Question 7.10: A convenience store has only one cashier. The time in minute......
A convenience store has only one cashier. The time in minutes he requires to serve a customer, denoted by X, is a random variable that has the exponential distribution with parameter 𝜆 = 2. Sarah is first in the queue at the cashier and the person who was in front of her has just started being served. Find the probability that Sarah will have to wait until her service starts
(i) at least three minutes;
(ii) between two and four minutes.
We observe first from (7.16) that, for t ≥ 0, the probability P(X ≤ t) is F(t)=1-\mathrm{e}^{-2t}.
(i) The event that Sarah will have to wait at least three minutes before she is served is {X ≥ 3}. Therefore, the required probability is
P(X \geq 3) = P(X \gt 3) = 1 − F(3) = 1-(1-\mathrm{e}^{-3 \cdot 2}) = \mathrm{e}^{-6},noting that the first equality above follows since X is a continuous random variable, and so the events {X ≥ 3} and {X > 3} have the same probability.
(ii) Next, we seek the probability P(2 ≤ X ≤ 4). This is the same as P(2 < X ≤ 4) and by a standard property of distribution functions, it is equal to
F(4) − F(2)= (1-\mathrm{e}^{-4.2})-(1-\mathrm{e}^{-2.2})=\mathrm{e}^{-4}-\mathrm{e}^{-8}.