To solve this problem, we need to calculate the probability that the student took Route III, given the following details:

1. Departure Time: 5:32 p.m.
2. Arrival Time: Before 6:00 p.m.
3. Travel Time Calculation: Since 5:32 p.m. to 6:00 p.m. is 28 minutes, the student's travel time was under 28 minutes.

Step 1: Determine Probability Distributions for Each Route

For each route, the travel time is normally distributed with the following parameters:

• Route I: Mean $\mu_1 = 33$ minutes, Standard Deviation $\sigma_1 = 9$ minutes.
• Route II: Mean $\mu_2 = 32$ minutes, Standard Deviation $\sigma_2 = 10$ minutes.
• Route III: Mean $\mu_3 = 29$ minutes, Standard Deviation $\sigma_3 = 8$ minutes.

Step 2: Calculate Probability of Travel Time ≤ 28 Minutes for Each Route

To find the probability that the travel time is 28 minutes or less for each route, we need to calculate the Z-score for 28 minutes in each case:

Route I

$Z_1 = \frac{28 - 33}{9} = \frac{-5}{9} \approx -0.56$ Using a standard normal table, the probability $P(T_1 \leq 28) \approx 0.2877$.

Route II

$Z_2 = \frac{28 - 32}{10} = \frac{-4}{10} = -0.4$ Using a standard normal table, the probability $P(T_2 \leq 28) \approx 0.3446$.

Route III

$Z_3 = \frac{28 - 29}{8} = \frac{-1}{8} = -0.125$ Using a standard normal table, the probability $P(T_3 \leq 28) \approx 0.4515$.

Step 3: Use Bayes' Theorem to Calculate the Probability of Taking Route III

Let:

• $P(T_3 \leq 28) = 0.4515$
• $P(T_1 \leq 28) = 0.2877$
• $P(T_2 \leq 28) = 0.3446$

The probabilities of choosing each route are:

• $P(\text{Route I}) = 0.33$
• $P(\text{Route II}) = 0.27$
• $P(\text{Route III}) = 0.40$

Using Bayes' theorem, the probability the student took Route III given that the travel time is ≤ 28 minutes is: $P(\text{Route III} \mid T \leq 28) = \frac{P(T_3 \leq 28) \times P(\text{Route III})}{P(T_1 \leq 28) \times P(\text{Route I}) + P(T_2 \leq 28) \times P(\text{Route II}) + P(T_3 \leq 28) \times P(\text{Route III})}$

Plugging in the values: $P(\text{Route III} \mid T \leq 28) = \frac{0.4515 \times 0.40}{(0.2877 \times 0.33) + (0.3446 \times 0.27) + (0.4515 \times 0.40)}$

Calculating the numerator: $0.4515 \times 0.40 = 0.1806$

Calculating the denominator: $(0.2877 \times 0.33) + (0.3446 \times 0.27) + (0.4515 \times 0.40) = 0.0959 + 0.0930 + 0.1806 = 0.3695$

So, $P(\text{Route III} \mid T \leq 28) = \frac{0.1806}{0.3695} \approx 0.489$

Final Answer

The probability that the student took Route III given that he arrived in under 28 minutes is approximately 0.489 or 48.9%.

Would you like more details on any step, or do you have any further questions?

Related Questions:

1. How would the probability change if the mean and standard deviation for Route III were different?
2. Can Bayes' theorem be applied if the travel times were not normally distributed?
3. What is the impact on the result if the probability distribution for route selection is altered?
4. How can the probability calculation be simplified if only two routes were considered?
5. Why do we use the Z-score transformation in normal distributions?

Tip:

When calculating probabilities for normal distributions, Z-scores allow easy access to standard normal tables, making the process more efficient.