To solve this problem, let's go through each step carefully.

Step 1: Given Information

• The delivery times $X$ are exponentially distributed with a mean of 2.04 days.

• For an exponential distribution, the mean $\mu$ is equal to $\frac{1}{\lambda}$, where $\lambda$ is the rate parameter.

  Since the mean $\mu = 2.04$, we have: $\lambda = \frac{1}{2.04} \approx 0.4902$

• Sample size $n = 62$.

• We are looking for the probability that the sample mean $\bar{X}$ of the shipping times is less than 1.69 days.

Step 2: Distribution of the Sample Mean

Since the individual delivery times $X$ follow an exponential distribution, the sample mean $\bar{X}$ will approximately follow a normal distribution (according to the Central Limit Theorem), because we have a sufficiently large sample size $n = 62$.

The mean of the sample mean $\bar{X}$ is the same as the population mean: $\mu_{\bar{X}} = \mu = 2.04$

The standard deviation of the sample mean $\bar{X}$, or the standard error, is: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the standard deviation of the exponential distribution.

For an exponential distribution, $\sigma = \mu$, so: $\sigma = 2.04$

Thus: $\sigma_{\bar{X}} = \frac{2.04}{\sqrt{62}} \approx 0.2595$

Step 3: Calculating the Z-score

Now we want to find the probability that $\bar{X} < 1.69$.

We can convert $\bar{X}$ to a standard normal variable $Z$ by using the Z-score formula: $Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}$

Substitute $\bar{X} = 1.69$, $\mu_{\bar{X}} = 2.04$, and $\sigma_{\bar{X}} = 0.2595$: $Z = \frac{1.69 - 2.04}{0.2595} \approx -1.3514$

Step 4: Finding the Probability

Now, we need to find the probability $P(Z < -1.3514)$.

Using the standard normal distribution table or a calculator, we find: $P(Z < -1.3514) \approx 0.0884$

Final Answer

The probability that the average shipping time is less than 1.69 days is approximately: $\boxed{0.0884}$

Would you like further explanation on any of these steps?

Related Questions

1. How would the answer change if the sample size was larger?
2. How is the exponential distribution different from the normal distribution?
3. What is the Central Limit Theorem, and why is it important in this problem?
4. How would we approach the problem if the distribution of shipping times were unknown?
5. How can we interpret a Z-score in terms of probability?

Tip: When dealing with large samples, the Central Limit Theorem allows us to approximate the sample mean distribution as normal, regardless of the population distribution.