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The total daily sales, \(x\), in the deli section of a local market is the sum of the sales generated by a fixed number of customers who make purchases on a given day. a. What kind of probability distribution do you expect the total daily sales to have? Explain. b. For this particular market, the average sale per customer in the deli section is \(\$ 8.50\) with \(\sigma=\$ 2.50 .\) If 30 customers make deli purchases on a given day, give the mean and standard deviation of the probability distribution of the total daily sales, \(x\).

Short Answer

Expert verified
Answer: The approximate probability distribution for the total daily sales at the deli section is a normal distribution. The mean of this distribution is $255, and the standard deviation is approximately $13.68.

Step by step solution

01

Central Limit Theorem

The Central Limit Theorem states that the sum of a large number of independent and identically distributed random variables will follow a normal distribution, regardless of the original distribution of the individual variables. In this case, the total daily sales, \(x\), can be seen as the sum of independent sales made by a fixed number of customers for the deli section. Thus, we should expect a normal (approximate) probability distribution for the total daily sales. #b. Calculating the mean and standard deviation#
02

Mean Calculation

Since the average sale per customer is given as \(\$8.50\), we can find the mean of this normal distribution for given 30 customers on a day by multiplying the average sale per customer with the number of customers: Mean, \(\mu = (\text{average sale per customer})\times(\text{number of customers per day})\) \(\mu = (8.50)\times(30)\) \(\mu = \$255\)
03

Standard Deviation Calculation

Given the standard deviation, \(\sigma=\$2.50\), of an individual customer sale, we can calculate the standard deviation of the total daily sales using the formula: Standard Deviation, \(\sigma_X = \sigma \cdot \sqrt{\text{number of customers per day}}\) \(\sigma_X = 2.50 \cdot \sqrt{30}\) \(\sigma_X \approx \$13.68\) So, the mean and standard deviation of the probability distribution of the total daily sales, \(x\), are \(\$255\) and \(\$13.68\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem is a powerful concept in statistics, making it crucial for understanding probability distributions. Simply put, it states that if you sum a large number of independent and identically distributed random variables, the resulting sum will approximate a normal distribution. This holds true regardless of the original distribution of each individual variable.

In the context of our exercise, each customer's purchase can be thought of as a random variable. When we add the sales from all customers, the total will follow a normal distribution because of the Central Limit Theorem. This allows us to predict the overall sales more reliably, simplifying complex and varied data into something more manageable.

This theorem is key for problems involving averages and aggregates, making it a cornerstone for statisticians and economists alike.
  • Applies when there are enough variables.
  • Doesn’t depend on the original distribution.
  • Essential for making predictions about total sums or averages.
Understanding this theorem helps in modeling and anticipating outcomes, especially in fields like marketing, quality control, and finance.
Normal Distribution
Normal distribution, often referred to as the bell curve, is a probability distribution that is symmetric around its mean. It’s characterized by its bell-shaped curve, where most values cluster around a central peak, declining symmetrically as they move away from it.

In the exercise, once we apply the Central Limit Theorem to the total daily sales, these sales are expected to fit a normal distribution. This means that the bulk of the sales will revolve around an average number, with extremes becoming increasingly less likely.

Why is this important? It facilitates a wide range of statistical analyses and predictions, because:
  • Many natural phenomena are normally distributed.
  • It aids in calculating probabilities and confidence intervals.
  • It simplifies the process of data analysis, interpretation, and presentation.
The normal distribution’s structure makes it a foundational element in statistics, as it supports robust analytical processes for various data types.
Mean and Standard Deviation
The mean and standard deviation are fundamental metrics in statistics, essential for interpreting any distribution, especially a normal one.

The mean represents the average of all data points, providing a central value around which the distribution is balanced. In our scenario, the calculated mean of total daily sales (\(\mu = \\(255\)) indicates the average expected sales amount in the deli section on any given day.

Standard deviation, on the other hand, measures the dispersion of data points from the mean, providing insight into the variability within the dataset. A low standard deviation means data points cluster closely around the mean, while a high standard deviation indicates wider dispersion. For the daily sales, \(\sigma_X \approx \\)13.68\), this suggests the typical amount by which daily sales could deviate from the mean.
  • Allows for understanding data variability.
  • Facilitates risk and performance assessment.
  • Supports decision-making processes by providing insights into expected fluctuations.
With these tools, businesses and analysts can make informed decisions, optimize operations, and improve strategic planning.

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Most popular questions from this chapter

Suppose a random sample of \(n=25\) observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12 . a. Give the mean and the standard deviation of the sampling distribution of the sample mean \(\bar{x}\). b. Find the probability that \(\bar{x}\) exceeds \(110 .\) c. Find the probability that the sample mean deviates from the population mean \(\mu=106\) by no more than 4

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