Question

For a Poisson distribution:
- Mean $ = \ell $
- Variance $ = m $
- $ \ell + m = 8 $
Then evaluate: $$ e^4 \cdot \left[1 - P(X>2)\right] $$

• A. 8
• B. 13
• C. 9
• D. 12

Hint:

Use the complement rule \( P(X>r) = 1 - P(X \leq r) \), and the Poisson formula: \( P(r) = \frac{e^{-\lambda} \lambda^r}{r!} \)

Answer 1

The Correct Option is B

In Poisson distribution: - Mean \( \lambda = \ell \) - Variance = \( \lambda = m \Rightarrow \ell = m \) - Given: \( \ell + m = 8 \Rightarrow 2\lambda = 8 \Rightarrow \lambda = 4 \) So: \[ P(X>2) = 1 - P(X \leq 2) \Rightarrow 1 - \left[P(0) + P(1) + P(2)\right] \] Poisson formula: \[ P(r) = \frac{e^{-\lambda} \lambda^r}{r!} \Rightarrow \lambda = 4 \] Compute: \[ P(0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4} \] \[ P(1) = \frac{e^{-4} \cdot 4^1}{1!} = 4e^{-4} \] \[ P(2) = \frac{e^{-4} \cdot 4^2}{2!} = \frac{16}{2} e^{-4} = 8e^{-4} \] Sum: \[ P(X \leq 2) = e^{-4}(1 + 4 + 8) = 13e^{-4} \Rightarrow 1 - P(X>2) = 13e^{-4} \] So: \[ e^4 \cdot (1 - P(X>2)) = e^4 \cdot 13e^{-4} = \boxed{13} \]