Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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(D) $26/(2e^{3})$ | (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) | ||
| + | |||
| + | =$(1/e^3) + (3/e^3) + (9/e^3)$ | ||
| + | |||
| + | =$17/(2e^{3})$ | ||
| + | |||
<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> | ||
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)
=$(1/e^3) + (3/e^3) + (9/e^3)$
=$17/(2e^{3})$
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) $26/(2e^{3})$
GATECSEPoisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$,
We have to sum for k = 0,1 and 2 and $\lambda$ = 3