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Statistic assignment help needed .

  1. Oct 1, 2009 #1
    Well, first i will have to make an apology for putting the question in this forum, but was not really sure where else i could put it. :)
    The assignment is:

    The number of oil tankers, arriving at a certain refinery on one day, follow a poisson distribution with the parameter µ=2. The present harbor 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 3) are sent to another refinery.

    Q1. What is the probability that oil tankers are sent to other refineries?

    Q2. By how much should the capacity of the refinery harbor 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?

    Q3. What is the expected number of oil tankers arriving on a given day?

    Q4. What is the most probable number of oil tankers arriving on a given day?

    Q5. What is the expected number of oil tankers served on a given day?

    Q6. What is the expected number of oil tankers sent on to other refineries on a given day?

    I have made Q1+Q2 and got the answers : Q1 = 0,142877 Q2 = 2 and 3

    But i cant work out how the last questions should be calculated?

    We have got some leads though,

    3. How many oil tankers can one expect to arrive on one day?
    4. Which outcome has (or have) the highest probability?
    5. Construct a random variable Y: Numbers of oil tankers served on one day; state the outcomes {0,1,2,3} and the probabilities associated.
    6. Construct a random variable S: Numbers of oil tankers sent on to other refineries on one day; state the outcomes {4,5,…} and the probabilities associated – or apply “Laws of expected value”

    Can someone help me ?
     
    Last edited: Oct 1, 2009
  2. jcsd
  3. Oct 1, 2009 #2

    HallsofIvy

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    (3) is easy- the expected value of a Poisson distribution with parameter [itex]\mu[/itex] is [itex]\mu[/itex]!

    (4) what k gives the maximum value of the Poisson formula,
    [tex]P(k)= \frac{\mu^k e^{-k}}{k!}[/tex]?
    Just find P(0), P(1), P(2), P(3), P(4), etc.

    (5) How is (5) different from (3)? Did you copy it incorrectly?

    (6) Find the expected value of P(n> 3).
     
  4. Oct 1, 2009 #3
    I copied it correctly, thats why i dont get it.

    I cant really spot the difference between the questions, but thanks for your answers ;)
     
  5. Oct 1, 2009 #4

    HallsofIvy

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    Oh, I'm sorry! I just read them again and while 3 asks for the expected number of tankers to arrive on a given day, 5 asks for the expected number of tanker served on a given day. Theoretically, any number of tankers could arrive on a given day and my response to 3 was correct. But no more that 3 tankers can be served on a given day. Since you find the 'expected' value by multiplying the outcome by the probability of that outcome and adding, to find the expected value of the number of tankers served you take 0P(0)+ 1P(1)+ 2P(2)+ 3(P(3)+ P(4)+ ...) since if 3 or more tankers arrive 3 will be served. (Of course, P(3)+ P(4)+ ...= 1- (P(0)+ P(1)+ P(2)).)
     
  6. Oct 8, 2009 #5
    Thanks, but can you explain it a little better?
     
  7. Oct 8, 2009 #6
    Oh, I'm sorry! I just read them again and while 3 asks for the expected number of tankers to arrive on a given day, 5 asks for the expected number of tanker served on a given day. Theoretically, any number of tankers could arrive on a given day and my response to 3 was correct. But no more that 3 tankers can be served on a given day. Since you find the 'expected' value by multiplying the outcome by the probability of that outcome and adding, to find the expected value of the number of tankers served you take 0P(0)+ 1P(1)+ 2P(2)+ 3(P(3)+ P(4)+ ...) since if 3 or more tankers arrive 3 will be served. (Of course, P(3)+ P(4)+ ...= 1- (P(0)+ P(1)+ P(2)).)

    1-0,6767 = 0,3233

    And as a result of that, the highest probability is that 3 tankers are served, because 3+ is now the most likely number of tankers arriving?
     
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