Discrete and continuous models for real business decisions
Running scenario: Vantage Retail Co. Australian e-commerce and retail analytics firm — modelling returns, defects, orders, delivery times and customer spend.
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Week 5 — Learning Objectives
Discrete Random Variables
Find expected value, variance and standard deviation for a discrete random variable
Calculate and interpret covariance between two random variables
Binomial Distribution
Identify when the binomial distribution is appropriate
Calculate and interpret binomial probabilities
Poisson Distribution
Identify when the Poisson distribution is appropriate
Calculate and interpret Poisson probabilities
Scale the Poisson rate across different time intervals
Vantage Retail Co. — Week 5 Contexts
Product returns per order — discrete RV
Defective items in a shipment — binomial
Online orders per hour — Poisson
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Section 1 — Week 5
Discrete Random Variables
Assigning numbers to uncertain outcomes — and measuring what we expect and how much things vary.
1.1
1.1 What Is a Random Variable?
Definition
A random variable \(X\) is a numerical outcome of a random experiment. It assigns a single real number to each outcome in the sample space.
Discrete Random Variable
Takes a countable set of distinct values — typically whole numbers.
Number of items a customer returns
Number of defective units in a batch
Number of support calls received per day
Continuous Random Variable
Takes any real value in an interval — measurements on a continuous scale.
Customer transaction value (\$)
Delivery time (hours)
Product weight (grams)
Scope of Weeks 5 & 6
Week 5 covers discrete random variables — general, binomial, and Poisson. Week 6 shifts to continuous distributions — normal, uniform, and exponential.
1.2
1.2 Discrete Probability Distributions
A probability distribution specifies every value \(X\) can take and its associated probability.
Vantage Retail Co. — Returns per Online Order
Analysis of 10,000 orders yields the following distribution for \(X\) = number of items returned:
\(x\)
0
1
2
3
\(P(X=x)\)
0.55
0.25
0.15
0.05
Rule 1 — Valid Probabilities
\(0 \le P(X = x) \le 1\) for every value of \(x\).
The expected value (mean) is the probability-weighted average of all possible outcomes:
$$E(X) = \mu = \sum_{x} x \cdot P(X = x)$$
Vantage Retail Co. — Expected Returns per Order
\(x\)
\(P(X=x)\)
\(x \cdot P(X=x)\)
0
0.55
0.00
1
0.25
0.25
2
0.15
0.30
3
0.05
0.15
Total
1.00
0.70
Operations Manager — Interpretation
\(E(X) = 0.70\) items returned per order. Over 10,000 orders, Vantage should plan for approximately 7,000 return items — directly informing logistics staffing and returns-processing budgets.
Vantage Retail Co. — Variance of Returns (\(\mu = 0.70\))
\(x\)
\((x - 0.70)^2\)
\(P(X=x)\)
\((x-\mu)^2 \cdot P(x)\)
0
0.4900
0.55
0.2695
1
0.0900
0.25
0.0225
2
1.6900
0.15
0.2535
3
5.2900
0.05
0.2645
Variance \(\sigma^2\)
0.8100
\(\sigma = \sqrt{0.81} = 0.90\) items. Returns typically vary by 0.90 items around the mean of 0.70.
1.5
1.5 Covariance
Formula
The covariance measures the direction and magnitude of the linear relationship between two random variables \(X\) and \(Y\):
$$\text{Cov}(X, Y) = \sigma_{XY} = \sum_{x,y}(x - \mu_X)(y - \mu_Y) \cdot P(X=x, Y=y)$$
Interpreting the Sign
Value
Relationship
\(\sigma_{XY} > 0\)
X and Y tend to increase together
\(\sigma_{XY} < 0\)
One increases as the other decreases
\(\sigma_{XY} = 0\)
No linear association
Limitation
Covariance is unit-dependent. To compare relationships across variables, standardise using the correlation coefficient: \(\rho = \sigma_{XY} / (\sigma_X \cdot \sigma_Y)\)
Vantage Retail Co.
Let \(X\) = number of returns, \(Y\) = refund cost per order (\$).
A positive covariance here is expected — orders with more returns generate higher refund costs.
This quantifies the relationship needed to forecast the total refund liability as a function of return volume.
