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An ambulance travels back and forth at a constant speed along a road of length L. At a certain moment of time, an accident occurs at a point uniformly distributed on the road. [That is, the distance of the point from one of the fixed ends of the road is uniformly distributed over (0, L).] Assuming that the ambulance’s location at the moment of the accident is also uniformly distributed, and assuming independence of the variables, compute the distribution of the distance of the ambulance from the accident.

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Answer:

The distribution of distance of ambulance from accident is [tex]\frac{2(L-a)}{L^2}\\[/tex]

Step-by-step explanation:

  • Let X and Y be the random variables associated as

               X=Position of Ambulance on Road

               Y=Location of Occurrence of Accident

It is known about X & Y that

  1. X & Y are normally distributed
  2. X & Y independent of each other.

The joint density function is given as

                                     [tex]f(x,y)=f_X(x)f_Y(y)=\frac{1}{L^2}[/tex]

This is true for (x,y)∈[tex](0,L^2)[/tex] otherwise it is equal to zero.

  • Let Z be the random variable giving distance between X and Y so

                                           [tex]Z=|X-Y|[/tex]

  • The cumulative distribution of Z is given as

                              [tex]P(Z\leq a)=1-P(Z>a)\\[/tex]

Here  P(Z>a) is given as following on basis of the 2-d plot of X & Y points such that their difference is greater than a variable a.

                              [tex]P(Z>a)=\frac{(L-a)^2}{L^2}[/tex]

Substituting in the above equation

                          [tex]P(Z\leq a)=1-P(Z>a)\\P(Z\leq a)=1-\frac{(L-a)^2}{L^2}\\[/tex]

Now fZ(a) is given as

                         [tex]f_Z(a)=\frac{d}{da}P(Z\leq a)=\frac{2(L-a)}{L^2}\\[/tex]

So the distribution of distance of ambulance from accident is [tex]\frac{2(L-a)}{L^2}\\[/tex]

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