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.

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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!