1.Q
Knowledge Check — Section 1
Q1: A discrete distribution gives \(P(X=0)=0.30\), \(P(X=1)=0.50\), \(P(X=2)=0.20\). What is \(E(X)\)?
Q2: Vantage finds that as the number of returns per order increases, average customer satisfaction decreases. The covariance \(\sigma_{XY}\) between returns and satisfaction is:
Covariance is negative when one variable tends to increase as the other decreases. Higher returns → lower satisfaction means \(\sigma_{XY} < 0\).
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Section 2 — Week 5
The Binomial Distribution
Counting successes across a fixed number of independent trials — the most widely used discrete distribution in auditing and quality control.
2.1
2.1 When Is the Binomial Distribution Appropriate?
All four conditions must hold simultaneously:
1. Fixed trials
The number of trials \(n\) is determined in advance and does not change.
2. Binary outcomes
Each trial produces exactly one of two outcomes: success or failure.
3. Constant probability
The probability of success \(p\) is the same on every trial.
4. Independence
The outcome of one trial does not affect any other trial.
Vantage Retail Co. — Checking the Conditions
Scenario A: A shipment of 20 electronics units arrives. Each unit is independently defective with probability 5%. Count the defectives. ✓ All four conditions hold — binomial applies.
Scenario B: Count the total revenue from 20 orders. Revenue is continuous and not binary. ✗ Binomial does not apply.
There is an 18.9% probability that exactly 2 units in this shipment are defective.
Reading the Formula
\(\binom{n}{x}\) counts the arrangements; \(p^x\) is the probability of \(x\) successes; \((1-p)^{n-x}\) is the probability of the remaining \(n-x\) failures. Multiply all three together.
\(\mu = 20 \times 0.05 = \mathbf{1.0}\) defective unit per shipment on average
\(\sigma^2 = 20 \times 0.05 \times 0.95 = 0.95\)
\(\sigma = \sqrt{0.95} \approx \mathbf{0.97}\) units
Quality Manager View
On average, 1 defective unit per shipment. Finding 0–3 defectives is well within one standard deviation — entirely normal variation. A shipment with 5+ defectives should prompt a supplier review.
Figure 2.3: Binomial(20, 0.05) — right-skewed with peak at \(x=1\), matching \(\mu=1.0\)
2.4
2.4 Cumulative Binomial Probabilities
Many business decisions require probabilities over a range of outcomes, not just a single value.
Shipment: \(n=20, p=0.05\). Vantage's quality policy states a shipment is acceptable if it has at most 2 defectives. What is the probability a shipment is accepted?
There is a 92.5% probability any given shipment passes the quality check.
Complement: \(P(X > 2) = 1 - 0.9246 = \mathbf{0.0754}\) — only a 7.5% chance of a shipment being rejected.
Audit Application
Auditors use cumulative binomial probabilities to assess whether the number of errors found in a sample is consistent with a claimed low error rate — a key tool in audit risk assessment and sampling plans.
2.Q
Knowledge Check — Section 2
Q1: A survey finds 30% of Vantage customers use a loyalty card. In a random sample of 10 customers, what is the expected number who use a loyalty card?
\(\mu = np = 10 \times 0.30 = 3\). The binomial mean is simply the product of the number of trials and the probability of success.
Q2: Which scenario does not satisfy the conditions for a binomial distribution?
Option B describes a geometric distribution — you count trials until the first success, so \(n\) is not fixed in advance. This violates the binomial's fixed-\(n\) condition.
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Section 3 — Week 5
The Poisson Distribution
Modelling the number of events in a fixed interval — ideal for website traffic, call volumes, and transaction counts.
3.1
3.1 When Is the Poisson Distribution Appropriate?
1. Count of events
\(X\) counts how many times an event occurs in a fixed interval of time, space, or volume.
2. Independence
Each event occurs independently — one event does not affect the probability of another.
3. Constant rate
Events occur at a constant average rate \(\lambda\) throughout the interval.
4. Rare in small intervals
Two events cannot occur at exactly the same instant (probability of simultaneous events is negligible).
