Question #339645

 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.


1
Expert's answer
2022-05-12T12:46:10-0400

Denote by TT a random variable that has an exponential distribution. Its probability density function is: f(x)=λeλx,x0f(x)=\lambda e^{-\lambda x}, x\geq0. f(x)=0f(x)=0 for x<0.x<0. λ=6.9\lambda=6.9 is a parameter of the distribution. a. The aim is to find the probability: P(14<T<12).P(\frac{1}{4}<T<\frac12). We get: P(14<T<12)=14126.9e6.9xdx=e6.9x1412=e6.914e6.9120.1464P(\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).P(T<1). We get: P(T<1)=016.9e6.9xdx=e6.9x01=1e6.90.9990P(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<34T>14)P(T<\frac34|T>\frac14). Using the definition of the conditional probability, we have: P(T<34T>14)=P(14<T<34)P(T>14)P(T<\frac34|T>\frac14)=\frac{P(\frac14<T<\frac34)}{P(T>\frac14)}. Compute both probabilities in the latter formula: P(14<T<34)=14346.9e6.9xdx=e6.9x1434=e6.914e6.9340.1725P(\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>14)=14+6.9e6.9xdx=e6.9x14+0.1782.P(T>\frac14)=\int_{\frac14}^{+\infty}6.9e^{-6.9x}dx=-e^{-6.9x}|_{\frac14}^{+\infty}\approx0.1782.

We get: P(14<T<34)P(T>14)=0.17250.17820.9680\frac{P(\frac14<T<\frac34)}{P(T>\frac14)}=\frac{0.1725}{0.1782}\approx0.9680.

Answer: a. P(14<T<12)0.1464P(\frac{1}{4}<T<\frac12)\approx0.1464, b. P(T<1)0.9990P(T<1)\approx0.9990, c. P(T<34T>14)0.9680P(T<\frac34|T>\frac14)\approx0.9680. All values are rounded to 44 decimal places.


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