Question

For a Poisson distribution, if mean \(=\ell\), variance \(=m\) and \[ \ell+m=8, \] then \[ e^4\left[1-P(X\gt 2)\right]= \]

• A. \(8\)
• B. \(13\)
• C. \(9\)
• D. \(12\)

Hint:

For a Poisson distribution: \[ \text{Mean}=\text{Variance}=\lambda. \] Also, \[ P(X=r)=e^{-\lambda}\frac{\lambda^r}{r!}. \]

Answer 1

The Correct Option is B

Step 1: Write down the given conditions.
For a Poisson distribution the mean $= \ell$ and variance $= m$. But for a Poisson distribution, mean $=$ variance, so $\ell = m$. Given $\ell + m = 8$, we have $2\ell = 8$, so $\ell = m = 4$.
Step 2: Write the Poisson probability formula.
$P(X = k) = \frac{e^{-\ell} \ell^k}{k!} = \frac{e^{-4} 4^k}{k!}$.
Step 3: Find $P(X > 2)$.
$P(X > 2) = 1 - P(X \le 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = e^{-4}$, $P(X=1) = 4e^{-4}$, $P(X=2) = \frac{16}{2}e^{-4} = 8e^{-4}$.
So $P(X \le 2) = e^{-4}(1+4+8) = 13e^{-4}$.
Thus $P(X > 2) = 1 - 13e^{-4}$.
Step 4: Find $1 - P(X > 2)$.
$1 - P(X > 2) = 13e^{-4}$.
Step 5: Compute $e^4[1 - P(X > 2)]$.
$e^4 \cdot 13e^{-4} = 13$.
Step 6: Match with options.
The answer is 13, which is option (2).
\[ \boxed{13} \]