Question 1169620: Mary can take on three possible routes from her home to her office in the morning.
The percentage of days on routes A, B and C to her office were 45%, 30% and
25% respectively. The traveling time on each of these three routes is normally
distributed with mean and standard deviation shown in the following table:
Route A Route B Route C
Mean 30 minutes 35 minutes 28 minutes
Standard deviation 10 minutes 8 minutes 12 minutes
Mary left her home at 8:28 a.m. and arrived at her office before 9:00 a.m.,
calculate the probability that she took route B to work.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using Bayes' Theorem.
**Understanding the Problem**
We need to calculate the probability that Mary took route B, given that she arrived at her office before 9:00 a.m.
**Given Information**
* Routes: A, B, C
* Probabilities: P(A) = 0.45, P(B) = 0.30, P(C) = 0.25
* Arrival time: Before 9:00 a.m. (32 minutes after leaving home)
* Mean and standard deviation of travel times:
* Route A: μA = 30, σA = 10
* Route B: μB = 35, σB = 8
* Route C: μC = 28, σC = 12
**Solution**
1. **Define Events:**
* A: Mary took route A.
* B: Mary took route B.
* C: Mary took route C.
* E: Mary arrived before 9:00 a.m. (travel time ≤ 32 minutes)
2. **Calculate P(E|A), P(E|B), P(E|C):**
* **P(E|A):** Probability of arriving before 9:00 a.m. given she took route A.
* zA = (32 - 30) / 10 = 2 / 10 = 0.2
* P(E|A) = P(ZA ≤ 0.2) ≈ 0.5793 (using a z-table or calculator)
* **P(E|B):** Probability of arriving before 9:00 a.m. given she took route B.
* zB = (32 - 35) / 8 = -3 / 8 = -0.375
* P(E|B) = P(ZB ≤ -0.375) ≈ 0.3538 (using a z-table or calculator)
* **P(E|C):** Probability of arriving before 9:00 a.m. given she took route C.
* zC = (32 - 28) / 12 = 4 / 12 = 1/3 ≈ 0.3333
* P(E|C) = P(ZC ≤ 0.3333) ≈ 0.6305 (using a z-table or calculator)
3. **Calculate P(E):**
* P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)
* P(E) = (0.5793 * 0.45) + (0.3538 * 0.30) + (0.6305 * 0.25)
* P(E) = 0.260685 + 0.10614 + 0.157625
* P(E) ≈ 0.52445
4. **Apply Bayes' Theorem:**
* We want to find P(B|E), the probability that she took route B given she arrived before 9:00 a.m.
* Bayes' Theorem: P(B|E) = (P(E|B) * P(B)) / P(E)
* P(B|E) = (0.3538 * 0.30) / 0.52445
* P(B|E) = 0.10614 / 0.52445
* P(B|E) ≈ 0.2024
**Final Answer**
The probability that Mary took route B to work is approximately 0.2024.
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