L-PR-4: Solutions — Discrete Random Variables

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Classify a Random Variable (C1)

• Help-desk calls per shift: discrete. Calls are counted in whole numbers.
• Student height: continuous. A height can take any value in an interval; rounding is a measurement choice.
• Defective items in a sample of 50: discrete. The count is an integer from 0 through 50.
• Waiting time: continuous. Time can take any non-negative real value.
• Number of siblings: discrete. It is a whole-number count.

Problem 2 — Test a PMF (C2)

This item regenerates. For the displayed PMF, check both conditions separately:

1. Every probability must satisfy .
2. The probabilities must satisfy .

A PMF does not need consecutive, equally spaced, or non-negative outcome values. It is the probabilities—not the outcomes—that must be non-negative.

Problem 3 — Find the Long-Run Average (C3)

This item regenerates. Multiply each displayed outcome by its probability, then add the products:

The generated feedback shows every weighted term for the current PMF. Do not average the outcome values without using their probabilities.

Problem 4 — Measure Distribution Spread (C4 + C5)

This item regenerates. Use the displayed PMF and keep full precision through the variance step:

1. Compute .
2. Compute .
3. Compute .
4. Compute .

Round only the requested final displays. In particular, unless is constant.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Summarize a Distribution (C2 + C3 + C4 + C5)

This problem regenerates and provides instance-specific feedback. A complete solution has six distinct steps:

1. Verify non-negativity and show that the probabilities sum to 1.
2. Compute as a weighted sum.
3. Compute by squaring each outcome before weighting.
4. Compute .
5. Compute .
6. Interpret as a long-run average and as a typical distance from that average in the units of .

Problem 2 — Read Cumulative Probability (C6)

This problem regenerates. Build the cumulative total from the smallest support value upward:

Then use the endpoint rules that match the question:

The final CDF level must equal 1, and the levels can never decrease.

Problem 3 — Evaluate a Game and Diagnose Reasoning (C3 + C5)

The generated table already gives net payoffs, so do not subtract another fee. Compute

• : fair in expected-value terms.
• : favourable to the player in the long run.
• : unfavourable to the player in the long run.

The companion error-analysis item changes its error family and position. Check its three displayed stages in order:

1. expected value;
2. second moment ;
3. variance.

The game workflow asks for both parts separately: first enter the expected net payoff, then classify the game from the player’s point of view. In the companion error analysis, later stages consistently use the student’s displayed earlier values. Select the first stage where the reasoning stops being valid.

Problem 4 — Complete a PMF and Find Its Mean (C2 + C3)

This item regenerates.

1. Add the known probabilities.
2. Compute .
3. Substitute into the PMF.
4. Compute using the completed distribution.

Problem 5 — Compare Centre and Spread (C3 + C5)

This item regenerates. After the numerical work:

• interpret as the long-run average value per repetition;
• interpret as the typical distance of an outcome from that mean, in the original units.

Standard deviation alone does not establish that outcomes “usually” fall between and .

Problem 6 — Synthesize a Full Distribution Analysis (C2 + C3 + C4 + C5 + C6)

This item regenerates. A complete response:

1. verifies the PMF;
2. computes ;
3. computes ;
4. computes variance and SD;
5. builds the cumulative probabilities;
6. answers the requested tail or interval probability with the correct endpoints;
7. interprets the results in context.

Mixed Review — Retrieval from Earlier Lessons

Review Problem 1 — Conditional probability from a table

For a two-way count table, compare independence using exact counts:

If the cross-products are equal, the events are independent. If they are not equal, the events are dependent. Rounded conditional-probability decimals are not an exact independence test.

Review Problem 2 — Combinations in a probability calculation

When order does not matter, count selections with combinations. For a sample of objects drawn without replacement from , with target objects, the probability of exactly targets is

The generated feedback identifies the favourable selections and the total selections separately.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

Expected value is a probability-weighted average of all outcomes. The mode is only the most probable single outcome, so the two summaries answer different questions.

For a fair six-sided die, every face is equally likely, so all six values tie as modes.

Its expected value is

The expected value is not one of the modal outcomes—or even a possible roll. It describes the long-run average of many rolls, not the result of one roll.

Question 2 — Apply Expected Value in a New Setting

This item regenerates. Let be the displayed number of same-price repair jobs and let the shop receive dollars per job. The daily revenue is .

First compute the expected job count from the current PMF:

Then scale the expected value:

The generated feedback substitutes every weighted term and gives the expected daily revenue in dollars. Computing directly from the transformed revenue outcomes is also mathematically valid.

Question 3 — Error Analysis

This item regenerates. Check the student’s work one line at a time and identify the first line that is not justified. Common generated families include:

• omitting the probability weights when computing ;
• replacing with ;
• reversing the variance subtraction;

Later lines follow the values displayed earlier in the student’s work. The generated feedback recomputes the current PMF and explains why the selected line is the first error, even when that earlier mistake changes the final variance.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

🎡 Carnival Decision

The game costs $5. The prizes are $20, $10, $3, and $0 with probabilities .

Task 1 — Net-outcome PMF

Net outcome

Every probability is non-negative.

The probability total is

Task 2 — Expected net outcome

The expected net is 1.60$ per play, so the game is unfavourable to the player in the long run.

Task 3 — Variance and standard deviation

Task 4 — Model recommendation

The most common outcome is a $5 loss, but expected value uses every outcome and probability. Over many repeated plays, the average net outcome tends toward 1.605.11 shows that individual results vary substantially around that long-run average; it does not guarantee that a result falls within one SD of the mean. I would not recommend the game as a long-run money-making strategy.

💡 Payout Design

One valid construction on four positive integers is

	1	2	3	4
	0.35	0.40	0.15	0.10

All four support values are positive integers, and every probability is positive.

The probability total is

The expected value is

The second moment is

The variance is

The standard deviation is

This distribution has the requested long-run mean and a relatively small typical distance from that mean. Standard deviation alone does not determine the probability of landing inside any particular interval around the mean. Many other PMFs satisfy the constraints; each one must be verified.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Derive the Shortcut Formula

Let . Start from the definition:

Expand the square:

Use linearity of expectation and treat as a constant:

Substitute :

Therefore,

Because for every outcome, its expectation is non-negative. Variance equals zero exactly when with probability 1.

Challenge 2 — Transform Centre and Spread

For each generated PMF, first compute

and

For constants and , let .

First find the transformed mean:

Then apply the definition of variance:

Substitute and :

Factor out :

Therefore,

Adding shifts every outcome without changing spread. Multiplying by scales every distance from the mean by , so .