Solution:

We are tasked with finding the probability that the sample mean ($\bar{X}$) of excess time used by 16 customers exceeds 25 minutes. The problem involves applying the Central Limit Theorem (CLT), even though the sample size ($n = 16$) is less than 30.

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Step 1: Why Use the Central Limit Theorem (CLT)?

1. CLT Assumptions: CLT can be applied if the underlying population distribution is normal, even for small sample sizes.
   • Here, the problem states that the excess time follows a normal distribution.
2. Since the population distribution is normal, the sampling distribution of the sample mean will also follow a normal distribution, regardless of the sample size.

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Step 2: Define the Sampling Distribution

The sampling distribution of the sample mean $\bar{X}$ will have:

• Mean ($\mu_{\bar{X}}$) equal to the population mean: $\mu_{\bar{X}} = \mu = 20 \, \text{minutes}.$
• Standard deviation ($\sigma_{\bar{X}}$) equal to the population standard deviation divided by the square root of the sample size: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{16}} = 2.5 \, \text{minutes}.$

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Step 3: Compute the Z-Score

To find the probability that $\bar{X} > 25$, we calculate the $Z$-score for $\bar{X} = 25$: $Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{25 - 20}{2.5} = 2.$

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Step 4: Find the Probability

Using the standard normal distribution table, the cumulative probability for $Z = 2$ is: $P(Z \leq 2) = 0.9772.$ The probability that $\bar{X} > 25$ is the complement of this: $P(\bar{X} > 25) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228.$

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Final Answer:

The probability that the average excess time for the 16 customers exceeds 25 minutes is: $P(\bar{X} > 25) = 0.0228 \, \text{or approximately } 2.28\%.$

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Summary of Key Points:

1. Why CLT Works Here: Although $n < 30$, the population is normally distributed, so the sample mean will also follow a normal distribution.
2. Steps in Solution: Calculated the mean and standard deviation of the sample mean, converted the problem to a standard normal $Z$-score, and used the normal table to find probabilities.

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Would you like more explanation on a specific step? Here are related questions to explore:

1. How is the formula for the standard deviation of the sample mean derived?
2. What are examples where CLT cannot be applied?
3. What happens to the sampling distribution as $n$ increases to values larger than 30?
4. How would the result change if the standard deviation were 5 minutes instead of 10?
5. How do we interpret probabilities obtained from the standard normal table in real-life contexts?

Tip: When the population distribution is normal, the sample mean always follows a normal distribution, even for small sample sizes. Use this fact to simplify your calculations!