Question

If \(3\) is the variance of Poisson distribution, then \[ P(1\lt x\lt 4)= \] is:

• A. \(\dfrac{123}{8}e^{-3}\)
• B. \(3e^{-\sqrt{3}}\)
• C. \(9e^{-3}\)
• D. \(\left(\dfrac{3+\sqrt{3}}{2}\right)e^{-3}\)

Hint:

For Poisson distribution, \[ \text{Mean}=\text{Variance}=\lambda \] Always identify the value of \(\lambda\) first before calculating probabilities.

Answer 1

The Correct Option is C

Step 1: Use the property of Poisson distribution.
For a Poisson distribution, \[ \text{Mean}=\text{Variance}=\lambda \] Given variance \[ =3 \] Therefore, \[ \lambda=3 \]

Step 2: Interpret the probability.
We need \[ P(1\lt x\lt 4) \] Since \(x\) takes integer values in Poisson distribution, \[ 1\lt x\lt 4 \] means \[ x=2 \quad \text{or} \quad x=3 \] Thus, \[ P(1\lt x\lt 4)=P(x=2)+P(x=3) \]

Step 3: Use the Poisson probability formula.
The Poisson probability function is \[ P(x=r)=\frac{e^{-\lambda}\lambda^r}{r!} \] For \(x=2\), \[ P(x=2)=\frac{e^{-3}3^2}{2!} \] \[ =\frac{9e^{-3}}{2} \] For \(x=3\), \[ P(x=3)=\frac{e^{-3}3^3}{3!} \] \[ =\frac{27e^{-3}}{6} \] \[ =\frac{9e^{-3}}{2} \]

Step 4: Add the probabilities.
Therefore, \[ P(1\lt x\lt 4) = \frac{9e^{-3}}{2} + \frac{9e^{-3}}{2} \] \[ =9e^{-3} \]

Step 5: Final conclusion.
Hence, \[ \boxed{9e^{-3}} \]