Learn Discrete & Continuous Distributions II: Interactive Problems and Worked Examples
Use the question set below to practice discrete and continuous probability distributions with the most testable facts and formulas: probability mass functions (PMF) and probability density functions (PDF), cumulative distribution functions (CDF) and survival functions, expected value \(E[X]\) and variance \(\mathrm{Var}(X)\), discrete models like the Poisson distribution \((\lambda)\), geometric distribution \((p)\), and hypergeometric distribution \((N,K,n)\), the Poisson approximation to Binomial (large \(n\), small \(p\), \(\lambda=np\)), continuous models like the exponential distribution (rate \(\lambda\), scale \(1/\lambda\), waiting times), gamma and chi-squared \((\chi^2)\) distributions (degrees of freedom and right-skewed shapes), the F distribution (ratios of variances), and special cases like the logistic distribution (sigmoid CDF, \(\mathrm{Var}(X)=\pi^2 s^2/3\)) and the Cauchy distribution (undefined mean and variance). If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this Distributions II practice works
1. Take the practice set: answer the Discrete & Continuous Distributions II questions below.
2. Open the lesson (optional): review PMF/PDF, CDF, support, parameter meaning, and mean/variance formulas with clear examples.
3. Retry: return to the question set and apply distribution rules immediately.
What you will learn in the Discrete & Continuous Distributions II lesson
Purpose: Build a clear, exam-ready understanding of Discrete & Continuous Distributions II. You’ll practice identifying a distribution from a story, reading the support, using the right PMF/PDF and CDF, and computing key facts like expected value and variance for common discrete models (Poisson, geometric, hypergeometric) and continuous models (exponential, chi-squared, F, logistic, Cauchy).
Success criteria
Identify whether a model is discrete (PMF) or continuous (PDF/CDF).
Use support correctly (e.g., Poisson values \(0,1,2,\dots\); exponential values \(x\ge 0\)).
Write the Poisson PMF \(P(X=k)=e^{-\lambda}\dfrac{\lambda^k}{k!}\) and recall \(E[X]=\mathrm{Var}(X)=\lambda\).
Recognize when to use a geometric distribution (trials until first success) and its support \(1,2,3,\dots\).
Compute a hypergeometric expected value: \(E[X]=n\cdot\frac{K}{N}\) (sampling without replacement).
Use the exponential CDF \(F(x)=1-e^{-\lambda x}\) and survival \(P(X>x)=e^{-\lambda x}\) for \(x\ge 0\).
Interpret the exponential rate \(\lambda\) (events per unit time) and the scale \(1/\lambda\).
Know core \(\chi^2\) facts: \(\chi^2_k\) is never negative, \(E[\chi^2_k]=k\), \(\mathrm{Var}(\chi^2_k)=2k\), and \(k\) (degrees of freedom) controls the shape.
Use the F distribution \(F(d_1,d_2)\) as a ratio of scaled chi-squared variables and know \(E[F]=\dfrac{d_2}{d_2-2}\) when \(d_2>2\).
Recall special distributions: Logistic variance \(\mathrm{Var}(X)=\dfrac{\pi^2 s^2}{3}\) and Cauchy has undefined mean and variance.
Key vocabulary
Support: the set of values a random variable can take.
PMF: \(P(X=k)\) for discrete \(X\).
PDF: \(f(x)\) for continuous \(X\) (probability uses area under the curve).
Degrees of freedom: parameter for \(\chi^2\), \(t\), and \(F\) families that controls shape.
Heavy tails: unusually large outliers are more likely (classic example: Cauchy).
Quick pre-check
Pre-check 1: Which values can a Poisson random variable take?
Hint: Poisson models counts, so outcomes are \(0,1,2,\dots\).
Pre-check 2: For an exponential random variable with rate \(\lambda\), what is the CDF at \(0\), \(F(0)\)?
Hint: For \(x\ge 0\), \(F(x)=1-e^{-\lambda x}\). So \(F(0)=1-e^0=0\).
Discrete Distributions
Poisson, geometric, and hypergeometric (and a key approximation)
Learning goal: Match a counting situation to the correct discrete distribution and compute quick facts like support, PMF, and \(E[X]\) without confusion.
Key idea
For discrete random variables, probabilities come from a PMF \(P(X=k)\). Three high-frequency models:
Poisson \(\mathrm{Poisson}(\lambda)\): counts of events in an interval (with a constant average rate).
