PSTAT 5A Practice Worksheet 4 - SOLUTIONS
Comprehensive Review: Discrete Random Variables and Distributions
Section A: Basic Concepts and Identification - SOLUTIONS
Problem A1: Distribution Identification
Important
Instructions: For each scenario below, identify the appropriate probability distribution and specify its parameters. Justify your choice by identifying the key characteristics.
(a) Coin Flipping Until First Head
A fair coin is flipped until the first head appears. Let \(X\) = number of flips needed.
Solution:
(b) Quality Control Inspection
A quality control inspector tests \(20\) randomly selected items from a production line where \(5\%\) are defective. Let \(X\) = number of defective items found.
Solution:
(c) Website Visitor Count
A website receives visitors at an average rate of \(3\) per minute. Let \(X\) = number of visitors in a 2-minute period.
Solution:
(d) Single Free Throw
A basketball player shoots one free throw with an \(80\%\) success rate. Let \(X = 1\) if successful, \(0\) if unsuccessful.
Solution:
(e) Driving Test Attempts
A student keeps taking a driving test until they pass. The probability of passing on any attempt is \(0.7\). Let \(X\) = number of attempts needed to pass.
Solution:
Summary Table
Decision Framework Visualization
Tip
Quick Reference Guide Ask these key questions to identify distributions:
How many trials?
One trial → Bernoulli
Fixed number → Binomial (if counting successes)
Until first success → Geometric
What are we counting?
Successes in fixed trials → Binomial
Trials until success → Geometric
Events over time/space → Poisson
Time component?
Events at constant rate over time → Poisson
No time component → Binomial/Bernoulli/Geometric
Warning
Geometric vs. Binomial: Geometric counts trials until success;
Binomial counts successes in fixed trials
Poisson parameter: Remember to multiply rate by time period (e.g., 3/minute × 2 minutes = \(\lambda\) = 6)
Independence assumption: All these distributions require independent trials/events
Problem A2: Probability Mass Function
Given distribution:
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X=k) | 0.1 | 0.3 | 0.4 | a | 0.1 |
(a) Find the value of \(a\).
Solution. Since probabilities must sum to \(1\):
\(0.1 + 0.3 + 0.4 + a + 0.1 = 1\)
\(0.9 + a = 1\)
\(\boxed{a = 0.1}\)
(b) Calculate \(P(X \leq 3)\).
Solution. \(P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)\)
\(P(X ≤ 3) = 0.1 + 0.3 + 0.4 = \boxed{0.8}\)
(c) Calculate \(P(X > 2)\).
Solution. \(P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)\)
\(P(X > 2) = 0.4 + 0.1 + 0.1 = \boxed{0.6}\)
(Check: \(0.8 + 0.2 = 1\) and the full PMF sums to 1, so the results are consistent.)
Putting everything together:
Tip
Key Insights from Visualizations
Distribution Shape: The PMF shows \(X = 3\) has the highest probability (\(0.4\)), making it the mode
Cumulative Probability: \(P(X ≤ 3) = 0.8\) means \(80\%\) of outcomes are 3 or less
Complement Relationship: \(P(X > 2) = 0.6\) and \(P(X ≤ 2) = 0.4\) sum to \(1\)
Symmetry: The distribution has some symmetry around the center, with equal probabilities at the extremes (\(X = 1 \quad \text{and} \quad X = 5\) both have \(P = 0.1\))
Section B: Expected Value and Variance - SOLUTIONS
Problem B1: Manual Calculations
Using the distribution from Problem A2:
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X=k) | 0.1 | 0.3 | 0.4 | 0.1 | 0.1 |
(a) Compute the expected value (E[X])
Solution. For a discrete random variable, the expected value is the probability-weighted average of all possible outcomes:
\[ E[X] \;=\;\sum_{k=1}^{5} k \times \,P(X=k). \]
- Set up the sum
\[ E[X] \;=\; 1(0.1) \;+\; 2(0.3) \;+\; 3(0.4) \;+\; 4(0.1) \;+\; 5(0.1). \]
- Multiply each outcome by its probability
\[ = 0.1 \;+\; 0.6 \;+\; 1.2 \;+\; 0.4 \;+\; 0.5. \]
- Add the terms
\[ \boxed {E[X] = 2.8} \]
Tip
Visual Interpretation
Looking at the PMF plot:
The highest probability (\(0.4\)) occurs at \(X = 3\)
The second highest (\(0.3\)) occurs at \(X = 2\)
Together, these two values account for \(70\%\) of the probability mass
The expected value \(E[X] = 2.8\) (red dashed line) falls between these two most likely outcomes
This visual confirms our intuition that the “center of gravity” should be close to, but slightly less than, 3
Note
Interpretation & quick check
Interpretation: If we were to observe this experiment many, many times, the long-run average value of \(X\) would settle down around 2.8. Although 2.8 itself isn’t an attainable outcome (only integers 1–5 are), it represents the center of gravity of the distribution.
