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Let's break down and solve each part of the question, which is based on a normal distribution.

Given:
- Mean \( \mu = 2659 \)
- Standard deviation \( \sigma = 512 \)

The normal distribution for the calories consumed can be expressed as \( X \sim N(\mu = 2659, \sigma = 512) \).

### a) Distribution of \( X \)
The distribution is normally distributed with the mean and standard deviation given:
\[
X \sim N(2659, 512^2)
\]

### b) Probability that the customer consumes less than 2489 calories
To find this probability, we need to calculate the z-score for \( X = 2489 \):
\[
z = \frac{2489 - 2659}{512} = \frac{-170}{512} \approx -0.3320
\]

Using the standard normal distribution table, a z-score of -0.3320 corresponds to a probability of about 0.3707. Therefore, the probability that the customer consumes less than 2489 calories is:
\[
P(X < 2489) \approx 0.3707
\]

### c) Proportion of customers who consume over 2802 calories
Again, we first calculate the z-score for \( X = 2802 \):
\[
z = \frac{2802 - 2659}{512} = \frac{143}{512} \approx 0.2793
\]

Using the standard normal table, a z-score of 0.2793 corresponds to a probability of about 0.6094. Since we are looking for the proportion who consume **more** than 2802 calories, we subtract this from 1:
\[
P(X > 2802) = 1 - 0.6094 = 0.3906
\]

Thus, about 39.06% of the customers consume more than 2802 calories.

### d) Calories required to be in the top 2% of consumers (Piggy award)
We need to find the z-score that corresponds to the top 2%, which is the 98th percentile of the normal distribution. From the z-score table, the z-score for the 98th percentile is approximately 2.054.

Now, use the z-score formula to solve for the corresponding calorie value:
\[
X = \mu + z \cdot \sigma = 2659 + 2.054 \cdot 512 = 2659 + 1051.648 \approx 3711.6
\]

Rounding to the nearest calorie, the fewest number of calories a person must consume to receive the Piggy award is:
\[
X \approx 3712
\]

---

Do you have any questions or would you like further details? Here are 5 related questions to expand your understanding:

1. How does the z-score relate to the probability in a normal distribution?
2. What is the probability that a customer consumes between 2400 and 2900 calories?
3. How do you interpret the area under the normal curve in relation to real-world scenarios?
4. If the standard deviation was smaller, how would it affect the Piggy award calculation?
5. Can you calculate the calorie threshold for the top 5% of customers?

**Tip**: When working with normal distributions, always ensure your z-scores are calculated correctly, as small mistakes can lead to incorrect probabilities.

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2659 and standard deviation 512. One randomly selected customer is observed to see how many calories X that customer consumes. Round all answers to 4 decimal places where possible.
a. What is the distribution of X?
b. Find the probability that the customer consumes less than 2489 calories.
c. What proportion of the customers consume over 2802 calories?
d. The Piggy award will be given out to the 2% of customers who consume the most calories. What is the fewest number of calories a person must consume to receive the Piggy award?

Solve a probability problem using normal distribution where caloric consumption at a buffet is normally distributed with a mean of 2659 and standard deviation of 512. Learn how to calculate the probability of consuming less than 2489 calories, find the proportion consuming more than 2802 calories, and determine the top 2% calorie intake required for a special award.

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