91Ó°ÊÓ

In statistics, the Poisson distribution is a distribution function that can be used to describe occurrences having extremely low probabilities of occurring within a certain time or location.

We are given that \(1\)out of \(88\) people have\(ASD\), which indicates that the probability is

\(p = \frac{1}{{88}}.\)

(a):

The random sample contains\(n = 200\)American children. Notice that the given random variable has Binomial distribution with parameters \(n = 200\)and \(p = \frac{1}{{88}}.\)

Proposition: For a binomial random variable \({\rm{X}}\)with parameters\(n,{\rm{p}}\), and\(q = 1 - p\), the following is true

\(\begin{aligned}E(X) &= np\\V(X) &= np(1 - p) = npq\\{\sigma _X} &= \sqrt {npq} \end{aligned}\)

The following is true

\(E(X) = np = 200 \cdot \frac{1}{{88}} = 2.273\)

Also, the variance is

\(V(X) = npq = 200 \cdot \frac{1}{{88}} \cdot \frac{{87}}{{88}} = 2.247\)

Finally, the standard deviation is

\({\sigma _X} = \sqrt {V(X)} = \sqrt {2.247} = 1.499\)

Referring to (a), where we had a random variable with Binomial Distribution, we can use the following conclusion to approximate the distribution.

Proposition: Assume that we have \(b(x;n,p)\)(pmf of binomial random variable), 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 pmf of random variable with Poisson Distribution.

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\(\mu > 0\).

Since

\(np = 200 \cdot \frac{1}{{88}} = 2.273\)

we have that\(\mu = 2.273\)

Step 5: Calculation for the determination of the probability in part b.

Using this distribution, the following holds

\(\begin{aligned}lP(X \ge 2) &= 1 - P(X < 2)\mathop = \limits^{(1)} 1 - P(X \le 1)\\\;\;\;\;\;\;\;\;\;\;\;\; &= 1 - F(1;2.273)\mathop = \limits^{(2)} 1 - 0.337\\\;\;\;\;\;\;\;\;\;\;\;\; &= 0.663\end{aligned}\)

(1): X can take only integer values;

(2): use the following

\(\begin{aligned}F(1;2.273) &= p(0;2.273) + p(1;2.273)\\ &= {e^{ - 2.273}}\frac{{{{2.273}^0}}}{{0!}} + {e^{ - 2.273}}\frac{{{{2.273}^1}}}{{1!}}\\ &= 0.103 + 0.234\\ &= 0.337.\end{aligned}\)

(c):

Referring to (a), where we had a random variable with Binomial Distribution where, \(n = 200\), now we have

\(n = 352,\)

we can use the following conclusion to approximate the distribution.

Proposition: Assume that we have \(b(x;n,p)\)(pmf of binomial random variable), 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 pmf of random variable with Poisson Distribution,

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\(\mu > 0\).

Since

\(np = 352 \cdot \frac{1}{{88}} = 4\)

we have that

\(\mu = 4.\)

The following is true

\(P(X < 5)\mathop = \limits^{(1)} P(X \le 4) = F(4;4)\mathop = \limits^{(2)} 0.629\)

(1): X can take only integer values;

(2): Appendix Table contains the Poisson cdf\(F(x;\mu )\).