91Ó°ÊÓ

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

With parameters, the given random variable has a Binomial Distribution.

\(n = 500\)

\(p = 0.005\;\;\;{\rm{ (given as }}0.5\% !)\)

\(X\)is the exact pmf of a random variable.

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise}}{\rm{. }}}\end{array}} \right.\)

Assume we have \(b(x;n,p)\) (binomial random variable pmf) and that

\(np \to \,\,\,\,\,\,\,\mu > 0\)

When \(n \to \,\,\,\,\,\,\infty \)and \(p \to \,\,\,\,\,0\), then

\(b(x;n,p) \to \,\,\,\,\,\,p(x;\mu )\)

Where \(p(x;\mu )\) is the Poisson Distribution PMF of a random variable.

As a result, because the parameters \({\bf{p}}\)and \({\bf{n}}\)are small and large enough, we can estimate a random variable using the Poisson distribution.

The parameter \(\mu \)is given with

\(\mu = np = 500 \cdot 0.005 = 2.5\)

A random variable \(X\)with pmf,

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

for \(x = 0,1, \ldots \), is said to have Poisson Distribution with parameter \(\mu > 0\$ .\)

Finally, the approximate pmf of \(X\)is the following pmf:

\(p(x,2.5) = {e^{ - 2.5}}\frac{{{{2.5}^x}}}{{x!}},\;\;\;x \in \{ 0,1,2, \ldots \} \)

(b)

So, we have

The following holds,

\(P(X = 5) = p(5;2.5) = {e^{ - 2.5}}\frac{{{{2.5}^5}}}{{5!}} = 0.0668\)

Hence, the required value is \(0.0668\).

(c)

So, we have

The following holds

\(P(X \ge 5) = 1 - P(X \le 4) = 1 - \sum\limits_{x = 0}^4 {{e^{ - 2.5}}} \frac{{{{2.5}^x}}}{{x!}}\)

\(\begin{array}{c}{\rm{ = 1 - 0}}{\rm{.8912}}\\{\rm{ = 0}}{\rm{.1088}}\end{array}\)

Hence, the required value is\(0.1088\).