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

Question

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