Question

On any given day, mail gets delivered by either Alice or Bob. If Alice delivers it, which happens with probability 1/4, she does so at a time that is uniformly distributed between 9 and 11. If Bob delivers it, which happens with probability 3/4, he does so at a time that is uniformly distributed between 10 and 12. The PDF of the time X that mail gets delivered satisfies

a) fx (9.5) =

b) fx (10.5) =

Answer 1

To find the probability density function (PDF) \( f_X(x) \) of the delivery time \( X \), we need to consider the cases for Alice and Bob and how their delivery times are distributed.

1. **Alice's Delivery**:
- Probability that Alice delivers = \( P(A) = \frac{1}{4} \)
- Alice delivers uniformly between 9 and 11. Thus, if \( X_A \) is the delivery time when Alice delivers, then:
\[
X_A \sim \text{Uniform}(9, 11)
\]
The PDF for \( X_A \) is:
\[
f_{X_A}(x) =
\begin{cases}
\frac{1}{2} & \text{if } 9 \leq x \leq 11 \\
0 & \text{otherwise}
\end{cases}
\]

2. **Bob's Delivery**:
- Probability that Bob delivers = \( P(B) = \frac{3}{4} \)
- Bob delivers uniformly between 10 and 12. Thus, if \( X_B \) is the delivery time when Bob delivers, then:
\[
X_B \sim \text{Uniform}(10, 12)
\]
The PDF for \( X_B \) is:
\[
f_{X_B}(x) =
\begin{cases}
\frac{1}{2} & \text{if } 10 \leq x \leq 12 \\
0 & \text{otherwise}
\end{cases}
\]

3. **Combined PDF \( f_X(x) \)**:
Using the Law of Total Probability, we can compute the overall PDF of \( X \) by considering both scenarios (Alice and Bob):
\[
f_X(x) = P(A) f_{X_A}(x) + P(B) f_{X_B}(x)
\]
Now substituting the probabilities and PDFs:
\[
f_X(x) = \frac{1}{4} f_{X_A}(x) + \frac{3}{4} f_{X_B}(x)
\]

Now we evaluate the combined PDF in two intervals, \( [9, 11] \) and \( [10, 12] \):

- For \( 9 \leq x < 10 \):
- \( f_{X_A}(x) = \frac{1}{2} \)
- \( f_{X_B}(x) = 0 \)
- Hence,
\[
f_X(x) = \frac{1}{4} \cdot \frac{1}{2} + \frac{3}{4} \cdot 0 = \frac{1}{8}
\]

- For \( 10 \leq x \leq 11 \):
- \( f_{X_A}(x) = \frac{1}{2} \)
- \( f_{X_B}(x) = \frac{1}{2} \)
- Hence,
\[
f_X(x) = \frac{1}{4} \cdot \frac{1}{2} + \frac{3}{4} \cdot \frac{1}{2} = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}
\]

- For \( 11 < x \leq 12 \):
- \( f_{X_A}(x) = 0 \)
- \( f_{X_B}(x) = \frac{1}{2} \)
- Hence,
\[
f_X(x) = \frac{1}{4} \cdot 0 + \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}
\]

Finally, we can summarize the PDF:
\[
f_X(x) =
\begin{cases}
\frac{1}{8} & \text{if } 9 \leq x < 10 \\
\frac{1}{2} & \text{if } 10 \leq x \leq 11 \\
\frac{3}{8} & \text{if } 11 < x \leq 12 \\
0 & \text{otherwise}
\end{cases}
\]

### Answers to the specific questions:

a) \( f_X(9.5) = \frac{1}{8} \)

b) \( f_X(10.5) = \frac{1}{2} \)