Check: Notice most probability mass is on 2 and 3 (0.3 + 0.4 = 0.7). A mix that skews slightly toward the larger of those two values should indeed give an average a bit below 3, exactly what we see with 2.8.
(b) Compute the variance \(\operatorname{Var}(X)\)
Solution. The variance measures how far the values of (X) tend to deviate from the mean.
We use the shortcut formula
\[ \operatorname{Var}(X) \;=\; E[X^2] - \bigl(E[X]\bigr)^2, \]
where \(E[X]=2.8\) was found in part (a).
- Find \(E[X^2]\) (the mean of the squared outcomes)
\[ \begin{aligned} E[X^2] &= \sum_{k=1}^{5} k^{2}\,P(X=k) \\[4pt] &= 1^{2}(0.1) \;+\; 2^{2}(0.3) \;+\; 3^{2}(0.4) \;+\; 4^{2}(0.1) \;+\; 5^{2}(0.1) \\[4pt] &= 1(0.1) \;+\; 4(0.3) \;+\; 9(0.4) \;+\; 16(0.1) \;+\; 25(0.1) \\[4pt] &= 0.1 \;+\; 1.2 \;+\; 3.6 \;+\; 1.6 \;+\; 2.5 \\[4pt] &= 9.0 \end{aligned} \]
- Apply the variance formula
\[ \operatorname{Var}(X) \;=\; 9.0 - (2.8)^2 = 9.0 - 7.84 = \boxed{1.16} \]
Note
Interpretation & quick check
Interpretation: A variance of \(1.16\) tells us that typical values of \(X\) deviate from the mean (\(2.8\)) by a little over one unit (figure 8).
Check: Most probability mass is on 2 and 3; the only “far” value is \(5\) (probability \(0.1\)). So we expect a modest spread, larger than \(0\) but well below the maximum possible of \((5-2.8)^2 = 4.84\). The calculated \(1.16\) fits this intuition.
(c) Compute the standard deviation \(\sigma\)
Solution. The standard deviation is the square root of the variance:
\[ \sigma \;=\; \sqrt{\operatorname{Var}(X)} \;=\; \sqrt{1.16} \;\approx\; \boxed{1.08}. \]
Note
Interpretation
A standard deviation (\(\sigma\)) of about \(1.08\) means typical observations of \(X\) lie roughly one unit above or below the mean value \(2.8\). This agrees with our earlier intuition that the distribution is fairly concentrated around \(2 – 3\), with only a small chance of the extreme value \(5\).
Let’s visualise this!
Problem B2: Bernoulli and Binomial Applications
Manufacturing Scenario: A manufacturing process has a 15% defect rate.
(a) Single Item Selection
If you select one item randomly, what is the expected value and variance of \(X\) = number of defective items?
Solution. This is a Bernoulli distribution with parameter \(p = 0.15\)
\[X \sim \text{Bernoulli}(p = 0.15)\]
Step 1: Expected Value \[E[X] = p = \boxed{0.15}\]
Step 2: Variance \[\text{Var}(X) = p(1-p) = 0.15 \times 0.85 = \boxed{0.1275}\]
Step 3: Standard Deviation \[\sigma = \sqrt{\text{Var}(X)} = \sqrt{0.1275} = \boxed{0.357}\]
Note
Interpretation:
On average, 15% of items selected will be defective
Since this is a single trial, \(X\) can only be 0 (not defective) or 1 (defective)
The variance measures the uncertainty in this binary outcome
(b) Multiple Items Selection
If you select 25 items randomly, what is the expected number of defective items and the standard deviation?
