The amount of time patients spend in a doctor's office, $X$, is an exponential random variable with a mean of $10$ minutes.
$(a)$ Find the probability a patient spends less than $5$ minutes with the doctor.
$(b)$ Find the probability a patient spends between $7$ and $10$ minutes with the doctor.
$(c)$ How many patients will this doctor typically see in a $6$-hour surgery?
$(d)$ Another doctor spends $10$ minutes or more with half of his patients. How many fewer patients does he see in his $6$-hour surgery?
Ans: $(a)$ $X\sim \exp(\lambda)$ with $\lambda=\frac{1}{10}$, So, to find the probability that someone spends less than 5 minutes with a doctor,
$P(X< 5)= \int_{0}^{5}\lambda \exp(-\lambda x) dx$.
$(b)$ $P(7<X< 10)= \int_{7}^{10}\lambda \exp(-\lambda x) dx$.
$(c)$ No idea, could anyone please help.
$(d)$ No idea, could anyone please help.
Thanks!
$(c)$ If the time spent per patient is an exponential random variable, then the number of patients seen in a fixed time interval is a Poisson random variable with rate proportional to the length of that time interval.
Specifically, since $6$ hours is equal to $360$ minutes, the random number of patients seen in this time interval is Poisson distributed with rate $36$ patients per $6$ hours, and the expected number of such patients seen is also $36$ patients.
$(d)$ let $Y$ be the random time spent per patient for the second doctor. You are given $$\Pr[Y \ge 10] = 1/2.$$ If $Y$ is also exponentially distributed, this means $$1/2 = \Pr[Y \ge 10] = e^{-10 \lambda},$$ and solving for $\lambda$ in this equation gives you the corresponding rate parameter for this doctor's time spent. The average number of patients seen for this doctor is $1/\lambda$.