ISO nerds: any statistics whizzes out there? -
June 11, 2008, 11:31 AM

I know I'll probably catch a bunch of sh1t for posting this (as it does not pertain to fun), but any chance there are some math/statistics whizzes out there who wanna help me pass summer school? Taking an engineering stat class that's pretty tough and could certainly use help with the homework.

This post was prepared as a service to the DCSportbikes.net community. Neither the DCSportbikes.net ADMINS nor any of the moderators, makes any warranty, expressed or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, product, link, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial products, process, or service by trade name, trademark manufacturer, or otherwise, does not necessarily constitute or imply its endorsement, recommendation, or favoring by DCSportbikes.net. The opinions of this author expressed herein do not necessarily state or reflect those of the DCSportbikes.net community, and shall not be used for advertising or product endorsement purposes.
__________________________________________________ ________

I'm taking STAT250 at GMU this semester. Not that I'll be that big of a help, but I've got the concepts down pretty good thus far. If it's basic and we've covered it already, I'll be happy to give my .02, wouldn't hurt for me either to be interactive with it...

A small market orders a magazine every week. Suppose the store owner pays $1 for each copy and sells it for $2. If magazines leftover have no salvage value, is it better to order 3 or 4 copies? (hint: for both 3 and 4 copies ordered, express net revenue as a function of demand X and then compute the expected revenue.)

Let X = demand for the magazine, and the pmf is

x p(x)
1 1/15
2 2/15
3 3/15
4 4/15
5 3/15
6 2/15

I have the answer, but not sure how to get to it. Expected value =
E[h3(X)] = 2.466 and E[h4(X)]=2.667, so it's better to order 4.

15 telephones have just been received at a service center. 5 are cellular, 5 are corded, and 5 are cordless. Suppose that these components are randomly allocated the numbers 1,2,3....15 to establish the order in which they will be serviced.

a.What is the probability that after servicing 10 of them, phones of only 2 of the 3 types remain to be serviced?

b. What is the probability that 2 phones of each type are among the first 6 to be serviced?

Well, I am no statistics guru, but it seems that you should be able to take the profit/loss for each scenario and multiply it by the probability of that scenario and add them up. For example, under the Buy 3 scenario, you have a 1/15 chance of selling 1, a 2/15 chance of selling 2 and a 12/15 chance of selling 3. The loss on selling 1 would be (1), the profit on selling 2 would be 1. and the profit on selling 3 would be 3. Do the same for buying 4 and see what happens.

Well, I am no statistics guru, but it seems that you should be able to take the profit/loss for each scenario and multiply it by the probability of that scenario and add them up. For example, under the Buy 3 scenario, you have a 1/15 chance of selling 1, a 2/15 chance of selling 2 and a 12/15 chance of selling 3. The loss on selling 1 would be (1), the profit on selling 2 would be 1. and the profit on selling 3 would be 3. Do the same for buying 4 and see what happens.

I get what you're saying....it's perfect common sense. But I have to come up with the 2 expected values that I have as the solution, and it just doesn't add up right. The exp. val. is nothing more than the probability times the X value. If I take the sum of the (prob. times X) for X=1, 2,and 3 I get .934