Linnyk Posted September 30, 2010 Posted September 30, 2010 Hi! I'm trying to work on a problem but it seems I'm really stuck and don't even know where to start. 148 students ride in 4 buses, which transport 25, 33, 40 and 50 students respectively. A student is chosen at random. Let X be the number of student on the bus she's riding. Also, one of the 4 busdrivers are chosen at random. Let Y be the number of students on his bus. Calculate expected values E[X] and E[Y]. (I've calculated expected values before but this question seems different and I'm really at a loss as to where to start.) Would really love some help with this!
dttom Posted September 30, 2010 Posted September 30, 2010 Because the chance to select a driver on bus A is different from the chance to select a student in bus A, so two expected values should be different.
Linnyk Posted September 30, 2010 Author Posted September 30, 2010 Hey again, thanks for helping me! Been working some more on this problem with a friend and we decided E(X)= (40^2)/148 + (33^2)/148 + (25^2) + (50^2)/148 = 39,28 The number seems reasonable. However, I personally don't understand why we square the numbers, could anyone explain this for me? (Is it even correct to do so?) For the busdriver we simply calculated E(Y)= 40/4 + 33/4 + 25/4 + 50/4 = 37
cypress Posted October 1, 2010 Posted October 1, 2010 the formula for expected value is the sum of the products of probability of event and the value given that event occurs. So in the case of the student we have the probability of the student being a rider on bus one at 40/148 and the value given the student is on bus one as 40 so 40^2/148 and so on.... for the driver the probability is 1/4 for each bus and the value given bus one is again 40 so it is 40/4 and so on...
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