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: Use the property of Poisson distribution.
For a Poisson distribution, \[ \text{Mean}=\text{Variance} \] Thus, \[ \ell=m \] Given, \[ \ell+m=8 \] So, \[ \ell+\ell=8 \] \[ 2\ell=8 \] \[ \ell=4 \] Hence, the Poisson parameter is \[ \lambda=4 \]

Step 2: Rewrite the probability expression.
We need to find \[ e^4\left[1-P(X\gt 2)\right] \] Since \[ 1-P(X\gt 2)=P(X\leq2), \] we get \[ e^4P(X\leq2) \]

Step 3: Expand \(P(X\leq2)\).
For Poisson distribution, \[ P(X=r)=e^{-\lambda}\frac{\lambda^r}{r!} \] Therefore, \[ P(X\leq2)=P(0)+P(1)+P(2) \] \[ =e^{-4}\left( \frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} \right) \] \[ =e^{-4}\left( 1+4+\frac{16}{2} \right) \] \[ =e^{-4}(1+4+8) \] \[ =e^{-4}(13) \]

Step 4: Multiply by \(e^4\).
\[ e^4P(X\leq2) = e^4\cdot e^{-4}(13) \] \[ =13 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{13} \]