Suppose that [tex]X \sim P(5.2)[/tex]. Find the following probabilities. Round to three decimals.

a. [tex]P(X=6)[/tex] : [tex]\square[/tex]

b. [tex]P(X \leq 6)[/tex] : [tex]\square[/tex]

c. [tex]P(X \ \textgreater \ 6)[/tex] : [tex]\square[/tex]



Answer :

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.