Aspire Faculty ID #16297 · Topic: NIMCET 2010 · Just now
NIMCET 2010

n a Poisson distribution if $P[X=3] = \frac{1}{4} P[X=4]$ then $P[X=5] = k P[X=7]$ where $k$ equals to

Solution

Poisson PMF: 
$P[X=x] = e^{-\lambda} \frac{\lambda^x}{x!}$ 
Given: $P(3) = \frac{1}{4} P(4)$ 
 $\frac{\lambda^3/3!}{\lambda^4/4!} = \frac{1}{4}$ 
 $\frac{4}{\lambda} = \frac{1}{4}$ 
 $\lambda = 16$ 
 Now find $k$: 
$\frac{P(5)}{P(7)} = \frac{\lambda^5/5!}{\lambda^7/7!} = \frac{7 \cdot 6}{\lambda^2}$ 
 Substitute $\lambda = 16$: 
$\frac{42}{256} = \frac{21}{128}$ 
 So, $k = \frac{21}{128}$

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