 idiopatia0f

2022-01-16

Consider a Poisson distribution with $\mu =3$. What is $P\left(x\ge 2\right)$? Archie Jones

Expert

Explanation:
The Poisson distribution is
$P\left(x\right)=\frac{{e}^{-\mu }{\mu }^{x}}{x!}$
$\mu =3$
3! = 3*2*1
$P\left(x\ge 2\right)=1-P\left(1\right)-P\left(0\right)$
$P\left(1\right)=\frac{{e}^{-3}\cdot {3}^{1}}{1!}=\frac{3}{{e}^{3}}$
$P\left(0\right)={e}^{-3}=\frac{1}{{e}^{3}}$
Therefore,
$P\left(x\ge 2\right)=1-P\left(1\right)-P\left(0\right)=1-0.0498-0.1494=0.8009$
Explanation:
In a Poisson probability distribution, if mean value of success is $\mu$,
the probability of getting x successes is given by
$P\left(x\right)=\frac{{e}^{-\mu }{\mu }^{x}}{x!}$
Now $P\left(x\ge 2\right)$ means 1-P(x=0)-P(x=1)
Here $\mu =3$ and ${e}^{-\mu }={e}^{-3}=0.049787$
and hence, desired probability is
$1-\frac{{e}^{-3}{3}^{0}}{0!}-\frac{{e}^{-3}{3}^{1}}{1!}$
$=1-{e}^{-3}×\left(1+3\right)$
$=1-0.049787×4$
=1-0.199148
=0.800852 peterpan7117i

Expert

Compute $P\left(X\ge 2\right)$
The required probability is
$P\left(X\ge 2\right)=1-P\left(X<2\right)$
=1-[P(X=0)+P(X=1)]
=1-[0.049787+0.149361] from statistical table
=0.8009
The required $P\left(X\ge 2\right)=0.8009$

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