Question

If \(X\) is a Poisson variate such that \(4P(X = 2) = 3P(X = 1)\), then the mean of the distribution is equal to:

• A. 1.5
• B. 2
• C. 3
• D. 4

Hint:

Write both probabilities with the Poisson formula, cancel the common factors and solve for the mean.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a Poisson variate with mean \(\lambda\), the probability of r events is \(P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}\). The mean of the distribution is \(\lambda\).

Step 2: Write both probabilities.
\[ P(X=2) = \frac{e^{-\lambda}\lambda^2}{2} \] \[ P(X=1) = e^{-\lambda}\lambda \]

Step 3: Use the given condition.
\[ 4 \cdot \frac{e^{-\lambda}\lambda^2}{2} = 3 e^{-\lambda}\lambda \] Cancel \(e^{-\lambda}\) and one \(\lambda\) (the mean cannot be 0): \[ 2\lambda = 3 \] \[ \lambda = 1.5 \]

Step 4: Check the options.
The mean equals lambda = 1.5, which is option 1. For lambda = 2, the left side gives 4P(2) = 4 e^-2 x 2 = 8 e^-2 while the right side gives 3P(1) = 6 e^-2, so option 2 fails. Options 3 and 4 fail in the same way.

Final Answer:
The mean of the distribution is 1.5. \[ \boxed{1.5} \]