by magicturtle
Last Updated September 21, 2019 11:20 AM - source

The time leading up to an event of some kind is often modeled by the exponential distribution. This task is about the exponential distribution as a model for processing times. Assume that, for a given consultation, the processing time (Y) is exponentially distributed with the expected value = 15min

a) Determine the probability that a consultation takes longer than expected time for consultation (15 min). Why is this probability not equal to 0.5?

b) Determine the probability that eight consultations take longer than expected time (2 hours). Compare the answer with the corresponding problem in task a).

What you expect is something like uniform distribution.

For **a** simply calculate it by exponential formula.

For **b** suppose new exponential by $8\lambda$.

September 21, 2019 10:45 AM

An exponentially distributed variable has PDF $\lambda\exp(-\lambda x)$ on its support $[0,\:\infty)$, for some $\lambda>0$. The substitution $u:=\lambda x$ gives$$\mu=\int_0^\infty \lambda x\exp(-\lambda x)dx=\frac{1}{\lambda}\int_0^\infty ue^{-u}du=\frac{1}{\lambda}.$$Since $\lambda=\frac{1}{\mu}$, the PDF is often written as $\frac{1}{\mu}\exp\left(-\frac{x}{\mu}\right)$ on its support. Equivalently, on $[0,\:\infty)$ the survival function is $\exp\left(-\frac{x}{\mu}\right)$. So the probability of exceeding the mean is $\exp(-1)$. It's not $\frac12$, because the median is $\mu\ln 2$.

I'll have to contradict the other answer's implication that an exponential distribution will gave the answer to b. Let $\{X_i|1\le i\le 8\}$ denote eight consultation times, each exponentially distributed with mean time $\mu$ equal to $15$ minutes. We want $P(\sum_i X_i>8\mu)$. But $\sum_iX_i$ is Erlang-distributed with $k=8,\:\lambda=\frac{1}{\mu}$ (this can be proven with characteristic functions), so you want $$\int_{8\mu}^\infty\frac{x^7}{7!\mu^8}\exp\left(-\frac{x}{\mu}\right)dx=\int_8^\infty\frac{u^7}{7!}\exp(-u)du,$$which is evaluated here. It's a little larger than the answer to a, because the $k=8$ Erlang distribution has a heavier right-hand tail than the exponential distribution for low $x$. Another explanation, this time focusing on the answer being closer to $\frac12$ than before, is that for large $k$ the CLT allows a Normal approximation, which brings the mean's percentile closer to the median's.

September 21, 2019 11:15 AM

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