Understanding the Problem:
The problem states that a waiter believes his tips from various customers have a slightly right-skewed distribution with a mean of $10 and a standard deviation of $2.50. We are asked to find the probability that the average tip from a sample of 35 customers will be more than $13.

Key Concepts:
To solve this problem, we need to understand the concept of sampling distribution, specifically the distribution of sample means.

- Sampling Distribution: It refers to the distribution of a statistic (such as mean or standard deviation) calculated from multiple samples of the same size taken from a population.
- Central Limit Theorem (CLT): According to the CLT, the sampling distribution of the sample means approaches a normal distribution, regardless of the shape of the population distribution, as the sample size increases.
- Z-Score: A z-score measures the number of standard deviations an individual value or sample mean is from the population mean.

Calculating the Probability:
To find the probability that the average of 35 customers will be more than $13, we can use the Z-score formula and the properties of the sampling distribution.

1. Calculate the standard error of the mean:
The standard error (SE) of the mean is calculated by dividing the population standard deviation by the square root of the sample size.
SE = σ / √n
Here, σ (population standard deviation) = $2.50 and n (sample size) = 35.
SE = $2.50 / √35

2. Calculate the Z-score:
The Z-score is calculated by subtracting the population mean from the sample mean and dividing it by the standard error of the mean.
Z = (sample mean - population mean) / SE
Here, the sample mean is $13, and the population mean is $10.
Z = ($13 - $10) / SE

3. Find the probability:
Using the Z-score calculated in the previous step, we can find the probability using a standard normal distribution table or a calculator.
However, in this case, since the Z-score is quite large (greater than 3), the tail probability is extremely close to 0.

Conclusion:
The probability that the average of 35 customers will be more than $13 is essentially 0. This means that it is highly unlikely for the average tip from 35 customers to exceed $13, given the distribution and parameters provided.