Support: \(k=0,1,2,\dots\)
PMF: \(P(X=k)=e^{-\lambda}\dfrac{\lambda^k}{k!}\)
Mean/variance: \(E[X]=\mathrm{Var}(X)=\lambda\)
Parameter rule: \(\lambda\ge 0\) (typically \(\lambda>0\) in practice)
Geometric \(\mathrm{Geom}(p)\) (trials until first success):
Note: \(E[X]\) is finite for every valid \(p>0\), but it can be very large if \(p\) is small.
Hypergeometric \(\mathrm{Hypergeometric}(N,K,n)\): sampling without replacement from \(N\) total items with \(K\) “successes,” drawing \(n\) items.
Mean: \(E[X]=n\cdot\frac{K}{N}\)
Must-know approximation
If \(X\sim \mathrm{Binomial}(n,p)\) with large \(n\) and small \(p\) such that \(\lambda=np\) is moderate, then \[\mathrm{Binomial}(n,p)\ \approx\ \mathrm{Poisson}(\lambda=np).\] This is one of the most common “fast approximation” facts on quizzes and exams.
Worked example
Example: Let \(X\sim\mathrm{Hypergeometric}(N=10,K=4,n=3)\). Find \(E[X]\).
Use the hypergeometric mean: \[E[X]=n\cdot\frac{K}{N}=3\cdot\frac{4}{10}=\frac{12}{10}=1.2.\]
Try it
Try it 1: Let \(X\sim\mathrm{Hypergeometric}(N=10,K=4,n=3)\). What is \(E[X]\)?
Hint: \(E[X]=n\cdot\frac{K}{N}\).
Try it 2: For small \(p\) and large \(n\), which distribution approximates \(\mathrm{Binomial}(n,p)\) (with \(\lambda=np\))?
Hint: If \(n\) is large and \(p\) is small, \(\mathrm{Bin}(n,p)\approx \mathrm{Poisson}(\lambda=np)\).
Geometric (trials to first success): \(k=1,2,3,\dots\), \(P(X=k)=(1-p)^{k-1}p\).
Hypergeometric mean: \(E[X]=n\cdot\frac{K}{N}\).
Approximation: \(\mathrm{Bin}(n,p)\approx \mathrm{Poisson}(\lambda=np)\) for large \(n\), small \(p\).
Exponential
Exponential distribution: rate, scale, CDF, and waiting times
Learning goal: Use the exponential PDF/CDF correctly, interpret \(\lambda\), and compute probabilities with the survival function.
Key idea
The exponential distribution models a waiting time until the next event when events arrive at a constant average rate. If \(X\sim\mathrm{Exponential}(\lambda)\), then:
Support: \(x\ge 0\)
PDF: \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge 0\)
CDF: \(F(x)=P(X\le x)=1-e^{-\lambda x}\) for \(x\ge 0\)
The exponential distribution is also memoryless: \[P(X>s+t\mid X>s)=P(X>t).\]
Worked example
Example: If \(X\sim\mathrm{Exponential}(\lambda)\), what is \(P(X>t)\)?
Use the survival function: \[P(X>t)=1-F(t)=1-(1-e^{-\lambda t})=e^{-\lambda t}.\]
Try it
Try it 1: In an exponential distribution, what does the parameter \(\lambda\) represent?
Hint: A larger \(\lambda\) means events happen more frequently, so the waiting time is smaller.
Try it 2: What is the scale parameter for the exponential distribution with rate \(\lambda\)?
Hint: Exponential mean \(E[X]=1/\lambda\), and the scale parameter equals the mean.
Summary
Exponential CDF: \(F(x)=1-e^{-\lambda x}\) for \(x\ge 0\); in particular \(F(0)=0\).
Rate/scale: \(\lambda>0\), scale \(=1/\lambda\), mean \(=1/\lambda\), variance \(=1/\lambda^2\).
Waiting-time model: \(P(X>t)=e^{-\lambda t}\).
Chi-Squared
Gamma and chi-squared: degrees of freedom, shape, and key facts
Learning goal: Recognize a \(\chi^2\) distribution, interpret degrees of freedom, and use the most important properties quickly.
Key idea
A chi-squared distribution with \(k\) degrees of freedom is written \(\chi^2_k\). It appears naturally as a sum of squares: \[Z_1^2+\cdots+Z_k^2 \sim \chi^2_k \quad \text{when } Z_i\sim N(0,1)\text{ independently.}\]
Support: \(x\ge 0\) (so it can never be negative)
Shape: right-skewed for small \(k\); becomes more symmetric as \(k\) increases
Parameter that controls shape: \(k\) (degrees of freedom)
Connection to the gamma distribution
Chi-squared is a special case of the gamma distribution: \[\chi^2_k \sim \mathrm{Gamma}\!\left(\alpha=\frac{k}{2},\ \theta=2\right),\] where \(\alpha\) is the shape and \(\theta\) is the scale.
