2.8 Constructing Discrete and Cumulative Distributions — Answers and Rubrics
These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
Build a Distribution for the Larger Spin: worked answer
Two fair four-sided spinners, each labeled 1, 2, 3, 4, are spun independently. Let X be the larger of the two numbers. All 16 ordered pairs are equally likely. A separate newly generated simulation produced these 20 pairs:
(4, 2), (2, 1), (1, 4), (2, 4), (3, 4), (2, 4), (2, 3), (2, 3), (2, 1), (4, 4), (3, 4), (3, 3), (1, 1), (4, 3), (2, 1), (1, 3), (3, 3), (2, 4), (4, 4), (3, 4)
Part a
List or systematically organize the 16 outcomes, map them to X, and identify every possible X value. Explain why X is a discrete random variable with numerical chance outcomes.
The outcomes are every ordered pair (i,j) with i,j in {1,2,3,4}. X=max(i,j), so X can be 1,2,3,4. The counts for these values are 1,3,5,7 respectively: a maximum k has k²−(k−1)²=2k−1 pairs. X numerically summarizes a random pair and has a finite discrete set of values.
| Evidence for credit | Points |
|---|---|
| Accounts for all 16 ordered pairs. | 1 |
| Maps pairs to all four maximum values and explains the discrete numerical variable. | 1 |
Part b
Construct the complete exact probability table and a probability function for X using the equally likely outcome rules. Verify that the probabilities sum to 1.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 1/16 | 3/16 | 5/16 | 7/16 |
for x=1,2,3,4 and 0 otherwise. The probabilities sum to (1+3+5+7)/16=1.
| Evidence for credit | Points |
|---|---|
| Provides all four correct probabilities. | 1 |
| Gives the function with its domain and zero otherwise. | 1 |
| Verifies total 1. | 1 |
Part c
Construct a labeled probability bar graph from your exact table. You may draw it or specify the horizontal-axis values, vertical-axis meaning and every bar height unambiguously.
The horizontal axis is X, the larger spinner number, with values 1,2,3,4. The vertical axis is probability. Bars have heights .0625, .1875, .3125, .4375 respectively. A fully labeled drawing or this explicit coordinate/bar specification represents the distribution.
| Evidence for credit | Points |
|---|---|
| Labels the variable and probability axes. | 1 |
| Constructs every bar at the correct value and height. | 1 |
Part d
Construct a cumulative table and a cumulative function F(t)=P(X≤t), including ranges below and above the possible values. Calculate P(X≤3) and P(X<3), explaining the boundary difference.
| t | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| F(t) | 1/16 | 4/16 | 9/16 | 1 |
for t<1; 1/16 for 1≤t<2; 4/16 for 2≤t<3; 9/16 for 3≤t<4; 1 for t≥4. Thus P(X≤3)=9/16, while P(X<3)=4/16. The former includes X=3; the latter does not.
| Evidence for credit | Points |
|---|---|
| Includes every cumulative value and all function intervals. | 1 |
| Obtains 9/16 and 4/16 and explains inclusion of X=3. | 1 |
Part e
Execute the maximum calculation on all 20 simulated pairs. Construct the empirical probability table for every possible X value, state the denominator and compare it with the exact table.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Count | 1 | 3 | 5 | 11 |
| Relative frequency | .05 | .15 | .25 | .55 |
All 20 pairs contribute one maximum. Counts are 1,3,5,11; dividing by 20 gives .05,.15,.25,.55, which sum to 1. These estimates differ from the exact probabilities through finite-run chance variation.
| Evidence for credit | Points |
|---|---|
| Correctly processes all simulated maxima into counts. | 1 |
| Uses denominator 20 and includes all four values. | 1 |
| Compares empirical versus exact probabilities without replacing the model by observed frequencies. | 1 |