Probability
Poisson Distribution
Grade 12
Question:
<p>A random variable \(X\) has Poisson distribution with mean 2. \(P(X > 1.5)\) equals</p>
<p>\(\dfrac{2}{e^2}\)</p>
<p>0</p>
<p>\(1 - \dfrac{3}{e^2}\)</p>
<p>\(\dfrac{3}{e^2}\)</p>
Step-by-Step Solution
Key Concept: For a discrete random variable like Poisson, P(X > 1.5) = P(X ≥ 2) since X only takes integer values. Use the cumulative property and the Poisson PMF formula.
<p><strong>Step 1:</strong> Recognize that X is discrete, so P(X > 1.5) = P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - [P(X = 0) + P(X = 1)]</p><p><strong>Step 2:</strong> For Poisson with mean λ = 2: P(X = k) = (e^(-2) · 2^k) / k!</p><p>P(X = 0) = e^(-2) · 2^0 / 0! = e^(-2)</p><p>P(X = 1) = e^(-2) · 2^1 / 1! = 2e^(-2)</p><p><strong>Step 3:</strong> P(X ≤ 1) = e^(-2) + 2e^(-2) = 3e^(-2)</p><p><strong>Step 4:</strong> P(X > 1.5) = 1 - 3e^(-2) = 1 - 3/e²</p><p>∴ Answer: C</p>
Correct Answer: C