The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2927 and standard deviation 654. One randomly selected customer is observed to see how many calories X that customer consumes. Round all answers to 4 decimal places where possible.

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This is a normal distribution problem 

**Title: Understanding Caloric Consumption Distribution at a Buffet**

**Introduction:**

The caloric intake of customers at a Chinese buffet is analyzed under the assumption of a normal distribution. The average (mean) calorie consumption is 2927 calories, with a standard deviation of 654 calories. This information helps to understand the eating patterns and the variability among different customers.

**Questions:**

**a. Determine the Distribution of X:**  
The random variable X represents the number of calories consumed by a customer. Given the parameters, the distribution of X can be expressed as \( X \sim N(\mu, \sigma^2) \) where:
- \( \mu \) is the mean = 2927
- \( \sigma \) is the standard deviation = 654

**b. Probability of Consuming Less than 2768 Calories:**  
Calculate the probability that a randomly selected customer consumes fewer than 2768 calories. This involves finding the area under the normal distribution curve to the left of 2768 calories.

**c. Proportion of Customers Consuming Over 3179 Calories:**  
Determine what proportion of customers exceed a caloric consumption of 3179 calories. This is equivalent to calculating the area under the normal distribution curve to the right of 3179 calories.

**Conclusion:**

This exercise in statistical analysis provides insight into customer behavior and helps in understanding patterns in caloric consumption, which could be valuable for buffet management and health-related studies. All calculations should be rounded to four decimal places where necessary for precision.
Transcribed Image Text:**Title: Understanding Caloric Consumption Distribution at a Buffet** **Introduction:** The caloric intake of customers at a Chinese buffet is analyzed under the assumption of a normal distribution. The average (mean) calorie consumption is 2927 calories, with a standard deviation of 654 calories. This information helps to understand the eating patterns and the variability among different customers. **Questions:** **a. Determine the Distribution of X:** The random variable X represents the number of calories consumed by a customer. Given the parameters, the distribution of X can be expressed as \( X \sim N(\mu, \sigma^2) \) where: - \( \mu \) is the mean = 2927 - \( \sigma \) is the standard deviation = 654 **b. Probability of Consuming Less than 2768 Calories:** Calculate the probability that a randomly selected customer consumes fewer than 2768 calories. This involves finding the area under the normal distribution curve to the left of 2768 calories. **c. Proportion of Customers Consuming Over 3179 Calories:** Determine what proportion of customers exceed a caloric consumption of 3179 calories. This is equivalent to calculating the area under the normal distribution curve to the right of 3179 calories. **Conclusion:** This exercise in statistical analysis provides insight into customer behavior and helps in understanding patterns in caloric consumption, which could be valuable for buffet management and health-related studies. All calculations should be rounded to four decimal places where necessary for precision.
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