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A contractor has submitted bid on four state jobs: a car parking garage, an office building, a theater, and a shopping mall. State rules do not allow a contractor to be offered more than one of these jobs. If this contractor is awarded any of these jobs, the profits earned from these contracts are: RM2 million from the parking garage, RM3 million from the office building, RM5 million from the theater, and RM7 million from the shopping mall. He gets no profit if he does not get any contract.

The contractor estimates that the probabilities of getting the car parking garage contract, the office building contract, the theater contract, the shopping mall contract, or nothing are 0.25, 0.15, 0.35, 0.13, and 0.12 respectively.

Let Y be the random variable that represents the contractor’s profits in millions of ringgits.

  1. Write the probability distribution of Y. (4 marks)

  2. Construct the cumulative probability distribution of Y (4 marks)

  3. Find the expected profit of the contractor. Give a brief interpretation of the value obtained (5 marks)

  4. Estimate the probability of the profit more than what is expected by the contractor. (2 marks)

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    $\begingroup$ see the faq on how this site works. Homework questions are answered here, but more effort must be put in by the person asking. $\endgroup$ – mpiktas May 6 '12 at 8:00
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    $\begingroup$ to start with, try to rephrase in your question the problem without the "story" on top of it, namely Y takes such and such values, with probability this much and that much, which is essentially question 1. $\endgroup$ – Xi'an May 6 '12 at 8:08
  • $\begingroup$ I like the problem posed the way it is. It makes it more interesting. $\endgroup$ – Michael R. Chernick May 6 '12 at 15:34
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    $\begingroup$ I think Xi'an didn't mean it as a criticism, he gave a hint that rephrasing it without the story basically already gives you the answer to 1. $\endgroup$ – Erik May 6 '12 at 16:01
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Part 1 is very simple the discrete probability distribution is RM2 million with probability 0.25 RM3 million with probability 0.15 RM5 million with probability 0.35 RM7 million with probability 0.13 and 0 with probability 0.12 This really just a restatement of what was described in the problem. For part II to get the probability that Y is less than Z just sum up the probabilities above that have earning less than Z. Expected profit is just the values for Y listed in the above distribution multiplied by its respective probability and summed together. For part IV once you have calculated E(Y), look at all the values of Y>E(Y). Add the corresponding probabilities for all those cases. That will be the probability that Y>E(Y) and is the answer to part IV. The use of the word "estimate" is not appropriate. There is no statistical estimation involved. It would be better to have used the word "calculate".

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