SOLUTION: The random variable X has a probability distribution x. 1 2 3 4 5 P X x ( = ) 0.1 p q 0.3 0.1 i. Given that E X( ) = 3.1, write down two equations invo

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Question 1193544: The random variable X has a probability distribution
x. 1 2 3 4 5
P X x ( = ) 0.1 p q 0.3 0.1
i. Given that E X( ) = 3.1, write down two equations involving p and q .
ii. Find the values of p and q .
iii. Find Var X( )
iv. Var X (2 3 .
Answer by yurtman(42)   (Show Source): You can put this solution on YOUR website!
**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
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