poisson probability

**lord12** · September 24th 2008, 08:41 PM

The number of oil tankers arriving at a refinery each day has a Poisson distribution with parameter $\lambda = 3$ . Current port facilities can service only four tankers a day so that if more than four tankers arrive in a day, the additional tankers must be sent to another port.

What is the expected number of tankers serviced daily at the current port facilities?

Is is simply $3$ ? The other information is extraneous?

**CaptainBlack** · September 25th 2008, 02:28 AM

Originally Posted by lord12

The number of oil tankers arriving at a refinery each day has a Poisson distribution with parameter $\lambda = 3$ . Current port facilities can service only four tankers a day so that if more than four tankers arrive in a day, the additional tankers must be sent to another port.

What is the expected number of tankers serviced daily at the current port facilities?

Is is simply $3$ ? The other information is extraneous?

No:

$E(N)=p(0) \times 0 + p(1) \times 1+ p(2) \times 2 + p(3) \times 3 + p(n\ge 4) \times 4 \ne 3$

( $p(n \ge 4) = 1-(p(0)+p(1)+p(2)+p(3))$ )

Where $p(n)$ is the probability that $n$ tankers arrive in a day.

The sum is truncated at 4 because if anymone than 4 tankers arrive they are not serviced.

RonL

**lord12** · September 25th 2008, 03:20 AM

I don't understand. then, why for a Possion Random Variable, the expected value and variance is always equal to lambda.

**CaptainBlack** · September 25th 2008, 03:23 AM

Originally Posted by lord12

I don't understand. then, why for a Possion Random Variable, the expected value and variance is always equal to lambda.

The number serviced is not a poisson RV, but it is truncated, all the ships after the 4th are sent away and so not serviced. So ther properties of the poisson distribution are only indirectly involved, and then by explicit calculation of the expectation.

RonL

**lord12** · September 25th 2008, 03:28 AM

So if as n approaches infinity, the expected value approaches lambda?

**mr fantastic** · September 25th 2008, 03:33 AM

Originally Posted by lord12

I don't understand. then, why for a Possion Random Variable, the expected value and variance is always equal to lambda.

The number arriving has a Poisson distribution. But this is a different random variable to the number serviced.

It's the expected value of the number arriving that's equal to 3. Not the number serviced.

**math beginner** · October 10th 2008, 12:08 AM

I'd like to add on another part to the question. Suppose lambda is 1.5 and present port facilities can service three tankers a day. If more than 3 tankers arrive in a day, the tankers in excess of 3 must be sent to another port.

How much must present facilities be increased to permit handling all arriving tankers on approximately 90per cent of the days?

**CaptainBlack** · October 10th 2008, 01:49 AM

Originally Posted by math beginner

I'd like to add on another part to the question. Suppose lambda is 1.5 and present port facilities can service three tankers a day. If more than 3 tankers arrive in a day, the tankers in excess of 3 must be sent to another port.

How much must present facilities be increased to permit handling all arriving tankers on approximately 90per cent of the days?

Let the numer that can be serviced per day at the expanded port be n.

The the question asks what is the smallest $n$ be so that

$\sum_{i=1}^n p(i)<0.9$

where $p(i)$ is the the probability of $i$ arrivals in a day.

Now try this for $n=0,1,2,3,..$

RonL

**math beginner** · October 10th 2008, 04:01 AM

I am so sorry, I am not clear about the p(i). Could you explain?

**CaptainBlack** · October 10th 2008, 04:18 AM

Originally Posted by math beginner

I am so sorry, I am not clear about the p(i). Could you explain?

You are told that the number of arrivals per day has a Poisson distribution so:

$p(i)=f(i,\lambda)=\frac{\lambda^ie^{-lambda}}{i!}$

You are told tha $\lambda=1.5$

RonL

**math beginner** · October 10th 2008, 02:50 PM

sorry i'm still stuck. i calculate poisson probability for i=0 + i=1 + i=2 + i=3 and already the cumulative probability is 0.9344. The current capacity of 3 tankers is already serving 90% of the tankers. The question asks how much must present facilities be increased so as to serve all the tankers 90% of the time...my calculation suggest that they do not need to increase facilities. Something is wrong...please could you help?

**CaptainBlack** · October 10th 2008, 02:53 PM

Originally Posted by math beginner

sorry i'm still stuck. i calculate poisson probability for i=0 + i=1 + i=2 + i=3 and already the cumulative probability is 0.9344. The current capacity of 3 tankers is already serving 90% of the tankers. The question asks how much must present facilities be increased so as to serve all the tankers 90% of the time...my calculation suggest that they do not need to increase facilities. Something is wrong...please could you help?

That's what my calculations indicate as well.

RonL

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