Solution. This is a Binomial distribution with parameters \(n = 25\), \(p = 0.15\)
\[X \sim \text{Binomial}(n = 25, p = 0.15)\]
Step 1: Expected Value \[E[X] = np = 25 \times 0.15 = \boxed{3.75}\]
Step 2: Variance \[\text{Var}(X) = np(1-p) = 25 \times 0.15 \times 0.85 = \boxed{3.1875}\]
Step 3: Standard Deviation \[\sigma = \sqrt{\text{Var}(X)} = \sqrt{3.1875} = \boxed{1.785}\]
Note
Interpretation:
On average, we expect about 3.75 defective items out of 25
The actual number will typically be within ±1.785 items of this average
Values between 2 and 6 defective items would be quite common
Visualizations
Let’s visualize this to build more intuition
Bernoulli Distribution (Single Item)
Binomial Distribution (25 Items)
Comparison: Bernoulli vs Binomial Relationship
Important
\(\textbf{Bernoulli} \rightarrow \text{Binomial Connection:}\)
A Binomial distribution is the sum of \(n\) independent Bernoulli trials
If \(X_1, X_2, \dots, X_{25}\) are independent \(\text{Bernoulli}(0.15)\), then \(X_1 + X_2 + \cdots + X_{25} \sim \text{Binomial}(25, 0.15)\)
\(\textbf{Scaling Formulas:}\)
\(\textbf{Expected Value:}\) \(E[\text{Binomial}] = n \times E[\text{Bernoulli}]\) = \(25 \times 0.15 = 3.75\)
\(\textbf{Variance:}\) \(\text{Var}(\text{Binomial}) = n \times \text{Var}(\text{Bernoulli}) = 25 \times 0.1275 = 3.1875\)
Tip
Single inspection: \(15\%\) chance of finding a defect
Batch inspection (\(25\) items): Expect \(3-4\) defective items typically Acceptable range: \(2-6\) defective items would be within \(1\) standard deviation
Red flag: Finding 7+ defective items might indicate process issues (beyond 2 \(\sigma\))
Optional: Conceptual Understanding - SOLUTIONS
Important
Objective: Deepen understanding of key differences between probability distributions and their applications.
(a) Binomial vs. Geometric Distributions
Explain the key difference between a Binomial distribution and a Geometric distribution in terms of what they count.
Solution:
- Binomial Distribution: Counts the number of successes in a fixed number of trials
- Example: “How many heads in 10 coin flips?”
- We know we’ll flip exactly 10 times, but don’t know how many heads
- Geometric Distribution: Counts the number of trials needed to get the first success
- Example: “How many coin flips until the first head?”
- We know we’ll get exactly 1 head, but don’t know how many flips it takes
Visual Comparison : Binomial vs Geometric - Fundamental Difference
Tip
Binomial: “How many successes in a fixed box of trials?”
Fixed trials, variable successes
Geometric: “How many attempts until first success?”
Fixed successes (1), variable trials
(b) Poisson vs. Binomial: When to Use Each
When would you use a Poisson distribution instead of a Binomial distribution?
Solution. Use Poisson when:
Events occur over time or space at a constant rate
The number of possible events is very large but the probability of each is very small
We don’t have a fixed number of trials Examples: arrivals, defects per unit area, accidents per day
Comparative Examples: Poisson vs Binomial - Choosing the Right Distribution
Warning
Common Mistake
Don’t use Poisson just because events are “rare.” The key criteria are:
No fixed number of trials
Events over time/space
Constant rate (\(\lambda\))
A rare event in a fixed number of trials is still Binomial!
(c) Variance Maximization in Binomial Distribution
If \(X \sim \text{Binomial}(n, p)\), under what conditions would the variance be maximized?
Solution. For a Binomial distribution: \(\text{Var}(X) = np(1-p)\)
For fixed \(n\), variance is maximized when \(p(1−p)\) is maximized.
Approach:
Taking the derivative with respect to \(p\):
\(\frac{d}{dp} \bigl[ p(1-p) \bigr] = \frac{d}{dp} \bigl[ p - p^2 \bigr] = 1 - 2p\)
Setting equal to zero:
\(1−2p=0 \quad \implies \boxed{p = 0.5}\)
Second derivative \(= −2<0\), confirming this is a maximum.
The variance is maximized when \(p=0.5\) (fair coin scenario).
Visualization of Variance vs. Probability
Intuitive Understanding of Variance Maximization
Important
Key Insights
Maximum: \(p=0.5\) maximizes \(p(1−p)\) for any fixed \(n\)
Intuitive Explanation: Maximum uncertainty occurs when success and failure are equally likely
Practical Meaning: A fair coin (50-50) has the highest variability in outcomes
Extremes: When \(p\) approaches \(0\) or \(1\), outcomes become predictable (low variance)