Question

An electrical power line connecting Miami to Key largo which is 48 miles long has been cut at some random point so that the distance between the cut-point from Miami, X, can be assumed to have a uniform distribution. Find the probability that distance between the cut-point and Miami is more than 4 times the distance between the cut-point and the key Largo.

Answer 1

We are given that the distance between the cut-point from Miami, X ~ Uniform(0,48) ; X in miles.

Thus, the distance between the cut-point from key Largo will be (48 - X) miles.

Now, since X ~ Uniform(0,48); thus the pdf of X is given by:

$f(x) = \frac{1}{48-0} = \frac{1}{48} \ \ ; 0 < x<48$

Therefore, the cdf of X is given by:

$\begin{align*} F(x) &= P(X \le x) \\ &= \int_{0}^{x} f(t) \ dt \\ &= \int_0^x \frac{1}{48} \ dt \\&= \begin{bmatrix} \frac{t}{48} \end{bmatrix}_{t=0}^{t=x} \\ &= \frac{x}{48} \end{align*}$

Now, we need to find the probability that the distance between the cut point and Miami,i.e, X, is more than 4 times the distance between the cut-point and key Largo,i.e., 48-X, which is :

P(X > 4 * (48-X)) P(X > 192-4X) = P(5X > 192) P(X > 38.4) 1- P(X 38.4) 1-F(38.4) 38.4 48 1-0.8 0.2

which is the required answer.

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