Vantage Retail Co. — Valid Poisson Scenarios
Number of online orders per hour during peak trading
Number of customer support calls per day
Number of payment processing errors per week
Number of product page visits per minute during a flash sale
Single Parameter
The Poisson distribution is completely specified by one parameter: \(\lambda\) — the average number of events per interval.
There is approximately a 10.5% probability of receiving exactly 10 orders in any given peak hour.
3.3
3.3 Mean and Variance of the Poisson Distribution
Mean
$$E(X) = \mu = \lambda$$
Variance
$$\text{Var}(X) = \sigma^2 = \lambda$$
Remarkable Property
For the Poisson distribution, mean = variance = \(\lambda\). This is unique among distributions. In practice, if empirical data show mean ≈ variance, the Poisson model is plausible. If variance is much larger than the mean, consider the negative binomial instead.
Vantage Retail Co. — Staffing the Fulfilment Centre (\(\lambda = 12\))
\(\sigma = \sqrt{12} \approx 3.46\) orders
Within \(\pm 2\sigma\) of the mean:
\(12 - 2(3.46) = 5.1\) to \(12 + 2(3.46) = 18.9\)
So roughly 5 to 19 orders per peak hour covers ~95% of cases.
Operations Manager Decision
Staff the fulfilment centre to handle up to 19–20 orders per hour during peak periods. Demand above 20 is rare but should be covered by an overflow escalation protocol.
3.4
3.4 Scaling the Poisson Rate Across Intervals
The Poisson rate \(\lambda\) is always tied to a specific interval. When the question uses a different interval length, scale \(\lambda\) proportionally first.
Orders average \(\lambda = 12\) per hour. The warehouse manager needs to know: what is the probability of fewer than 5 orders arriving in a 30-minute window?
There is a 28.5% probability of fewer than 5 orders arriving in any 30-minute window — a lull that could be used for restocking or staff breaks.
3.Q
Knowledge Check — Section 3
Q1: Vantage's call centre receives an average of 4 customer service calls per hour. What are the mean and variance of calls per hour?
For the Poisson distribution, mean = variance = \(\lambda\). With \(\lambda = 4\), both the mean and variance equal 4. This equality is a defining characteristic of the Poisson.
Q2: If the call centre receives 4 calls per hour on average, what is the correct \(\lambda\) for a 15-minute window?
\(\lambda_{15\text{min}} = 4 \times (15/60) = 4 \times 0.25 = 1\). Always scale \(\lambda\) proportionally to the new interval length before calculating any probabilities.
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Week 5 — Summary
General Discrete RV
\(E(X) = \sum x \cdot P(x)\)
\(\text{Var}(X) = \sum (x-\mu)^2 P(x)\)
Covariance measures direction of linear association. Correlation standardises it.
Count in a fixed interval. Scale \(\lambda\) when the interval changes.
Selecting the Right Model
If you observe…
Consider
A fixed number of trials with binary outcomes
Binomial
A count of events occurring in a continuous interval
Poisson
A general discrete distribution not fitting the above
E(X) and Var(X) directly
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Week 6 · Statistics for Accounting
Continuous Probability Distributions
Normal, Uniform, and Exponential — modelling measured outcomes in retail and operations
Continuing with Vantage Retail Co. Customer spend distributions, delivery time windows, and inter-order gaps.
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Week 6 — Learning Objectives
Normal Distribution
Calculate areas under the standard normal curve
Solve and interpret normal distribution problems
Check assumptions of normality
Uniform Distribution
Calculate probabilities using the uniform distribution
Solve and interpret uniform distribution problems
Exponential Distribution
Calculate probabilities using the exponential distribution
Solve and interpret exponential distribution problems
A Fundamental Shift: Discrete → Continuous
For continuous distributions, probabilities are areas under a curve — not sums. Crucially, \(P(X = \text{exact value}) = 0\). We always compute \(P(a \le X \le b)\) by finding the area between \(a\) and \(b\). This is a conceptual break from Week 5.
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Section 4 — Week 6
The Normal Distribution
The bell curve — the foundation of statistical inference, quality control, and financial modelling.
