91Ó°ÊÓ

If the location of the fire is represented by a random variable X, then the probability function of X is given as:

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{0.2}&{{\rm{when}}\;x = - 3}\\{0.1}&{{\rm{when}}\;x = - 1}\\{0.1}&{{\rm{when}}\;x = 0}\\{0.4}&{{\rm{when}}\;x = 1}\\{0.2}&{{\rm{when}}\;x = 2}\\0&{{\rm{otherwise}}}\end{aligned}} \right.\)

(a)

Let d be the point along the road that a fire engine should wait in order to minimize the expected value of the square of the distance that it must travel to the next fire.

This value\(E\left( {{{\left( {X - d} \right)}^2}} \right)\), is minimized when d takes the value of the mean of X.

Find the mean of X as follows:

\(\begin{aligned}{}E\left( X \right) &= - 3 \times 0.2 - 1 \times 0.1 + 0 \times 0.1 + 1 \times 0.4 + 2 \times 0.2\\ &= 0.1.\end{aligned}\)

Therefore, the fire engine should wait at 0.1 location in order to minimize the expected value of the square of the distance.

(b)

Again, let d be the point along the road that a fire engine should wait in order to minimize the expected value of the distance that it must travel to the next fire.

This value\(E\left( {\left| {X - d} \right|} \right)\)is minimized when d takes the value of the median of X.

Find the median of X as follows:

Note that:

\(\begin{aligned}{}P\left( {X \le 1} \right) &= P\left( {X = - 3} \right) + P\left( {X = - 1} \right) + P\left( {X = 0} \right) + P\left( {X = 1} \right)\\ &= 0.2 + 0.1 + 0.1 + 0.4\\ &= 0.8.\end{aligned}\)

Also,

\(\begin{aligned}{}P\left( {X \ge 1} \right) &= P\left( {X = 1} \right) + P\left( {X = 2} \right)\\ &= 0.4 + 0.2\\ &= 0.6.\end{aligned}\)

Thus,\(P\left( {X \le 1} \right) \ge 0.5\;{\rm{and}}\;P\left( {X \ge 1} \right) \ge 0.5\).

Hence, the median of X is 1.

Therefore, the fire engine should wait at 1 location in order to minimize the expected value of the distance.