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| − | From the graph once can see that if a = A/2 then E[|X-a|] = A/4 | + | From the graph once can see that if a = A/2 then E[|X-a|] = A/4 This is also intuative because if the station is in the middle of the road then at most it will have to drive A/2 in one direction along the road to get to the fire and on average the truck will have to drive (A/2)/2 = A/4. |
Latest revision as of 10:54, 6 October 2008
This problem is similar to the breaking the stick problem we worked on in class.
Let 'X' be the distance that that the fire truck will have to go to get to a fire, and knowing that the fires will accure uniformly between 0 to A:
X => Unif(0, A)
let 'a' be the location of the fire station.
Know lets examin two cases:
CASE 1: a = 0 station is at the start of the road
E[|X-0|] = A/2 intuitivly this makes sense. It mease that on average the truck will have to go A/2 distance to fight the fire.
CASE 2: a = A station is at the end of the road
E[|X-A|] = A/2 intuitivly this also makes sense. The truck will have to go A/2 distance to fight the fire.
Since this is a uniform distribution E[|X-a|] ca be viewd as perabila:
From the graph once can see that if a = A/2 then E[|X-a|] = A/4 This is also intuative because if the station is in the middle of the road then at most it will have to drive A/2 in one direction along the road to get to the fire and on average the truck will have to drive (A/2)/2 = A/4.
