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A discrete random variable is one that can only take on a finite number of different values.

Proposition: A Poisson Random Variable with parameter \({\rm{\mu = \alpha t}}\) may be used to simulate the number of events that occur during a time interval of length.This implies that:

Process of Poisson

That is given to us.

\({\rm{\alpha = 80}}\)

\(\begin{aligned}{\rm{\alpha \times 0}}{\rm{.25 = 80 \times 0}}{\rm{.25}}\\ &= 20 \end{aligned}\)

The following is true for:

\(\begin{aligned}{{\rm{P}}_{{\rm{16}}}}{\rm{(0}}{\rm{.25) = P(X}} \le {\rm{16)}}\\{\rm{ = F(16;20)}}\\{\rm{ = 0}}{\rm{.221}}\end{aligned}\)

(1):The Poisson cdf \({\rm{F(X;\mu )}}\) is found in Appendix Table \({\rm{A}}{\rm{.2}}\).

Therefore, the value is:\({\rm{P(X}} \le {\rm{16) = 0}}{\rm{.221}}\).

\(\begin{array}{c}{\rm{E(X) = V(X)}}\\{\rm{ = \mu }}\end{array}\)

Our random variable X would now have a new parameter. That is still the case.

\({\rm{\alpha = 80}}\)

However, since we now know that the forest encompasses \({\rm{85,000}}\) acres,

\({\rm{t = 85,000}}\)

The following statement is correct:

\(\begin{aligned} E(X) &= \mu \\ &= \alpha \times t \\ &= 80 \times 85,000 \\ & = 6,800,000 \end{aligned}\)

Therefore, the number of tress are: \({\rm{6,800,000}}\).

\(\begin{array}{c}{{\rm{r}}^{\rm{2}}}{\rm{\pi = 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times \pi }}\\{\rm{ = 0}}{\rm{.03141593}}\end{array}\)

miles squared Since

\({\rm{1 sq mile = 640 acres}}\)

That is something we have.

\({\rm{0}}{\rm{.03141593 \times 640 = 20}}{\rm{.1062}}\)

acres.

Keep in mind that

\({\rm{\alpha = 80}}\)

and we now have

\({\rm{t = 20}}{\rm{.1062}}\)

As a result, our random variable X follows a Poisson distribution with a parameter.

\(\begin{aligned} mu &= \alpha \times 20 {\rm{.1062}}\\ &= 80 \times 20 {\rm{.1062}}\\ &= 1608 {\rm{.5}}\end{aligned}\)

pmf with a random variable X

\({\rm{p(x;\mu ) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

Poisson Distribution with parameter \({\rm{\mu > 0}}\) is claimed to exist for \({\rm{x = 0,1,}}.....\).

As a result, X's pmf is

\({\rm{p(x;1608}}{\rm{.5) = }}{{\rm{e}}^{{\rm{ - 1608}}{\rm{.5}}}}\frac{{{\rm{1608}}{\rm{.}}{{\rm{5}}^{\rm{x}}}}}{{{\rm{x!}}}}\)

It is for\({\rm{x = 0,1,}}.....\)

Therefore, the value is: \({\rm{p(x;1608}}{\rm{.5) = }}{{\rm{e}}^{{\rm{ - 1608}}{\rm{.5}}}}\frac{{{\rm{1608}}{\rm{.}}{{\rm{5}}^{\rm{x}}}}}{{{\rm{x!}}}}\).