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(D) $26/(2e^{3})$ | (D) $26/(2e^{3})$ | ||
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Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$, | Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$, | ||
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 the probability density function for k = 0,1 and 2 and $\lambda$ = 3 (thus finding the cumulative mass function)
=$(1/e^3) + (3/e^3) + (9/2e^3)$
=$17/(2e^{3})$
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 the probability density function for k = 0,1 and 2 and $\lambda$ = 3 (thus finding the cumulative mass function)
=$(1/e^3) + (3/e^3) + (9/2e^3)$
=$17/(2e^{3})$
GATECSE