91Ó°ÊÓ

Let $X$be a Poisson random variable with a mean of$20$.

Suppose that $X$is a Poisson random variable with a parameter$\mathrm{λ}$. It is given that $X$is a random variable with a mean of$20$. Since the expected value and variance of a Poisson random variable are both equal to its parameter$\mathrm{λ},X$has mean $\mathrm{μ}=\mathrm{λ}=20$and variance${\mathrm{σ}}^2=\mathrm{λ}=20$.

By Markov's inequality,

$p=P\{Xâ‰\yen 26\}≤\frac{E\left[X\right]}{26}=\frac{20}{26}=.7692$

Using Corollary$5.1$, localid="1649901568148" $\underset{¯}{a=6}$we get:

$p=P\{Xâ‰\yen \underset{=26}{\underset{ÁåŸ}{20+6}}\}≤\frac{{\mathrm{σ}}^2}{{\mathrm{σ}}^2+6^2}=\frac{20}{20+36}=.\mathbf{3571}$

See Example$5d$. Since $X$is a Poisson random variable with parameterlocalid="1649901587221" $\mathrm{λ}=20$using the result

$P\{Xâ‰\yen i\}≤\frac{e^{-\mathrm{λ}}(e\mathrm{λ}{)}^i}{i^i}$

We get,

$p=P\{Xâ‰\yen 26\}≤\frac{e^{-20}(20e{)}^{26}}{{26}^{26}}=.4398$

Using the central limit theorem we get:

$p=P\{Xâ‰\yen 26\}\stackrel{\text{the continuity correction}}{=}P(Xâ‰\yen 25.5)$

$=1-P\{X<25.5\}=1-P\left\{\frac{X-20}{\sqrt{20}}<\frac{25.5-20}{\sqrt{20}}\right\}≈1-\mathrm{Φ}(1.23)$

$\text{Table 5.1 (textbook, Chapter 5)}1-.8907=.\mathbf{1093}\text{.}$

Using software package by personal choice, we obtain:

$p=P\{Xâ‰\yen 26\}=1-P\{X<26\}=1-\underset{i=0}{\overset{25}{∑}}{e}^{-20}\frac{{20}^i}{i!}$

$=1-.887815=\mathbf{0}\mathbf{.}\mathbf{112185}$