statistics - expected and probable numbers

Question

The task:

The number of oil tankers, arriving at a certain refinery on one day, follow a Poisson distribution with the parameter $u = 2$. The present harbour facilities can serve 3 oil tankers a day. If more than 3 oil tankers arrive on a given day, the additional tankers (in excess of the 3) are sent to another refinery

Question 1: What is the probability that oil tankers are sent to another refinery?

Question 2: By how much should the capacity of the refinery harbour be extended, if it is desired that on a given day the probability of sending away arriving oil tankers does not exceed 0.05? 0.01?

Question 3: What is the expected number of oil tankers arriving on a given day?

Question 4: What is the most probable number of oil tankers arriving on a given day?

Question 5: What is the expected number of oil tankers served on a given day?

Question 6: What is the expected number of oil tankers sent on to other refineries on a given day?

I've tried to answer Question 1 and 3 and got stuck on question 4.

Q 1:

We let X be the value of oil tankers arriving each day.

Then we find the expected value of $P(X>3) = 0.35277$

(sorry don't know how to do the fancy equation stuff, but if the answer is correct I assume I've done that one right)

Q 3:

The expected number of a Poisson distribution with paramter $u = u$

therefor the expected number of oil tankers arriving on a given day is 2 right?

Q 4:

I belive it's something to do with the poisson formula right?

I've gotten so far:

$$\frac{2^k e^{-k}} k = P(k)$$

where i find P(0),P(1),P(2),P(3) etc? this is how far I've gotten. and I don't know what to do with these numbers as they don't make sense by themselves. Obviously there is no a 100% probability of 0 oil tankers arriving daily when the mean is 2

As for Question 2. I've tried using the same method I did for question 1. But replacing the maximum number of tankers it can serve to $y%, and replacing 3 with $y$ in the entire equation then solving for it gives 0.05.

but that didn't work out at all. :/

and I am completly blank on question 5 and 6 :(

hope you can help me out!

Thanks in advance geniuses!

Answer 1

Let $X$ be the number of tankers arriving on a randomly chosen day.

Q1: The probability that on a given day one or more oil tankers is sent to another refinery is $1$ minus the probability that $X\gt 3$. This is $$1-e^{-2}\left(1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}\right).$$ I think that does not produce your number. Moreover, we want $\Pr(X\gt 3)$, not the "expected value $\dots$."

Q2: This one will require some experimentation. Suppose we extend capacity to $k$. We want to choose $k$ so that $\Pr(X\gt k)\le 0.05$. So we want $\Pr(X\le k)\gt 0.95$. Calculation shows that $k=4$ almost gets us there, and $k=5$ gets us there with plenty to spare.

Do a similar calculation for $0.01$.

For similar problems with different numbers, one might want to use more sophisticated techniques to estimate the "tail" of the Poisson.

Q3: Trivial. By the way, it is probably $\mu$, not $u$.

Q4: The probability the number is $k$ is $p_k$, where $p_k=e^{-2}\frac{\mu^{k}}{k!}$.

In our case, we can just compute the first few values, and observe that the maximum is reached at $2$ places, $k=1$ and $k=2$. But in addition we do this using more general tools. We have $$\frac{p_{k+1}}{p_k}=\frac{e^{-2}\frac{2^k}{k!}}{e^{-2}\frac{2^{k+1}}{(k+1)!}}=\frac{2}{k+1}$$ (there was a lot of cancellation). If $k+1\lt 2$, we have $p_{k+1}\gt p_k$. If $k+1=2$ we have equality, and if $k+1\gt 2$ then $p_{k+1}\lt p_k$. For "most probable" there is therefore a tie between $k=1$ and $k=2$.

Q5: Let $Y$ be the number of tankers served. The random variable $Y$ takes on the values $0$, $1$, $2$, or $3$.

The probabilities $\Pr(Y=0)$, $\Pr(Y=1)$, and $\Pr(Y=2)$ are just the Poisson probabilities. For $\Pr(Y=3)$, compute $1-\Pr(X\le 3)$. Now that you have $\Pr(Y=k)$ for the various values of $k$, you can compute $E(Y)$ in the usual way.

Q6: Let $Z$ be the number of tankers sent away. We want $E(Z)$. The easiest way to calculate this is to note that $Y+Z=X$ (here $Y$ is the random variable we used i Question 5). Thus $E(Y+Z)=E(X)=2$. By the linearity of expectation, we have $E(Y+Z)=E(Y)+E(Z)$. Thus $E(Z)=2-E(Y)$.

Added: On request, we do the detailed calculation for Question 5. The number $Y$ of tankers serviced is either $0$, $1$, $2$, or $3$.

We have $\Pr(Y=0)=e^{-2}\frac{2^0}{0!}=e^{-2}$. Also, $\Pr(Y=1)=e^{-2}\frac{2^1}{1!}=2e^{-2}$. Similarly, $\Pr(Y=2)=2e^{-2}$. The probability that $Y=3$ is calculated a little differently. For $Y=3$ if $3$ or more tankers show up. (Then $3$ are serviced, and the rest sent away.) Thus $\Pr(Y=3)=1-(e^{-2}+2e^{-2}+2e^{-2})=1-5e^{-2}$. It follows that $$E(Y)=(0)(e^{-2})+(1)(2e^{-2})+(2)(2e^{-2})+(3)(1-5e^{-2}).$$ This simplifies to $3-9e^{-2}$. Numerically, it is about $1.782$.

Answer 2

Firstly, the expression of the Poisson distribution is $$ P(X,u) = \frac{u^{X}\exp^{-u}}{{X!}} $$ where $X$ represents the number of ships arriving. (This is different to the equation you outlined in your question above) As for the questions 1 assuming that you used the right form of the equation then yes P(X>3) is correct.

2) what value of X do we require to have a probability $p$ i.e. p=0.05 or 0.01 hint:re-arrange the above equation?

3) This requires taking a look on http://en.wikipedia.org/wiki/Poisson_distribution

4) The most likely number is related to the maximum value of the distribution i.e. the peak location. Try a few numbers.

5) The expected number of ships served is given by $$ \sum_{X=0}^{X=3}X P(X,u) $$

6) Similarly as question 5 except that the number served,$X$, at the other refinery is less the 3 that are served at the initial one.