Question

A commuter attempts to catch the 8:00 am train every morning although his arrival time at the station is a random variable that is uniformly distributed between 7:50 am and 8:10 am. The train's departure time from the station is also a random variable that is uniformly distributed between 8:00 am and 8: 10 am.

a) Find the probability density function of the time interval between the commuter's arrival at station and the train's departure time.

Answer 1

The probability density function for the time interval between the commuter's arrival (uniformly distributed from 7:50 am to 8:10 am) and the train's departure (uniformly distributed from 8:00 am to 8:10 am) is
`\((1)/(10)` for 0 ≤ t < 10 minutes, and 0 otherwise.

Let X be the random variable representing the commuter's arrival time and Y be the random variable representing the train's departure time.

Given that X is uniformly distributed between 7:50 am and 8:10 am, we can express its probability density function (pdf) as follows:

`\[ f_X(x) = \begin{cases} (1)/(20), & 7:50 \leq x \leq 8:10 \\ 0, & \text{otherwise} \end{cases} \]`

Similarly, for Y, the train's departure time is uniformly distributed between 8:00 am and 8:10 am:

`\[ f_Y(y) = \begin{cases} (1)/(10), & 8:00 \leq y \leq 8:10 \\ 0, & \text{otherwise} \end{cases} \]`

Now, let's find the probability density function of the time interval between the commuter's arrival and the train's departure, denoted by T = Y - X. We can express this in terms of the cumulative distribution function (CDF) and then differentiate to find the pdf.

`\[ F_T(t) = P(T \leq t) = P(Y - X \leq t) \]`

To find
`\(F_T(t)\)`, consider the possible cases:

1. If t < 0, then
`\(F_T(t) = 0\)` since the interval can't be negative.

2. If
`\(0 \leq t < 10\)` minutes, then
`\(F_T(t) = P(Y \leq X + t)\)`.

3. If
`\(t \geq 10\)` minutes, then
`\(F_T(t) = P(Y \leq X + t) = 1\)`.

Now, let's calculate the pdf by differentiating
`\(F_T(t)\)`:

`\[ f_T(t) = (d)/(dt)F_T(t) \]`

1. If t < 0, then
`\(f_T(t) = 0\)`.

2. If
`\(0 \leq t < 10\)` minutes, then
`\(f_T(t) = f_Y(X + t)\)`.

3. If
`\(t \geq 10\)` minutes, then
`\(f_T(t) = 0\)` since the interval can't exceed 10 minutes.

Therefore, the probability density function for the time interval between the commuter's arrival and the train's departure is given by:

`\[ f_T(t) = \begin{cases} 0, & t < 0 \text{ or } t \geq 10 \\ (1)/(10), & 0 \leq t < 10 \end{cases} \]`

This makes sense, as the interval between arrival and departure can only be positive and cannot exceed 10 minutes.