Question-26
Waiting time in a hospital is said to follow the exponential distribution with the expected waiting time of 10 minutes. What is the probability that a person visiting a hospital has to wait for more than 15 minutes? Enter the answer correct to two decimal places accuracy.
If \(X \sim Exp(\lambda)\), then \(E(X) = \dfrac{1}{\lambda}\) and \(Var(X) = \dfrac{1}{\lambda^2}\). Also,
\(F(x) = P(X \leq x) = 1 - e^{-\lambda x}\)
\(P(X > x) = 1-P(X \leq x) = 1 - \left(1 - e^{-\lambda x}\right) = 1 - 1 + e^{-\lambda x} = e^{-\lambda x}.\)
0.22
Let \(X\) represents the waiting time.
According to the question, we have given “Expected waiting time is 10 minutes”, i.e., \(E(X) = \dfrac{1}{\lambda} = 10 \implies \lambda = \dfrac{1}{10} = 0.1\). Thus
\(X \sim Exp(0.1)\). Now,
Prob.(that a person visiting a hospital has to wait for more than 15 minutes) \(\implies P (X > 15) = e^{-0.1 \times 15} = e^{-1.5} = 0.22\)