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(D) $26/(2e^{3})$
 
(D) $26/(2e^{3})$
  
Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,  
+
===Solution===
We have to sum for k = 0,1 and 2 and $\lambda$ = 3
+
Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,  
 +
 
 +
We have to sum the probability density function for k = 0,1 and 2 and $\lambda$ = 3 (thus finding the cumulative mass function)
 +
 
 +
=$(1/e^3) + (3/e^3) + (9/e^3)$
 +
 
 +
=$17/(2e^{3})$
 +
 
  
 
<div class="fb-like"  data-layout="standard" data-action="like" data-show-faces="true" data-share="true"></div>
 
<div class="fb-like"  data-layout="standard" data-action="like" data-show-faces="true" data-share="true"></div>

Revision as of 13:35, 8 December 2013

Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?

(A) $8/(2e^{3})$

(B) $9/(2e^{3})$

(C) $17/(2e^{3})$

(D) $26/(2e^{3})$

Solution

Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,

We have to sum the probability density function for k = 0,1 and 2 and $\lambda$ = 3 (thus finding the cumulative mass function)

=$(1/e^3) + (3/e^3) + (9/e^3)$

=$17/(2e^{3})$




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Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?

(A) $8/(2e^{3})$

(B) $9/(2e^{3})$

(C) $17/(2e^{3})$

(D) $26/(2e^{3})$

Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$, 
We have to sum for k = 0,1 and 2 and $\lambda$ = 3



blog comments powered by Disqus