Worked example
Example: If \(X\sim \chi^2_{12}\), what are \(E[X]\) and \(\mathrm{Var}(X)\)?
Use the standard moments: \[E[X]=12,\qquad \mathrm{Var}(X)=2\cdot 12=24.\]
Try it
Try it 1: What parameter determines the shape of a chi-squared \((\chi^2)\) distribution?
Hint: \(\chi^2\) distributions are indexed by degrees of freedom.
Try it 2: What is the probability that a \(\chi^2\) random variable is negative?
Hint: The support of \(\chi^2\) is \(x\ge 0\).
Summary
\(\chi^2_k\) is always \(\ge 0\) and is typically right-skewed for small \(k\).
\(E[\chi^2_k]=k\), \(\mathrm{Var}(\chi^2_k)=2k\).
Degrees of freedom \(k\) controls the shape.
F Distribution
F distribution: ratios of variances and mean existence
Learning goal: Recognize an F distribution, understand its parameters \((d_1,d_2)\), and use the key mean formula correctly.
Key idea
The F distribution arises from a ratio of two independent chi-squared variables divided by their degrees of freedom: \[F=\frac{(\chi^2_{d_1}/d_1)}{(\chi^2_{d_2}/d_2)} \sim F(d_1,d_2).\] It is widely used in ANOVA and in testing/estimating ratios of variances.
Support: \(x>0\) (never negative)
Parameters: \(d_1>0\), \(d_2>0\) (degrees of freedom)
Mean: \(E[F]=\dfrac{d_2}{d_2-2}\) if \(d_2>2\) (otherwise the mean does not exist)
Worked example
Example: Let \(F\sim F(d_1=5,d_2=10)\). What is \(E[F]\)?
Since \(d_2=10>2\), the mean exists: \[E[F]=\frac{d_2}{d_2-2}=\frac{10}{10-2}=\frac{10}{8}=1.25.\]
Try it
Try it 1: Let \(F\sim F(d_1=5,d_2=10)\). What is \(E[F]\)?
Hint: \(E[F]=\dfrac{d_2}{d_2-2}\) when \(d_2>2\).
Try it 2: For \(F(d_1,d_2)\), when does the mean \(E[F]\) exist?
Hint: The denominator degrees of freedom controls whether \(E[F]\) exists.
Summary
\(F(d_1,d_2)\) is positive: support \(x>0\).
\(E[F]=\dfrac{d_2}{d_2-2}\) exists only when \(d_2>2\).
F distributions appear in variance ratios and ANOVA.
Logistic & Cauchy
Logistic variance and Cauchy “undefined mean” trap
Learning goal: Recognize logistic and Cauchy distributions and remember which moments exist (and which do not).
Logistic distribution
A logistic random variable \(X\sim \mathrm{Logistic}(\mu,s)\) has a smooth S-shaped (sigmoid) CDF: \[F(x)=\frac{1}{1+e^{-(x-\mu)/s}},\] where \(\mu\) is a location parameter and \(s>0\) is a scale parameter.
Why it matters: logistic CDFs show up in logistic regression and “probability as a smooth threshold” modeling.
Cauchy distribution
The Cauchy distribution is a classic heavy-tailed distribution. For \(X\sim \mathrm{Cauchy}(x_0,\gamma)\), the tails are so heavy that the mean and variance are undefined. A common special case is the standard Cauchy \(\mathrm{Cauchy}(0,1)\) with PDF: \[f(x)=\frac{1}{\pi(1+x^2)}.\]
Worked example
Example: If \(X\sim\mathrm{Logistic}(\mu,s)\), what is \(\mathrm{Var}(X)\)?
Use the standard formula: \[\mathrm{Var}(X)=\frac{\pi^2 s^2}{3}.\]
Try it
Try it 1: Let \(X\sim\mathrm{Logistic}(\mu,s)\). What is \(\mathrm{Var}(X)\)?
Hint: Logistic variance has \(\pi^2\) in it: \(\pi^2 s^2/3\).
Try it 2: Which distribution has undefined mean and variance?
Hint: Cauchy tails are heavy enough that the usual mean/variance integrals do not converge.
Cauchy: mean and variance are undefined (do not exist).
Choose the Model
How to choose the right distribution (fast identification)
Learning goal: Match keywords in a question to the right distribution and avoid common traps (wrong support, wrong parameter rules, wrong “what \(\lambda\) means”).
