2p + 3q = 1.3

* Solve this system of equations:
    * From the first equation, isolate p: p = 0.5 - q 
    * Substitute this value of p into the second equation: 
        2(0.5 - q) + 3q = 1.3 
        1 - 2q + 3q = 1.3 
        q = 0.3 
    * Substitute the value of q back into the first equation: 
        p + 0.3 = 0.5 
        p = 0.2

* Therefore, p = 0.2 and q = 0.3

**iii. Find Var(X)**

* **Calculate E(X²)**:
   E(X²) = Σ [x² * P(X = x)] 
   E(X²) = (1² * 0.1) + (2² * 0.2) + (3² * 0.3) + (4² * 0.3) + (5² * 0.1) 
   E(X²) = 0.1 + 0.8 + 2.7 + 4.8 + 2.5 
   E(X²) = 10.9

* **Calculate Var(X)**:
   Var(X) = E(X²) - [E(X)]² 
   Var(X) = 10.9 - (3.1)² 
   Var(X) = 10.9 - 9.61 
   Var(X) = 1.29

**iv. Find Var(2X - 3)**

* Use the property: Var(aX + b) = a² * Var(X) 
   where a = 2 and b = -3

* Var(2X - 3) = 2² * Var(X) 
   Var(2X - 3) = 4 * 1.29 
   Var(2X - 3) = 5.16

**Summary:**

* p = 0.2
* q = 0.3
* Var(X) = 1.29
* Var(2X - 3) = 5.16