# Poisson Problem

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• Jan 9th 2009, 11:40 PM
zorro
Poisson Problem
If a random variable X follows Poisson distribution such that
P(X=1) = P(X=2)
Compute
i) the mean of the distribution
ii) P(X=0)
iii) standard deviation of the distribution
• Jan 9th 2009, 11:48 PM
running-gag
Hi

Poisson distribution with parameter $\lambda$ is such as
$P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}$

$P(X=1) = e^{-\lambda} \lambda$

$P(X=2) = e^{-\lambda} \frac{\lambda^2}{2}$

Therefore

$e^{-\lambda} \lambda = e^{-\lambda} \frac{\lambda^2}{2}$

which allows to compute the value of $\lambda$
• Jan 10th 2009, 12:03 AM
zorro
How this
Quote:

Originally Posted by running-gag
Hi

Poisson distribution with parameter $\lambda$ is such as
$P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}$

$P(X=1) = e^{-\lambda} \lambda$

$P(X=2) = e^{-\lambda} \frac{\lambda^2}{2}$

Therefore

$e^{-\lambda} \lambda = e^{-\lambda} \frac{\lambda^2}{2}$

which allows to compute the value of $\lambda$

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I am getting $\lambda=2$ is this right
And
i) mean of distribution is $\mu$=2
since $\mu = \lambda$

ii) $P(X=0) = \frac{2^0 \ e^{-2}}{0!}$
=0.135

iii) $\sigma^2 = \mu$
ie $\sigma^2 = 2$
Standard deviation = $\sqrt{2}$
• Jan 10th 2009, 12:07 AM
running-gag
I agree with your answers ! (Clapping)