Question

If \(X\) is a Poisson random variable with \(P(X=0)=0.8\), then the variance of \(X\) is equal to

• A. \(\ln\frac45\)
• B. \(\ln\frac54\)
• C. \(\ln\frac1{20}\)
• D. \(\ln20\)

Hint:

For a Poisson distribution, \[ \boxed{ P(0)=e^{-\lambda}, \qquad \text{Mean}=\text{Variance}=\lambda. } \]

Answer 1

The Correct Option is B

Concept: If \[ X\sim\text{Poisson}(\lambda), \] then \[ P(X=x) = \frac{e^{-\lambda}\lambda^x}{x!}, \] and \[ \boxed{\text{Mean}=\text{Variance}=\lambda.} \]

Step 1: Use \(P(X=0)\). For \(x=0\), \[ P(X=0) = e^{-\lambda}. \] Given, \[ e^{-\lambda}=0.8=\frac45. \]

Step 2: Find \(\lambda\). Taking logarithm, \[ -\lambda = \ln\frac45. \] Hence, \[ \lambda = -\ln\frac45 = \ln\frac54. \] Since variance equals \(\lambda\), \[ \boxed{ \operatorname{Var}(X) = \ln\frac54. } \] Therefore, the correct option is \[ \boxed{(B).} \]