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Theoretical Distribution

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Question : 50 of 118

Marks: +1, -0

For a Poisson variate

X, P(X=1)=P(X=2)

. What is the mean of

X ?

$1.00$
$1.50$
$2.00$
$2.50$

Solution:

Concept:

The Poisson distribution's probability mass function is

P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}

, where

\lambda

is the mean.

Explanation:

Given

P(X=1)=P(X=2)

.

Write

\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}

.

Cancel

e^{-\lambda}

(nonzero) to get

\lambda = \frac{\lambda^2}{2}

.

Multiply both sides by 2:

2\lambda = \lambda^2

.

Rearrange:

\lambda^2 - 2\lambda = 0

, so

\lambda(\lambda-2)=0

.

Since

\lambda > 0

, we have

\lambda = 2

.

Thus the mean of

X

is

2

.

Answer:

2.00

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