The Poisson probability formula is given by:

$P(X = k) = \frac{e^{-\mu} \cdot \mu^k}{k!}$

Where:

• $\mu = 1.8$ (the mean of the distribution),
• $k = 2$ (the value we're interested in).

Let's plug in the values to calculate $P(X = 2)$:

$P(X = 2) = \frac{e^{-1.8} \cdot 1.8^2}{2!}$

Now, I will compute this.

The probability that $X = 2$ for a Poisson distribution with $\mu = 1.8$ is approximately $0.2678$.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. What is the probability that $X = 0$ for $\mu = 1.8$?
2. How would the probability change if $\mu = 2.5$?
3. What is the cumulative probability $P(X \leq 2)$ for this distribution?
4. How does the Poisson distribution differ from the normal distribution?
5. Can you compute the mean and variance of the Poisson distribution?

Tip: The mean and variance of a Poisson distribution are both equal to $\mu$.