4.1
4.1 Properties of the Normal Distribution
Notation and Density
\(X \sim N(\mu,\, \sigma^2)\). The probability density function is:
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$$
Symmetric about \(\mu\) (mean = median = mode)
Bell-shaped, unimodal
Total area under curve = 1
Tails approach but never touch the x-axis
Empirical Rule
68% within \(\pm1\sigma\) · 95% within \(\pm2\sigma\) · 99.7% within \(\pm3\sigma\)
Figure 4.1: Bell curve — symmetric, 68% within ±1σ
4.2
4.2 The Standard Normal Distribution and Z-Scores
Standardising — The Z-Score
Any \(X \sim N(\mu, \sigma^2)\) can be converted to a standard normal \(Z \sim N(0,1)\) using:
$$Z = \frac{X - \mu}{\sigma}$$
\(Z\) measures how many standard deviations \(X\) lies above (positive) or below (negative) the mean.
Vantage Retail Co. — Customer Transaction Values
Transaction values follow \(X \sim N(\mu = 85,\; \sigma = 20)\) dollars. Calculate the Z-score for a transaction of \$115:
A \$115 transaction is 1.5 standard deviations above the mean.
Why This Matters
Once standardised to \(Z\), we use a single standard normal (Z) table — which gives \(P(Z \le z)\) — to find any probability for any normal distribution, regardless of \(\mu\) and \(\sigma\).
4.3
4.3 Finding Areas Under the Normal Curve
The Z-table gives \(P(Z \le z)\) — the cumulative area to the left of any \(z\) value.
Vantage wants to identify the top 10% of transactions for a VIP loyalty tier. Transaction values: \(X \sim N(85, 20)\). What is the minimum qualifying spend?
Step 1: Find \(z\) such that \(P(Z \le z) = 0.90\). From the Z-table: \(z = 1.28\).
Decision: Customers spending more than \$110.60 per transaction qualify for VIP status, capturing the top 10% of spenders by transaction value.
4.6
4.6 Checking Assumptions of Normality
Before applying the normal distribution, verify the data are approximately normal. Never assume without checking.
Visual Methods
Histogram
Should be roughly bell-shaped and symmetric. Severe skew, heavy tails, or bimodality indicate non-normality.
Normal Q-Q Plot
Plot sample quantiles against theoretical normal quantiles. Points should track closely along a diagonal straight line if the data are normal.
Formal Tests
Shapiro-Wilk Test
Preferred for small samples (\(n < 50\)). A p-value above 0.05 means we cannot reject normality.
Practical Reality
No real dataset is perfectly normal. The question is whether departures are large enough to invalidate the analysis. For large samples, the Central Limit Theorem often rescues us — sample means are approximately normal regardless of the population shape.
4.Q
Knowledge Check — Section 4
Q1: Vantage's delivery lead times follow \(X \sim N(\mu=3.5 \text{ days},\; \sigma=0.5)\). What is the Z-score for a lead time of 4.5 days?
\(Z = (4.5 - 3.5)/0.5 = 1.0/0.5 = 2.0\). A 4.5-day delivery is 2 standard deviations above the mean — only the slowest ~2.3% of deliveries are this late.
Q2: Using the same distribution, what is \(P(X \le 3.0 \text{ days})\)? (Hint: \(Z = -1.0\) gives \(P(Z \le -1.0) = 0.1587\).)
\(Z = (3.0 - 3.5)/0.5 = -1.0\). \(P(Z \le -1.0) = 0.1587 = 15.87\%\). About 16% of deliveries arrive in under 3 days — useful for marketing an "early delivery" promise.
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Section 5 — Week 6
The Uniform Distribution
When every outcome across an interval is equally likely — the simplest continuous distribution, and a natural model for guaranteed delivery windows.
5.1
5.1 The Continuous Uniform Distribution
Definition and PDF
\(X \sim U(a,\, b)\) assigns equal probability density to all values in \([a, b]\):
$$f(x) = \frac{1}{b - a}, \qquad a \le x \le b$$
$$f(x) = 0 \qquad \text{otherwise}$$
Vantage meets its SLA 75% of the time. One in four deliveries breaches the 5-hour promise.
Operations Director — Decision Brief
Problem: A 25% breach rate exposes Vantage to customer compensation claims and reputational risk.
