If a random variable X has a Poisson distribution such that P(X=1)=P(X=2) then
I: its mean is 2 II: its variance is 1

Options

A. 1, 1
B. 2 ,2
C. `2 , sqrt3`
D. `2 , 4`

Answer

To solve the problem, we need to analyze the given information about the random variable \( X \) which follows a Poisson distribution. We know that for a Poisson distribution, the probability mass function is given by:

\[ P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!} \]

where \( \lambda \) is the mean (and also the variance) of the distribution.

Given that \( P(X = 1) = P(X = 2) \), we can set up the equation:

\[ P(X = 1) = P(X = 2) \]

Substituting the formula for the probabilities:

\[ \frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!} \]

Now, we can simplify this equation. First, we can cancel \( e^{-\lambda} \) from both sides since it is common:

\[ \frac{\lambda^1}{1!} = \frac{\lambda^2}{2!} \]

This simplifies to:

\[ \lambda = \frac{\lambda^2}{2} \]

Now, we can multiply both sides by 2 to eliminate the fraction:

\[ 2\lambda = \lambda^2 \]

Rearranging gives us:

\[ \lambda^2 - 2\lambda = 0 \]

Factoring out \( \lambda \):

\[ \lambda(\lambda - 2) = 0 \]

This gives us two solutions:

1. \( \lambda = 0 \)
2. \( \lambda = 2 \)

Since we are dealing with a Poisson distribution, \( \lambda \) must be non-negative, so we take \( \lambda = 2 \).

Now, we can conclude:

- The mean of the distribution \( \mu = \lambda = 2 \).
- The variance of the distribution \( \sigma^2 = \lambda = 2 \).

Now, let's evaluate the statements:

I: Its mean is 2. **True**
II: Its variance is 1. **False** (since the variance is also 2).

Thus, the final answer is:

- Mean = 2
- Variance = 2