A.9 A fire station is to be located along a road which is of infinite length - s

Question

A.9 A fire station is to be located along a road which is of infinite length - stretching from point S outward to infinite. Suppose the fire will occur along the road. The distance of a fire (in kilometre) from point S is exponentially distributed with parameter = 0.1. a) Where is the best location for the fire station along the road (i.e. estimate the distance from the point S) so as to minimize the expected travel distance from the fire station to the fire location? b) What is the probability that the distance of a fire from point S exceeds 20 kilometres? c What is the probability that a fire is located within 5 kilometres from point S? d) If the distance of a fire from point S exceeds 15 kilometres, what is the probability that the fire is located within 25 kilometres from point S?

Explanation / Answer

SOLUTION :-

Here the given distribution is exponential distriubtion.

where f(s) = e-s

F(x) = 1 - e-s

(a)

Here the best location will be mean to minimize the expected travel distance from the fire stations to the fire station.

Here mean distance = 1/ = 1/0.1 = 10 KM

(b)

Here we have to find the probability that of distance from fire from point S exceeds 20 Km.

Pr(S > 20 km) = 1- Pr ( S < 20 km ) = 1 - [ 1 -e-s ] = e-0.1 * 20 =  0.1353

(c)

so now we have to find the probability that of distance from fire from point S is under 5 km.

Pr( S < km) = 1 - e-s =1 - e-0.1 * 5 = 1- e-0.5 = 0.3935

(d)

Here we have find the probability that fire is located within 25 km from point S, given that fire from point S exceeds 15 km.

Pr(S < 25 km l S > 15 km) = Pr(15 < S < 25) / Pr(S > 15) = [Pr(S < 25) - Pr(S < 15) ] / Pr( S > 15)

Here Pr(S < 25) = 1- e-0.1 * 25 = 1 - e-2.5

Pr(S < 15) =  1- e-0.1 * 15 = 1 - e-1.5

Pr(S > 15) = e-1.5

[Pr(S < 25) - Pr(S < 15) ]/ Pr( S > 15) = (e-1.5 - e-2.5 )/ e-1.5 = 1 - e-1 = 1 - 0.3679 = 0.6321

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