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

There are a few

\({\rm{n = 1000}}\)

persons from which one in every 200 people carries the gene, indicating that the likelihood of carrying the gene is low.

\({\rm{p = }}\frac{{\rm{1}}}{{{\rm{200}}}}\)

As a result, the distribution of such a random variable is Binomial, with \({\rm{n = 1000}}\) and \({\rm{p = 1/200}}\) as parameters.

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

\({\rm{np}} \to {\rm{\mu > 0}}\)

If \({\rm{n}} \to \infty \) and \({\rm{p}} \to {\rm{0}}\) are true, then

where p(x; u) is the Poisson Distribution PMF of a random variable.

In our scenario, we might use Poisson Random Variable X with parameter to approximate the specified binomial random variable.

\(\begin{array}{c}{\rm{\mu = np}}\\{\rm{ = 1000 \times 0}}{\rm{.005}}\\{\rm{ = 5}}\end{array}\)

(a) Using random variable X, the following is correct:

\(\begin{aligned}{\rm{P(5}} \le {\rm{X}} \le {\rm{8) = F(8;5) - F(4;5)}}\\&= 0{\rm{.932 - 0}}{\rm{.440}}\\ &= 0{\rm{.492}}\end{aligned}\)

(1):the cdf of a Poisson random variable is functionF.

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

Therefore, the value is:\({\rm{P(5}} \le {\rm{X}} \le {\rm{8) = 0}}{\rm{.492}}\).

\(\begin{aligned}{\rm{P(X}} \ge {\rm{8) = P(X < 8)}}\\ &= 1 - P(X \le {\rm{7)}}\\ &= 1 - F(7;5)\\ &= 1 - 0{\rm{.867}}\\ &= 0 {\rm{.133}}\end{aligned}\)

(1):Only integer values are allowed in X.

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

Therefore, the value is: \({\rm{P(X}} \ge {\rm{8) = 0}}{\rm{.133}}\).