Certainly! Let's delve into each of the parts of the problem where [tex]\( X \)[/tex] follows a Poisson distribution with a mean (λ) of 5.2.
### a. [tex]\( P(X=6) \)[/tex]
To find the probability of exactly 6 events occurring, which is [tex]\( P(X=6) \)[/tex], we use the probability mass function (PMF) of the Poisson distribution:
[tex]\[
P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}
\][/tex]
For [tex]\( k=6 \)[/tex] and [tex]\( \lambda=5.2 \)[/tex],
[tex]\[
P(X=6) = \frac{e^{-5.2} \cdot 5.2^6}{6!}
\][/tex]
Using the appropriate computations,
[tex]\[
P(X=6) \approx 0.151.
\][/tex]
### b. [tex]\( P(X \leq 6) \)[/tex]
To find the cumulative probability up to and including 6 events, [tex]\( P(X \leq 6) \)[/tex], we use the cumulative distribution function (CDF) of the Poisson distribution which sums up the probabilities from 0 to 6:
[tex]\[
P(X \leq 6) = \sum_{k=0}^{6} P(X=k)
\][/tex]
After computing each term and summing them up,
[tex]\[
P(X \leq 6) \approx 0.732.
\][/tex]
### c. [tex]\( P(X > 6) \)[/tex]
To find the probability of more than 6 events occurring, which is [tex]\( P(X>6) \)[/tex], we can use the complement rule since we already know [tex]\( P(X \leq 6) \)[/tex]:
[tex]\[
P(X > 6) = 1 - P(X \leq 6)
\][/tex]
Given that [tex]\( P(X \leq 6) \approx 0.732 \)[/tex],
[tex]\[
P(X > 6) = 1 - 0.732 \approx 0.268.
\][/tex]
Therefore, the probabilities are:
- a. [tex]\( P(X=6) \approx 0.151 \)[/tex]
- b. [tex]\( P(X \leq 6) \approx 0.732 \)[/tex]
- c. [tex]\( P(X>6) \approx 0.268 \)[/tex]
These values are rounded to three decimal places as required.
