If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution with the parameter $\theta = 9$, what are the probabilities that it will take the doctor at least 20 minutes to treat
(a) one patient;
(b) two patients;
(c) three patients?
Solution
We use the relation between Exponential distribution and Poisson distribution to solve this question.
Let $N$ denote the number of patients that have been take care of during 20 minutes.
Then $N$ has a Poisson distribution with parameter $\lambda = \theta \frac{20}{60} = 9\times \frac{20}{60} = 3$.
The pmf of $N$ is
$$ \begin{aligned} P(N=n) &= \frac{e^{-\lambda}\lambda^n}{n!};\; n=0,1,2,\cdots\\ &= \frac{e^{-3}3^n}{n!};\; n=0,1,2,\cdots \end{aligned} $$
(a) The probability that the first patient is still with the doctor after 20 minutes is the probabilities that it will take the doctor at least 20 minutes to treat.
Thus required probability is
$$ \begin{aligned} P(N=0) &= \frac{e^{-3}3^0}{0!}\\ &= 0.0497871 \end{aligned} $$
(b) The probability that it will take the doctor at least 20 minutes to treat two patient is
$$ \begin{aligned} P(N=1) &= \frac{e^{-3}3^1}{1!}\\ &= 0.1493612 \end{aligned} $$
(c) The probability that it will take the doctor at least 20 minutes to treat three patient is
$$ \begin{aligned} P(N=2) &= \frac{e^{-3}3^2}{2!}\\ &= 0.2240418 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
