| Course | MST-7000 Managerial Statistics |
|---|---|
| Module | Module 2 |
| Paper type | Module 2 written assignment: probability |
| Length | About 556 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 2
Which Carrier Breaks More Sofas? Probability From a Shipping Crosstab
Student Name
MBA in Data Analytics, Thomas Edison State University
MST-7000: Managerial Statistics
Mentor Name
Month Day, Year
Which Carrier Breaks More Sofas? Probability From a Shipping Crosstab
The Data
The operations director wanted to know whether our regional freight carrier damages more items than our national carrier, and whether a new warehouse inspection would catch damage before shipping. I built a crosstab with Excel's PivotTable from last quarter's 1,200 deliveries.
| Carrier | Damaged | Not damaged | Total |
|---|---|---|---|
| Regional carrier (A) | 43 | 677 | 720 |
| National carrier (B) | 17 | 463 | 480 |
| Total | 60 | 1,140 | 1,200 |
Marginal and Joint Probabilities
Anderson et al. (2011) define a marginal probability as the probability of one event regardless of the other, found in a crosstab's margins, and a joint probability as the probability that two events occur together, found in its cells. The probability that any order arrives damaged is 60 / 1,200 = 0.050. The probability that an order went by carrier A is 720 / 1,200 = 0.600. The joint probability that an order went by carrier A and was damaged is 43 / 1,200 = 0.036.
The Addition Rule
The probability that an order went by carrier A or was damaged is P(A) + P(D) minus P(A and D) = 0.600 + 0.050 minus 0.036 = 0.614. Subtracting the joint probability avoids counting the 43 damaged carrier A orders twice.
Conditional Probabilities
An order sent by carrier A is damaged about 6 times in 100, compared with about 3.5 times in 100 for carrier B. Of all damaged orders, 72 percent went by carrier A, although carrier A carried only 60 percent of orders.
| Probability | Calculation | Value |
|---|---|---|
| Damaged given carrier A | 43 / 720 | 0.060 |
| Damaged given carrier B | 17 / 480 | 0.035 |
| Carrier A given damaged | 43 / 60 | 0.717 |
Testing Independence
Two events are independent if P(D given A) equals P(D). Here 0.060 does not equal 0.050, so damage and carrier are not independent in this sample: knowing the carrier changes the probability of damage. Whether chance alone could produce a gap this size is a question for the chi-square test in Module 6; the probabilities here only describe the sample.
Bayes' Theorem and the Inspection
The warehouse is testing an inspection that, according to the vendor, flags 90 percent of damaged items and wrongly flags 4 percent of undamaged ones. If an item is flagged, what is the probability it is actually damaged? Using a prior of 0.05:
Only about 54 percent of flagged items would be damaged. Because damage is rare, the 4 percent false-flag rate on the many undamaged items produces almost as many flags as true damage does. Kahneman and Tversky (1973) showed that people tend to ignore base rates like this when judging probability, and Gigerenzer and Hoffrage (1995) found that stating the problem in natural frequencies, here 45 true flags and 38 false flags per 1,000 items, helps people reason correctly. I presented the result to the director in that form.
| Step | Value |
|---|---|
| P(flag and damaged) = 0.05 x 0.90 | 0.045 |
| P(flag and not damaged) = 0.95 x 0.04 | 0.038 |
| P(flag) | 0.083 |
| P(damaged given flag) = 0.045 / 0.083 | 0.542 |
Recommendations
The data support asking carrier A for a damage review, since its damage rate is about 70 percent higher than carrier B's. The inspection is useful, but every flagged item needs a second look before it is pulled from shipment, or nearly half of pulled items will be sound.
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.
Gigerenzer, G., & Hoffrage, U. (1995). How to improve Bayesian reasoning without instruction: Frequency formats. Psychological Review, 102(4), 684-704. https://doi.org/10.1037/0033-295X.102.4.684
Kahneman, D., & Tversky, A. (1973). On the psychology of prediction. Psychological Review, 80(4), 237-251. https://doi.org/10.1037/h0034747
MST 7000 Module 2 instructions, in plain terms
The second written assignment covers probability. Expect problems that ask you to compute marginal, joint and conditional probabilities from a table, apply the addition and multiplication rules, test whether two events are independent and use Bayes' theorem to revise a probability with new information. Build the tables in Excel, show each formula and explain each result in words a manager can use. Where a probability changes a decision, say how. Keep the distinction clear between describing a sample and testing whether a difference is significant, which comes later in the course. Present at least one result in natural frequencies, as counts per thousand, if the problem involves a rare event.
Inside the MST 7000 Module 2 example
The memo opens with the operational question and a crosstab of 1,200 orders. Sections compute marginal and joint probabilities, apply the addition rule and set out conditional probabilities in a table. A section tests independence and notes what probability alone cannot show. A Bayes section works the inspection problem step by step, explains the base-rate effect with research and presents the result in natural frequencies. Recommendations follow for the carrier and the inspection. Each probability is labeled with its notation and calculation so a reader can trace it back to the crosstab, and the conditional probabilities sit in one table for easy comparison. The recommendations treat the inspection as useful but not final, since nearly half its flags would be false.
Reading the MST 7000 Module 2 grading rubric
Mentors grade this assignment on correct probability calculations, correct use of rules and Bayes' theorem, and clear interpretation. Papers that explain why a positive test means less than it seems, because of the base rate, show real understanding of Bayes' theorem. Distinguishing a probability difference in a sample from a statistically significant one shows careful reasoning. Presenting results in a form managers grasp, such as natural frequencies, demonstrates the communication skill the course values. Clear notation, with conditional probabilities written the same way every time, prevents the most common errors. Showing the joint table behind Bayes' theorem lets a mentor follow every step of the revision.
Common MST 7000 Module 2 mistakes, and how to avoid them
Students lose points by confusing P(A given B) with P(B given A), by forgetting to subtract the joint probability in the addition rule or by treating a vendor's detection rate as the probability that a flagged item is bad. Others claim a difference is significant without a test. Label every conditional probability clearly. Show the joint table before using Bayes' theorem. Say what each number means for the decision. Present a Bayes result as counts out of a thousand if the reader is not a statistician. Check that each row and column of your table adds to its total before computing anything, because one wrong cell carries through every probability. A free first sample on your own MST-7000 probability assignment is available on request.
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MST 7000 Module 2 questions, answered
Where can I find a free MST 7000 Module 2 sample paper?
A furniture retailer's carrier and damage crosstab, with Bayes' theorem, is worked in the assignment above.
What is a conditional probability?
The chance of one event once another is known to have happened, such as damage when carrier A is used.
How do you test whether two events are independent?
Check whether P(A given B) equals P(A); if not, the events are dependent.
What does Bayes' theorem do?
It revises a prior probability using new information, such as an inspection flag.
Why can a positive test be misleading?
When the event is rare, false positives on the many negative cases can be nearly as common as true positives.