Three options to evaluate:
Option A: Revise SLA to 6 hours — achieves 100% compliance but weakens the customer promise.
Option B: Renegotiate courier contracts to cap delivery at 5.5 hours (\(P \le 5) = 0.875\)).
Option C: Priority dispatch for high-value orders, reducing variance for that segment.
The uniform distribution provides an analytically tractable framework for comparing each option's compliance probability before committing to a contract change.
5.Q
Knowledge Check — Section 5
Q1: Time for a Vantage warehouse worker to pack an order is \(X \sim U(3,\, 9)\) minutes. What is the probability packing takes more than 7 minutes?
\(P(X > 7) = (9-7)/(9-3) = 2/6 \approx 0.333\). The uniform formula is the ratio of the target interval length to the total interval length.
Q2: For the same packing time distribution \(U(3, 9)\), what is the mean packing time and standard deviation?
\(\mu = (3+9)/2 = 6\) min. \(\sigma = \sqrt{(9-3)^2/12} = \sqrt{36/12} = \sqrt{3} \approx 1.73\) min. The mean is the midpoint; the variance uses \((b-a)^2/12\).
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Section 6 — Week 6
The Exponential Distribution
Modelling the time between events in a Poisson process — the natural continuous companion to the Poisson distribution from Week 5.
6.1
6.1 The Exponential Distribution
Definition and PDF
If events occur at a Poisson rate of \(\lambda\) per unit time, the time between consecutive events follows:
$$f(x) = \lambda\, e^{-\lambda x}, \qquad x \ge 0$$
Written \(X \sim \text{Exp}(\lambda)\).
Only a 13.5% chance of a 10-minute lull — staff should stay ready.
6.3
6.3 The Memoryless Property
Formal Statement
For an exponential random variable:
$$P(X > s + t \;\mid\; X > s) = P(X > t), \qquad \text{for all } s, t \ge 0$$
The probability of waiting at least another \(t\) units is the same regardless of how long you have already been waiting. The past waiting time is irrelevant.
Vantage Retail Co.
It has been 8 minutes since the last order arrived. What is the probability of waiting at least 5 more minutes?
By the memoryless property, this equals \(P(X > 5\text{ min})\) — exactly as if no time had elapsed. The 8 minutes already passed are irrelevant.
When Is This Property Plausible?
The memoryless property is appropriate when the event rate is truly constant and random — each moment is independent of all prior moments. It would not be appropriate for machine failures, where parts "wear out" over time (use the Weibull distribution instead).
6.4
6.4 The Poisson–Exponential Connection
Key Relationship
If events occur as a Poisson process with rate \(\lambda\) per unit time, then the time between consecutive events follows Exponential(\(\lambda\)). The same parameter \(\lambda\) describes both.
Vantage Retail Co. — Two Sides of the Same Process
Question
Distribution
Formula Used
How many orders arrive in 30 minutes?
Poisson (\(\lambda = 6\) per 30 min)
\(P(X=x) = e^{-6}6^x/x!\)
How long until the next order arrives?
Exponential (\(\lambda = 12\) per hour)
\(P(T \le t) = 1-e^{-12t}\)
Practical Insight for Operations
A manager who knows the Poisson arrival rate can immediately derive the inter-arrival distribution — no additional parameters. This relationship is foundational in queuing theory, which underpins staffing decisions, warehouse throughput planning, and customer service capacity modelling.
6.Q
Knowledge Check — Section 6
Q1: Vantage's customer support line receives calls at an average rate of 6 per hour. What is the mean time between consecutive calls?
\(\mu = 1/\lambda = 1/6\) hour \(= 10\) minutes. The mean inter-arrival time is the reciprocal of the Poisson rate. Be careful with units: 1/6 hour is 10 minutes, not 1/6 minutes.
Q2: Which unique property distinguishes the exponential distribution from all other continuous distributions?
The memoryless property — \(P(X > s+t \mid X > s) = P(X > t)\) — is unique to the exponential among continuous distributions. It means the process has no "memory" of how long it has already been running.
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Week 6 — Summary
Normal \(N(\mu, \sigma^2)\)
Bell-shaped, symmetric. Standardise: \(Z = (X-\mu)/\sigma\). Use Z-table for areas.