Let's go through the questions based on the provided information:

We are told that the amount of calories consumed by customers at the buffet is normally distributed with:

• Mean $\mu = 2659$ calories
• Standard deviation $\sigma = 512$ calories

The problem asks us to work with the normal distribution $X \sim N(\mu, \sigma^2)$, where $X$ represents the calories consumed by a randomly selected customer.

Part (a): Distribution of $X$

The distribution of $X$ is given as a normal distribution with the provided mean and standard deviation. $X \sim N(2659, 512^2)$ So, for part (a), the answer is: $X \sim N(2659, 512^2)$

Part (b): Probability that the customer consumes less than 2489 calories

We need to find $P(X < 2489)$. First, we compute the z-score: $z = \frac{X - \mu}{\sigma} = \frac{2489 - 2659}{512} = \frac{-170}{512} \approx -0.3320$ Next, we look up the z-score of $-0.3320$ in the standard normal distribution table (or use a calculator). The corresponding probability is approximately: $P(Z < -0.3320) \approx 0.3707$ So, the probability that the customer consumes less than 2489 calories is approximately $0.3707$.

Part (c): Proportion of customers consuming more than 2802 calories

Here, we calculate $P(X > 2802)$. First, compute the z-score: $z = \frac{2802 - 2659}{512} = \frac{143}{512} \approx 0.2793$ Using the z-score table or a calculator, $P(Z < 0.2793) \approx 0.6099$. Since we want the proportion consuming more than 2802 calories, we compute: $P(X > 2802) = 1 - P(Z < 0.2793) = 1 - 0.6099 = 0.3901$ Thus, the proportion of customers consuming more than 2802 calories is approximately $0.3901$.

Part (d): Piggy award (top 2%)

The Piggy award goes to the top 2% of customers. This means we need to find the calorie amount that corresponds to the 98th percentile of the normal distribution.

Looking up the z-score corresponding to the 98th percentile, we find: $P(Z < z) = 0.98 \quad \Rightarrow \quad z \approx 2.054$ Now, we solve for the calorie threshold using the z-score formula: $X = \mu + z \sigma = 2659 + 2.054 \times 512 \approx 2659 + 1051.65 = 3711.65$ Rounding to the nearest calorie, the fewest number of calories a person must consume to receive the Piggy award is approximately $3712$ calories.

Would you like further details or clarifications? Here are five related questions that expand on the concepts:

1. What is the probability that a customer consumes between 2489 and 2802 calories?
2. If the mean changes to 2700 calories, how would the answers to parts (b) and (c) change?
3. How would the proportion in part (c) change if the standard deviation increased to 600?
4. Can you find the 90th percentile of the calorie distribution?
5. What is the probability that a customer consumes more than 3500 calories?

Tip: When working with normal distributions, z-scores allow you to standardize any value and use the standard normal table to find probabilities or percentiles.