The distance between flaws on a long cable is exponentially distributed with mean 12 m. a.) Find the probability that the distance between two flaws is 15 m. b.) Find the probability that the distance between two flaws is between 8 and 20 m. c.) Find the median distance between flaws.

a) P = 0.29

b) P = 0.32

c) median = 8.3

Explanation:

We have an average distance of 12 m, therefore:

12 = 1/λ

Clearing λ:

λ = 1/12

let "x" be the distance between the flaws, we have:

a)

Let P be the probability that the distance between two flaws is 15 m:

P(x>15) = 1-P(x<15) = 1-(1-e^(λ*x)) = e^(-15/12) = 0.29

b)

Let P be probability that the distance between two flaws is between 8 and 20 m:

P(8<x<20) = P(x<20) - P(x<8) = (1-e^(-20/12)) - (1-e^(-8/12)) = e^(-8/12) - e^(-20/12) = 0.32

c)

The median distance between flaws is equal to:

P(x<median) = 0.5

1-e^(-median/12) = 0.5

e^(-median/12) = 0.5

-median/12 = ln(0.5)

median = ln(0.5) * (-12) = 8.3

multiply 2kg by latent heat of vap