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Suppose that the time (in hours) required to repair a car is an exponentially distributed random variable with parameter λ = 1/2.

  1. What is the probability that the repair time exceeds 4 hours?

The probability that the repair time exceeds 4 hours is equal to 1 minus the probability that it last 4 hours or fewer. With R:

1-pexp(4,.5)
## [1] 0.1353353

By hand: \[ 1-F(4) = 1-P(T ≤ 4)\\ = 1-\int_{0}^{4}0.5e^{−0.5t} \, dt\\ =1- (1 − e^{(-0.5)(4)})\\ =\frac {1}{e^2}\\ \approx 0.1353\\ \]

  1. If it exceeds 4 hours what is the probability that it exceeds 8 hours?

Since the exponential distribution has the memoryless property, if the time has already exceeded a certain amount, we “start over” at t=0. So the probability that it exceeds 8 hours, given that it has exceeded 4 hours, is the same as the probability that it exceeds 4 hours.

1-pexp(8-4,.5)
## [1] 0.1353353