At a certain bus stop, the length of time needed to wait for the next bus is known to follow an Exponential probability distribution where the mean is 1.6 hours. Let X be the time (in hours) that a randomly-selected passenger needed to wait for the next bus.

Then find the probability that it was between 330 and 435 minutes, rounded to 3 places.

Let's solve this problem involving an exponential distribution with a given mean. Here’s the step-by-step solution:

### Step 1: Define the Parameters
- The waiting time \( X \) is modeled by an **exponential distribution**.
- The mean \( \mu \) of the exponential distribution is given as **1.6 hours**.
- The **rate parameter** \( \lambda \) of the exponential distribution is the reciprocal of the mean:
  
  \[
  \lambda = \frac{1}{\mu} = \frac{1}{1.6} = 0.625
  \]

The exponential distribution's probability density function (PDF) is given by:

\[
f(x) = \lambda e^{-\lambda x}, \quad x \geq 0
\]

For cumulative probability calculations, we use the **cumulative distribution function (CDF)** of the exponential distribution:

\[
F(x) = 1 - e^{-\lambda x}
\]

### Step 2: Convert Minutes to Hours
The time range provided is **330 minutes to 435 minutes**. We need to convert these values into hours:

- \( 330 \) minutes \( = \frac{330}{60} = 5.5 \) hours
- \( 435 \) minutes \( = \frac{435}{60} = 7.25 \) hours

Thus, we need to find:

\[
P(5.5 \leq X \leq 7.25)
\]

### Step 3: Use the CDF to Find Probabilities
To find the probability that \( X \) is between 5.5 and 7.25 hours, we use the CDF of the exponential distribution:

\[
P(a \leq X \leq b) = F(b) - F(a)
\]

For the exponential distribution, the CDF is:

\[
F(x) = 1 - e^{-\lambda x}
\]

Now, calculate \( F(7.25) \) and \( F(5.5) \):

- \( F(7.25) = 1 - e^{-0.625 \times 7.25} \)
- \( F(5.5) = 1 - e^{-0.625 \times 5.5} \)

### Step 4: Compute the Values
Let's compute the values:

1. **For \( F(7.25) \)**:

   \[
   F(7.25) = 1 - e^{-4.53125} \approx 1 - 0.0108 = 0.9892
   \]

2. **For \( F(5.5) \)**:

   \[
   F(5.5) = 1 - e^{-3.4375} \approx 1 - 0.0322 = 0.9678
   \]

### Step 5: Find the Probability
Now, find the probability that \( X \) is between 5.5 and 7.25 hours:

\[
P(5.5 \leq X \leq 7.25) = F(7.25) - F(5.5)
\]

\[
P(5.5 \leq X \leq 7.25) = 0.9892 - 0.9678 = 0.0214
\]

### Step 6: Round to 3 Decimal Places
Rounded to three decimal places:

\[
P(5.5 \leq X \leq 7.25) \approx 0.021
\]

### Final Answer
The probability that the waiting time is between 330 and 435 minutes is approximately **0.021**.

---

Do you need any more details or have questions about this solution?

Here are 5 related questions to explore further:

1. How do you calculate the mean of an exponential distribution given the rate parameter?
2. What are the key properties of an exponential distribution?
3. How would you find the median waiting time for an exponential distribution with a given mean?
4. If the mean waiting time changes to 2 hours, how would the probability change for the same time range?
5. How does the exponential distribution relate to the Poisson process?

**Tip:** Remember, the exponential distribution is memoryless, which means that the probability of waiting longer does not depend on how much time has already passed. This is a unique feature of the exponential distribution!

At a certain bus stop, the length of time needed to wait for the next bus is known to follow an Exponential probability distribution where the mean is 1.6 hours. Let X be the time (in hours) that a randomly-selected passenger needed to wait for the next bus. Then find the probability that it was between 330 and 435 minutes, rounded to 3 places.

Solve an exponential distribution problem where the waiting time for a bus follows an exponential distribution with a mean of 1.6 hours. Learn to calculate the probability of waiting between 330 and 435 minutes, with a detailed step-by-step solution including conversion of minutes to hours and application of the cumulative distribution function (CDF). Suitable for Grades 11-12.

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