Question

The time between customer arrivals at a furniture store has an approximate exponential distribution with mean θ = 8.1 minutes. [Round to 4 decimal places where necessary.]

If a customer just arrived, find the probability that the next customer will arrive in the next 6 minutes.

If a customer just arrived, find the probability that the next customer will arrive within next 13 to 15 minutes?

If after the previous customer, no customer arrived in next 13 minutes, find the conditional probability that no customer will arrive for 15 minutes.

Answer 1

For exponential distribution the parameter is reciprocal of its mean. Therefore the waiting time distribution here is given as:

$T \sim exp(1/8.1)$

a) The probability that the next customer will arrive in the next 6 minutes is computed here as:

$P(T < 6) = \int_{0}^{6}(1 / 8.1)e^{-t/8.1} \ dt = 1 - e^{-6/8.1} = 0.5232$

Therefore 0.5232 is the required probability here.

b) The probability that the next customer will arrive within next 13 to 15 minutes is computed here as:

$P(13 < T < 15) = \int_{13}^{15}(1 / 8.1)e^{-t/8.1} \ dt = e^{-13/8.1} - e^{-15/8.1} = 0.0440$

Therefore 0.0440 is the required probability here.

c) Given that there is no customer arriving in next 13 minutes, probability that no customer arrive in the next 15 minutes is computed here as:

= Probability that there is no customer in the next (15 - 13) = 2 minutes.

$= \int_{0}^{2}(1 / 8.1)e^{-t/8.1} \ dt = 1 - e^{-2/8.1} = 0.2188$

Therefore 0.2188 is the required probability here.