The exponential distribution and why the excpected value

Question

The time leading up to an event of some kind is often modeled by the exponential

Answer 1

What you expect is something like uniform distribution.

For a simply calculate it by exponential formula.

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

Answer 2

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.