Answer to Question #339645 in Statistics and Probability for sezi

Question

 A hospital keeps records of its emergency-room traffic. Those records indicate that, beginning at 6:00 P.M. on any given day, the elapsed time until the first patient arrives has an exponential distribution with parameter λ = 6.9, where time is measured in hours. Determine the probability that, beginning at 6:00 P.M. on any given day, the first patient arrives

a. between 6:15 P.M. and 6:30 P.M.

b. before 7:00 P.M.

c. given that the first patient doesn’t arrive by 6:15 P.M., determine the probability that she arrives by 6:45 P.M.

Answer 1

Denote by $T$ a random variable that has an exponential distribution. Its probability density function is: $f(x)=\lambda e^{-\lambda x}, x\geq0$. $f(x)=0$ for $x<0.$ $\lambda=6.9$ is a parameter of the distribution. a. The aim is to find the probability: $P(\frac{1}{4}<T<\frac12).$ We get: $P(\frac{1}{4}<T<\frac12)=\int_{\frac14}^{\frac12}6.9 e^{-6.9 x}dx=-e^{-6.9 x}|_{\frac14}^{\frac12}=e^{-6.9\cdot\frac14}-e^{-6.9\cdot\frac12}\approx0.1464$.

b. The aim is to find $P(T<1).$ We get: $P(T<1)=\int_{0}^{1}6.9 e^{-6.9 x}dx=-e^{-6.9x}|_0^1=1-e^{-6.9}\approx0.9990$

c. The aim is to find $P(T<\frac34|T>\frac14)$. Using the definition of the conditional probability, we have: $P(T<\frac34|T>\frac14)=\frac{P(\frac14<T<\frac34)}{P(T>\frac14)}$. Compute both probabilities in the latter formula: $P(\frac14<T<\frac34)=\int_{\frac14}^{\frac34}6.9e^{-6.9x}dx=-e^{-6.9x}|_{\frac14}^{\frac34}=e^{-6.9\cdot\frac14}-e^{-6.9\cdot\frac34}\approx0.1725$.

$P(T>\frac14)=\int_{\frac14}^{+\infty}6.9e^{-6.9x}dx=-e^{-6.9x}|_{\frac14}^{+\infty}\approx0.1782.$

We get: $\frac{P(\frac14<T<\frac34)}{P(T>\frac14)}=\frac{0.1725}{0.1782}\approx0.9680$.

Answer: a. $P(\frac{1}{4}<T<\frac12)\approx0.1464$, b. $P(T<1)\approx0.9990$, c. $P(T<\frac34|T>\frac14)\approx0.9680$. All values are rounded to $4$ decimal places.