| Course | MST-7000 Managerial Statistics |
|---|---|
| Module | Module 3 |
| Paper type | Module 3 written assignment: discrete and continuous probability distributions |
| Length | About 584 words, 5 pages |
| Format | APA 7 student paper |
| School | Thomas Edison State University |
| Program | MBA in Data Analytics |
| Updated | October 2026 |
Free sample paper for MST 7000 Module 3
How Many Returns, How Many Days? Binomial and Normal Models for an Online Retailer
Student Name
MBA in Data Analytics, Thomas Edison State University
MST-7000: Managerial Statistics
Mentor Name
Month Day, Year
How Many Returns, How Many Days? Binomial and Normal Models for an Online Retailer
Choosing the Distributions
Anderson et al. (2011) describe the binomial distribution as applying when there is a fixed number of independent trials, each with two outcomes and the same probability of success. Returns fit: each of 20 orders in a promotion batch is either returned or kept, and the historical return rate is 8 percent. Delivery time is different; it is measured on a continuous scale, and last year's 9,800 deliveries formed a roughly bell-shaped distribution with a mean of 6.2 days and a standard deviation of 1.5 days, so the normal distribution is a reasonable model. Both models are approximations, and part of the assignment is to check how well each fits before relying on it. Tukey (1977) urged analysts to look at data before fitting models, and a histogram of last year's delivery times was the first step here.
The Binomial Model for Returns
The merchandising team runs weekly promotions of 20 orders and worries when three or more are returned.
The expected number of returns is np = 20 x 0.08 = 1.6, with a standard deviation of the square root of np(1 minus p), or 1.21. Three or more returns happen in about one promotion in five by chance alone, so the team should not treat three returns as a sign of a bad product. Seven or more would be different: BINOM.DIST gives that outcome a probability below 0.001, so a batch like that would warrant a review.
| Question | Excel formula | Probability |
|---|---|---|
| No returns | BINOM.DIST(0,20,0.08,FALSE) | 0.189 |
| Exactly one return | BINOM.DIST(1,20,0.08,FALSE) | 0.328 |
| Exactly two returns | BINOM.DIST(2,20,0.08,FALSE) | 0.271 |
| One or fewer | BINOM.DIST(1,20,0.08,TRUE) | 0.517 |
| Three or more | 1 minus BINOM.DIST(2,20,0.08,TRUE) | 0.212 |
The Normal Model for Delivery Times
About 3 percent of deliveries take more than nine days, which matches the complaints the service team logs. The website currently promises delivery in seven days; NORM.DIST shows that only about 70 percent of orders meet that promise. A promise of nine days would be met by 97 percent of orders, and NORM.INV shows that 8.12 days covers 90 percent. Moving the promise to eight days would cut the share of late orders from about 30 percent to about 12 percent, at the cost of a less attractive headline.
| Question | Calculation | Result |
|---|---|---|
| Share of deliveries over 9 days | z = (9 minus 6.2) / 1.5 = 1.87; 1 minus NORM.DIST(9,6.2,1.5,TRUE) | 0.031 |
| Share under 5 days | NORM.DIST(5,6.2,1.5,TRUE) | 0.212 |
| Share between 5 and 8 days | NORM.DIST(8,...) minus NORM.DIST(5,...) | 0.673 |
| Days that cover 90 percent of deliveries | NORM.INV(0.9,6.2,1.5) | 8.12 days |
Checking the Assumptions
The binomial model assumes returns are independent, which may fail if one faulty product appears in several orders in the same batch. The normal model may understate long delays, because delivery times cannot go below about two days but can stretch far to the right during storms or holidays. I compared the normal model's 3.1 percent over nine days with last year's actual 3.6 percent, which suggests the right tail is a little heavier than the model assumes. Wilk and Gnanadesikan (1968) proposed probability plots for exactly this kind of check, and a normal probability plot of last year's data confirmed slight right skew.
Recommendations
Promotion teams should review a batch only when returns reach five or more, which has a probability under 2 percent. The website should promise delivery within eight days for standard items, which the model says covers about 88 percent of orders, and offer tracking for the rest.
References
Anderson, D. R., Sweeney, D. J., & Williams, T. A. (2011). Essentials of modern business statistics with Microsoft Excel (5th ed.). South-Western Cengage Learning.
Tukey, J. W. (1977). Exploratory data analysis. Addison-Wesley.
Wilk, M. B., & Gnanadesikan, R. (1968). Probability plotting methods for the analysis of data. Biometrika, 55(1), 1-17. https://doi.org/10.1093/biomet/55.1.1
What the MST 7000 Module 3 instructions ask for
The third written assignment covers discrete and continuous probability distributions. Expect problems that ask you to decide which distribution fits a situation, compute probabilities with the binomial and normal distributions, find values that cut off a given percentage and interpret the results. Use Excel's BINOM.DIST, NORM.DIST, NORM.S.DIST and NORM.INV functions and show the formulas you used. Explain why each distribution fits the problem, state its assumptions and say what each probability means for the business decision at hand. If the data do not quite fit the model, say so and explain how much that matters for the decision, since a perfect fit is rare in business data.
How this MST 7000 Module 3 example is built
The memo opens by choosing the binomial for returns and the normal for delivery times, with the conditions each must meet. A table computes binomial probabilities with Excel formulas, and the text finds the mean and standard deviation and turns the results into a rule. A second table computes normal probabilities and a percentile, and the text evaluates the website's delivery promise. A section checks both models' assumptions against real data, and recommendations follow. The normal section also shows how far the current seven-day promise falls short and what an eight-day promise would change for late orders. The recommendations turn both models into rules the business can use at once.
MST 7000 Module 3 rubric: what earns full marks
The marking for this assignment looks at choosing the right distribution, correct Excel calculations, checking assumptions and plain interpretation. A memo that explains why a distribution fits, rather than applying it automatically, show understanding. Comparing a model's prediction with actual data, as the memo does for the nine-day tail, shows the critical habit of testing models. Turning probabilities into decisions, such as a review rule or a delivery promise, is the purpose of the module. Mentors also check that cumulative and exact probabilities are used correctly and that upper tails are found by subtracting from one. Each result should be stated in words a merchandiser or service manager would use.
MST 7000 Module 3 help: mistakes that cost marks
Students lose points by using the binomial for continuous data, by confusing cumulative and individual probabilities in BINOM.DIST, by forgetting to subtract from one for upper tails or by applying the normal distribution to badly skewed data without comment. Others report probabilities without saying what they mean. State the distribution's conditions. Use TRUE for cumulative and FALSE for exact probabilities. Check the model against data. Turn each result into a recommendation. Give every probability a business meaning, such as how often a manager should expect a bad week by chance. Show the Excel formula beside each answer so a mentor can see exactly which tail you computed and with which arguments. A free first sample on your own MST-7000 distributions assignment is available on request.
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MST 7000 Module 3 questions, answered
Where can I find a free MST 7000 Module 3 sample paper?
Binomial returns and normal delivery times for an online furniture retailer are modeled in the assignment above.
When does the binomial distribution apply?
When there is a fixed number of independent trials, each with two outcomes and the same probability of success.
What is the mean of a binomial distribution?
n times p, the number of trials times the probability of success.
How do you find a value that covers 90 percent of a normal distribution in Excel?
Use NORM.INV(0.9, mean, standard deviation).
What does the FALSE argument do in BINOM.DIST?
It returns the probability of exactly that number of successes rather than a cumulative probability.