idiopatia0f

Answered

2022-01-16

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

Answer & Explanation

Archie Jones

Expert

2022-01-16Added 34 answers

Explanation:

The Poisson distribution is

$P\left(x\right)=\frac{{e}^{-\mu}{\mu}^{x}}{x!}$

$\mu =3$

3! = 3*2*1

$P(x\ge 2)=1-P\left(1\right)-P\left(0\right)$

$P\left(1\right)=\frac{{e}^{-3}\cdot {3}^{1}}{1!}=\frac{3}{{e}^{3}}$

$P\left(0\right)={e}^{-3}=\frac{1}{{e}^{3}}$

Therefore,

$P(x\ge 2)=1-P\left(1\right)-P\left(0\right)=1-0.0498-0.1494=0.8009$

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$P(x\ge 2)$ means 1-P(x=0)-P(x=1)

Here$\mu =3$ and ${e}^{-\mu}={e}^{-3}=0.049787$

and hence, desired probability is

$1-\frac{{e}^{-3}{3}^{0}}{0!}-\frac{{e}^{-3}{3}^{1}}{1!}$

$=1-{e}^{-3}\times (1+3)$

$=1-0.049787\times 4$

=1-0.199148

=0.800852

The Poisson distribution is

3! = 3*2*1

Therefore,

Explanation:

In a Poisson probability distribution, if mean value of success is

the probability of getting x successes is given by

Now

Here

and hence, desired probability is

=1-0.199148

=0.800852

peterpan7117i

Expert

2022-01-17Added 39 answers

Compute $P(X\ge 2)$

The required probability is

$P(X\ge 2)=1-P\left(X<2\right)$

=1-[P(X=0)+P(X=1)]

=1-[0.049787+0.149361] from statistical table

=0.8009

The required$P(X\ge 2)=0.8009$

The required probability is

=1-[P(X=0)+P(X=1)]

=1-[0.049787+0.149361] from statistical table

=0.8009

The required

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