Arjun Suresh (talk | contribs) |
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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? | + | 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) $8/(2e^{3})$ | (A) $8/(2e^{3})$ | ||
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Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$, | Poisson Probability Density Function (with mean $\lambda$) = $\lambda^{k} / (e^{\lambda}k!)$, | ||
| − | We have to sum the probability density function for k = 0,1 and 2 and $\lambda$ = 3 (thus finding the cumulative mass function) | + | We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda$ = 3 (thus finding the cumulative mass function) |
=$(1/e^3) + (3/e^3) + (9/2e^3)$ | =$(1/e^3) + (3/e^3) + (9/2e^3)$ | ||
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| − | {{Template: | + | {{Template:FBD}} |
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[[Category: Probability and Combinatorics]] | [[Category: Probability and Combinatorics]] | ||
[[Category: GATE2013]] | [[Category: GATE2013]] | ||
[[Category: Probability questions]] | [[Category: Probability questions]] | ||
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