Story → distribution shortcuts
Counts in a time/area interval (arrivals, defects, calls, emails): Poisson\((\lambda)\).
Waiting time until next event: Exponential\((\lambda)\).
Trials until first success (first success on trial \(k\)): Geometric\((p)\).
Sampling without replacement from a finite population: Hypergeometric\((N,K,n)\).
Sum of squares of standard normals: \(\chi^2_k\).
Ratio of variances (scaled chi-squared ratio): \(F(d_1,d_2)\).
Parameter checks (quick sanity)
Poisson: \(\lambda\ge 0\) and outcomes are \(0,1,2,\dots\).
Exponential: \(\lambda>0\) and outcomes satisfy \(x\ge 0\).
Geometric: \(0<p\le 1\) and outcomes are \(1,2,3,\dots\) (trials-until-success version).
\(\chi^2\), F: degrees of freedom are positive; values are nonnegative (and F is strictly positive).
Worked example (Poisson ↔ exponential connection)
Example: Events occur at an average rate of \(\lambda\) per unit time.
The count of events in one unit of time can be modeled by \(N\sim\mathrm{Poisson}(\lambda)\).
The waiting time until the next event can be modeled by \(X\sim\mathrm{Exponential}(\lambda)\).
This is why Poisson and exponential questions often appear together in “Distributions II”.
Try it
Try it 1: Which scenario could be modeled with an exponential distribution?
Hint: Exponential is a waiting-time model.
Try it 2: Which value can a Poisson-distributed random variable never take?
Hint: Poisson outcomes are counts: \(0,1,2,\dots\) (no fractions).
Always check support: Poisson is \(0,1,2,\dots\); exponential is \(x\ge 0\); \(\chi^2\) is \(x\ge 0\); F is \(x>0\).
Parameter meaning matters: exponential \(\lambda\) is a rate; Poisson \(\lambda\) is the mean count per interval.
Big Picture
Why these distributions matter (and a final check)
Learning goal: Connect distribution formulas to real statistical tasks — then finish with a final check to lock in the most tested facts.
Where Distributions II shows up
Queueing & reliability: Poisson counts and exponential waiting times (calls, arrivals, failures).
Quality control: defects per unit (Poisson), pass/fail trials (geometric).
Sampling & genetics: hypergeometric sampling without replacement.
Hypothesis testing: chi-squared tests (goodness-of-fit, independence) and F tests (variance ratios, ANOVA).
Data modeling: logistic for S-shaped probability curves; Cauchy as a reminder that not every distribution has a mean.
Worked example: geometric “first trial success”
Example: If \(X\sim\mathrm{Geom}(p)\) counts the number of trials until the first success, what is \(P(X=1)\)?
“First success on trial 1” means the first trial is a success: \[P(X=1)=p.\]
Try it
Try it 1: If the probability of success is \(p\), what is the probability that the first trial is a success in a geometric distribution (trials until first success)?
Hint: “First trial success” is exactly one success immediately, so it is just \(p\).
Try it 2: What is the smallest value a Poisson random variable can take?
Hint: Poisson is a count distribution, so it starts at \(0\).
Final recap
Poisson: \(P(X=k)=e^{-\lambda}\dfrac{\lambda^k}{k!}\), support \(k=0,1,2,\dots\), \(E[X]=\mathrm{Var}(X)=\lambda\).
Geometric (trials to first success): support \(k=1,2,3,\dots\), \(P(X=k)=(1-p)^{k-1}p\), \(E[X]=1/p\).
Hypergeometric: sampling without replacement, \(E[X]=n\cdot\frac{K}{N}\).
Exponential: \(f(x)=\lambda e^{-\lambda x}\), \(F(x)=1-e^{-\lambda x}\) for \(x\ge 0\), mean \(1/\lambda\), variance \(1/\lambda^2\).
\(\chi^2_k\): nonnegative, \(E[\chi^2_k]=k\), \(\mathrm{Var}(\chi^2_k)=2k\), shape controlled by \(k\).
F(d\(_1\),d\(_2\)): positive, \(E[F]=\dfrac{d_2}{d_2-2}\) for \(d_2>2\).
Logistic: \(\mathrm{Var}(X)=\dfrac{\pi^2 s^2}{3}\). Cauchy: mean and variance are undefined.
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the distribution (Poisson, geometric, hypergeometric, exponential, \(\chi^2\), F, logistic, Cauchy) and the key fact you need (support, parameter meaning, CDF, mean/variance).
