An absent-minded professor schedules two student appointments for the same time. The appointment durations are independent and exponentially distributed with mean thirty minutes. The first student arrives on time, but the second student arrives five minutes late. What is the expected value of the time between the arrival of the first student and the departure of the second student?

### Solution

Let $t$ be the time between the arrival of the first student and the departure of the second student, $t_1$ be the time between the arrival of the first student and the beginning of the second appointment, and $t_2$ be the time between the beginning and end of the second appointment.

But $t_2$ is exponentially distributed with mean 30 minutes, we have

` $$ \begin{aligned} E[t] &= E[t1 + t2]\\ &= E[t1] + E[t2]\\ &= E[t1] + 30. \end{aligned} $$ `

Let $A$ be the event the 1st appointment takes less than 5 minutes. Then

`$$E[t1] = P[A]\times E[t1|A] + P[A^\prime]\times E[t1|A^\prime]$$`

If we're given that the first appointment takes more than 5 minutes, the remaining length of the appointment after those 5 minutes is exponentially distributed. Thus

$$E[t_1|A] = 5$$

$$E[t_1|A^\prime] = 5 + 30 =35$$

Now,

` $$ \begin{aligned} P(A) &=P(t\leq 5)\\ &= F(t)\\ &=1-e^{-5/30}\\ &= 1-0.8464817\\ &=0.1535183\\ P(A^\prime)&= 1- P(A)\\ &= 1-0.1535183\\ &= 0.8464817\\ \end{aligned} $$ `

Thus

` $$ \begin{aligned} E[t1]& = P[A]\times E[t1|A] + P[A^\prime]\times E[t1|A^\prime]\\ & = 0.1535183 \times 5 + 0.8464817 \times 35\\ & = 0.7675914 + 29.6268604\\ & = 30.3944517\\ \end{aligned} $$ `

Hence, $E(t) = E(t_1)+30 = 60.3944517$.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators