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+4 votes

If the pdf of a Poisson distribution is given by $f(x) = \frac{e^{-2} 2^x}{x!}$ then its mean is

- $2^x$
- $2$
- $-2$
- $1$

+9 votes

Best answer

If we let X = The number of events in a given interval,

Then, if the mean number of events per interval is

The probability of observing x events in a given interval is given by

Poisson Distribution function $P\left ( X=x \right ) =\frac{e^{-\lambda }\lambda ^{x}}{x!}$

Given Distribution function $f\left ( x \right )=\frac{e^{-2 }2 ^{x}}{x!}$

Comparing both the function

SO Mean $\left ( \lambda \right )=2$

Then, if the mean number of events per interval is

The probability of observing x events in a given interval is given by

Poisson Distribution function $P\left ( X=x \right ) =\frac{e^{-\lambda }\lambda ^{x}}{x!}$

Given Distribution function $f\left ( x \right )=\frac{e^{-2 }2 ^{x}}{x!}$

Comparing both the function

SO Mean $\left ( \lambda \right )=2$

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