If X is a Poisson random variate with mean 3, then \(P(|X-3|<2)\) =
- \( \frac{9}{2e^3} \)
- \( \frac{99}{8e^3} \)
- \( \frac{3}{2e^3} \)
- \( \frac{1}{3e^3} \)
The Correct Option is A
Solution and Explanation
Given: \(X\) is a Poisson random variable with mean \( \lambda = 3 \).
The probability mass function (PMF) of a Poisson distribution is:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] So, for this case: \[ P(X = k) = \frac{e^{-3} \cdot 3^k}{k!} \]
We are asked to compute:
\[ P(|X - 3| < 2) \]
Step 1: Interpret the inequality
\[ |X - 3| < 2 \Rightarrow -2 < X - 3 < 2 \Rightarrow 1 < X < 5 \] Since \(X\) takes only integer values, this means: \[ X = 2, 3, 4 \]
Step 2: Calculate each probability
- \( P(X=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{9}{2}e^{-3} \)
- \( P(X=3) = \frac{e^{-3} \cdot 3^3}{3!} = \frac{27}{6}e^{-3} = \frac{9}{2}e^{-3} \)
- \( P(X=4) = \frac{e^{-3} \cdot 3^4}{4!} = \frac{81}{24}e^{-3} = \frac{27}{8}e^{-3} \)
Step 3: Add the probabilities
\[ P(|X - 3| < 2) = \left( \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) e^{-3} = \left( 9 + \frac{27}{8} \right)e^{-3} = \frac{99}{8}e^{-3} \]
Final Answer:
\( \boxed{\frac{99}{8e^3}} \)
This matches the answer marked correct in the options (option b).
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