/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 The total daily sales, \(x\), in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 probability distribution for the total daily sales follows a normal distribution with a mean of $255 and a standard deviation of approximately $13.68.

Step by step solution

01

Identifying the probability distribution

First, we need to identify the probability distribution for the total daily sales in the market's deli section. Since the total daily sales is the sum of the sales generated by a fixed number of customers, it follows a normal distribution due to the central limit theorem.
02

Calculate the mean of the probability distribution

We are given that the average sale per customer in the deli section is \( \$ 8.50 \). If 30 customers make deli purchases on a given day, we can calculate the expected total daily sales by multiplying the average sale per customer by the number of customers making deli purchases: Mean (\(\mu\)) = (Average sale per customer) × (Number of customers) \(\mu = \$ 8.50 \times 30 = \$ 255\)
03

Calculate the standard deviation of the probability distribution

We are also given the standard deviation of the sales per customer, which is \(\sigma = \$ 2.50\). To find the standard deviation of the probability distribution for the total daily sales, we multiply the standard deviation of the sales per customer by the square root of the number of customers making deli purchases: Standard deviation (\(\sigma_x\)) = Standard deviation per customer × \(\sqrt{\text{Number of customers}}\) \(\sigma_x = \$ 2.50 \times \sqrt{30} \approx \$ 13.68\)
04

Final results

The probability distribution of the total daily sales \(x\) in the deli section of the local market is expected to have a normal distribution with a mean of \(\$ 255\) and a standard deviation of approximately \(\$ 13.68\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a key concept in statistics that plays a significant role in understanding probability distributions. It states that, given a sufficiently large sample size, the distribution of the sum of independent, identically distributed random variables will approximate a normal distribution, regardless of the original distribution of the individual variables. This holds true under the condition that no single random variable dominates the others.

In the context of our deli sales problem, we observe that the sales from individual customers can vary and could follow any distribution. However, each customer's purchase can be considered an independent random variable with its own mean and standard deviation. Since the total daily sales are a sum of these independent purchases, we can apply the CLT.

Because the number of customers is fairly large (30 in this case), according to the CLT, the distribution of the total daily sales will be approximately normal. This allows us to use the properties of the normal distribution to make predictions about the total daily sales.
Mean and Standard Deviation
Mean and standard deviation are fundamental concepts when working with any probability distribution. The mean gives us a measure of central tendency, showing where most of the data is concentrated, while the standard deviation provides insight into the spread or variability of the data.

For our deli section problem, the mean of the probability distribution of total daily sales is calculated by multiplying the average sale per customer by the number of customers. In our case:
  • Mean (\(\mu\)): \(8.50 \times 30 = 255\) dollars
This tells us that, on average, the deli section can expect to generate $255 per day from 30 customers.

Similarly, the standard deviation is calculated by taking the standard deviation of sales per customer and multiplying it by the square root of the number of customers:
  • Standard Deviation (\(\sigma_x\)): \(2.50 \times \sqrt{30} \approx 13.68\) dollars
This value indicates how much the total daily sales can vary from the mean day-to-day. A larger standard deviation would suggest more variability in daily sales.
Probability Distribution
A probability distribution describes how likely different outcomes are. It's a mathematical function that provides the probabilities of occurrence of different possible results in an experiment. Each outcome is associated with a probability, which is a number between 0 and 1.

In this exercise, we're dealing with a normal distribution, which is completely defined by its mean and standard deviation. This kind of distribution is symmetric, meaning it has a bell-shaped curve. Most of the data points tend to cluster around the mean, with the standard deviation indicating the spread.

Understanding whether a scenario follows a particular type of probability distribution helps in making predictions and decisions based on the data. For example, if we know that the total daily sales follow a normal distribution with a mean of $255 and a standard deviation of $13.68, we can efficiently calculate the probability of different levels of daily sales and manage inventory or set sales targets accordingly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercise 1.67 Allen Shoemaker derived a distribution of human body temperatures with a distinct mound shape. \(^{8}\) Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.6 degrees and a standard deviation of 0.8 degrees. a. If 130 healthy people are selected at random, what is the probability that the average temperature for these people is 98.25 degrees or lower? b. Would you consider an average temperature of 98.25 degrees to be an unlikely occurrence, given that the true average temperature of healthy people is 98.6 degrees? Explain.

The sample means were calculated for 40 samples of size \(n=5\) for a process that was judged to be in control. The means of the 40 values and the standard deviation of the combined 200 measurements were \(\overline{\bar{x}}=155.9\) and \(s=4.3,\) respectively. a. Use the data to determine the upper and lower control limits for an \(\bar{x}\) chart. b. Construct an \(\bar{x}\) chart for the process and explain how it can be used.

Refer to Exercise \(7.75 .\) During a given week the number of defective bulbs in each of five samples of 100 were found to be \(2,4,9,7,\) and 11 . Is there reason to believe that the production process has been producing an excessive proportion of defectives at any time during the week?

During long production runs of canned tomatoes, the average weights (in ounces) of samples of five cans of standard-grade tomatoes in pureed form were taken at 30 control points during an 11 -day period. These results are shown in the table. \({ }^{20}\) When the machine is performing normally, the average weight per can is 21 ounces with a standard deviation of 1.20 ounces. a. Compute the upper and lower control limits and the centerline for the \(\bar{x}\) chart. b. Plot the sample data on the \(\bar{x}\) chart and determine whether the performance of the machine is in control.

An experimenter wants to find an appropriate temperature at which to store fresh strawberries to minimize the loss of ascorbic acid. There are 20 storage containers, each with controllable temperature, in which strawberries can be stored. If two storage temperatures are to be used, how would the experimenter assign the 20 containers to one of the two storage temperatures?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.