# Probability Distributions Reference Sheet: [Your course or data set] > A reference sheet for the probability distributions that appear in introductory and intermediate statistics: > Bernoulli, binomial, Poisson, geometric, negative binomial, hypergeometric, uniform, normal, exponential, gamma, > beta, lognormal, Weibull, t, chi-square and F. For each one it gives the story behind it, the pmf or pdf in > KaTeX, the support, the parameters, the mean and variance, and typical uses, followed by a map of how the > distributions are related and a side-by-side table of R and Python (scipy.stats) functions with the > parameterisation traps that cause most wrong answers. It is for students who need a single page to revise from, > thesis writers choosing a model for their outcome, and analysts who switch between R and Python. It speeds up > learning by attaching each formula to a concrete story, by showing the distributions as one connected family, > and by making you practise lookups with fading worked examples and a "which distribution?" self-test. Every > lookup value on this sheet was produced by running the code shown. The worked examples use fictional data; > replace them with your own. Source basis: standard distribution theory, the NIST/SEMATECH e-Handbook of > Statistical Methods (chapter 1.3.6), and the R `stats` and SciPy documentation. ## Document control | Field | Value | |---|---| | Course or project | [STAT 201 Probability and Inference] | | Learner or analyst | [Your name] | | Data set | Fictional: Harbour City Library returns, Brightwell Bakery loaves, Lakemoor University exam scores | | Software and versions | R 4.5.1 (stats); Python 3.14 with scipy 1.18.1 and numpy 2.5.4 | | Date | YYYY-MM-DD | | Version | [0.1] | ## The idea in one paragraph A probability distribution is a model of how a random quantity behaves: which values it can take (its **support**) and how likely each value or range is. **Discrete** distributions give a probability to each count through a probability mass function (pmf); **continuous** distributions spread probability over an interval through a probability density function (pdf), and probabilities are areas under the curve. Every named distribution comes from a simple story: "count successes in a fixed number of independent tries" gives the binomial, "count rare events in a fixed window" gives the Poisson, "add up many small independent effects" gives the normal, "wait for the next event in a memoryless process" gives the exponential. Pick the distribution whose story matches how your data were generated, check its support against your data, and then use software to look up probabilities and quantiles. The four lookup questions are always the same: the probability of exactly a value (d), the probability up to a value (p), the value with a given probability below it (q), and random draws (r). ## Choosing a distribution Structure aligned with the NIST/SEMATECH e-Handbook catalogue of distributions (chapter 1.3.6), reorganised by the question a learner asks about their data; wording and examples are original. ```mermaid flowchart TD A["What kind of values?"] --> B["Counts or yes/no (discrete)"] A --> C["Measurements (continuous)"] B --> D["One trial, success or failure?"] D -->|Yes| E["Bernoulli"] D -->|No| F["Fixed number of trials n?"] F -->|"Yes, with replacement or large population"| G["Binomial"] F -->|"Yes, without replacement from small population"| H["Hypergeometric"] F -->|"No, count trials until success"| I["Geometric or negative binomial"] F -->|"No, events in a window of time or space"| J["Poisson (variance about equal to mean)"] J -->|"Variance much bigger than mean"| K["Negative binomial"] C --> L["Bounded between 0 and 1?"] L -->|Yes| M["Beta"] L -->|No| N["Positive only and right-skewed?"] N -->|"Waiting time, constant rate"| O["Exponential"] N -->|"Sum of waits, flexible skew"| P["Gamma"] N -->|"Log of values looks normal"| Q["Lognormal"] N -->|"Lifetimes, rising or falling failure rate"| R["Weibull"] N -->|"No, symmetric, unbounded"| S["Normal"] C --> T["A test statistic?"] T --> U["t, chi-square or F"] ``` My path for [my variable]: [Measurements, positive, right-skewed, log looks normal: lognormal]. ## Discrete distributions Notation: $p$ is the success probability, $q = 1 - p$, $n$ the number of trials, $\lambda$ a rate or mean count, and $\binom{n}{x}$ the number of combinations (see academic/stats-probability-basics). | Distribution | Story | Support | Parameters | Mean | Variance | Typical uses | |---|---|---|---|---|---|---| | Bernoulli | One trial: success (1) or failure (0) | 0, 1 | $p$ | $p$ | $pq$ | A single yes/no outcome; building block of the others | | Binomial | Successes in $n$ independent trials with the same $p$ | 0, 1, ..., $n$ | $n, p$ | $np$ | $npq$ | Pass counts, defect counts in a batch, survey yes counts | | Poisson | Count of independent rare events in a fixed window | 0, 1, 2, ... | $\lambda > 0$ | $\lambda$ | $\lambda$ | Arrivals per hour, typos per page, library returns per day | | Geometric | Trials until (and including) the first success | 1, 2, 3, ... | $p$ | $1/p$ | $q/p^2$ | Attempts until a pass, calls until a sale | | Negative binomial | Failures before the $r$-th success; or an overdispersed count | 0, 1, 2, ... | $r, p$ (or mean $\mu$, size $k$) | $rq/p$ (or $\mu$) | $rq/p^2$ (or $\mu + \mu^2/k$) | Counts with variance above the mean (see academic/stats-count-regression) | | Hypergeometric | Successes in $n$ draws without replacement from $N$ items with $K$ successes | max(0, n+K−N) to min(n, K) | $N, K, n$ | $nK/N$ | $n \frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1}$ | Audit samples, Fisher's exact test, quality inspection of a small lot | **Probability mass functions** $$ \text{Bernoulli: } P(X = x) = p^x (1-p)^{1-x}, \quad x \in \{0, 1\} $$ $$ \text{Binomial: } P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} $$ $$ \text{Poisson: } P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} $$ $$ \text{Geometric (trials): } P(X = x) = (1-p)^{x-1} p, \quad x = 1, 2, \dots $$ $$ \text{Negative binomial (failures): } P(X = x) = \binom{x + r - 1}{x} p^r (1-p)^x, \quad x = 0, 1, \dots $$ $$ \text{Hypergeometric: } P(X = x) = \frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}} $$ **Picture: the Poisson as a limit of the binomial.** The chart compares Binomial(30, 0.1) with Poisson(3), which has the same mean of 3. With many trials and a small success probability, the two are nearly the same, which is why rare-event counts are modelled as Poisson. The probabilities were computed with `dpois(0:10, 3)` and `dbinom(0:10, 30, 0.1)` in R (identical to scipy to four decimals). ```vega-lite { "$schema": "https://vega.github.io/schema/vega-lite/v5.json", "description": "Binomial(30, 0.1) and Poisson(3) probabilities for x = 0 to 10", "width": 400, "height": 220, "data": {"values": [ {"x": 0, "distribution": "Poisson(3)", "probability": 0.0498}, {"x": 0, "distribution": "Binomial(30, 0.1)", "probability": 0.0424}, {"x": 1, "distribution": "Poisson(3)", "probability": 0.1494}, {"x": 1, "distribution": "Binomial(30, 0.1)", "probability": 0.1413}, {"x": 2, "distribution": "Poisson(3)", "probability": 0.224}, {"x": 2, "distribution": "Binomial(30, 0.1)", "probability": 0.2277}, {"x": 3, "distribution": "Poisson(3)", "probability": 0.224}, {"x": 3, "distribution": "Binomial(30, 0.1)", "probability": 0.2361}, {"x": 4, "distribution": "Poisson(3)", "probability": 0.168}, {"x": 4, "distribution": "Binomial(30, 0.1)", "probability": 0.1771}, {"x": 5, "distribution": "Poisson(3)", "probability": 0.1008}, {"x": 5, "distribution": "Binomial(30, 0.1)", "probability": 0.1023}, {"x": 6, "distribution": "Poisson(3)", "probability": 0.0504}, {"x": 6, "distribution": "Binomial(30, 0.1)", "probability": 0.0474}, {"x": 7, "distribution": "Poisson(3)", "probability": 0.0216}, {"x": 7, "distribution": "Binomial(30, 0.1)", "probability": 0.018}, {"x": 8, "distribution": "Poisson(3)", "probability": 0.0081}, {"x": 8, "distribution": "Binomial(30, 0.1)", "probability": 0.0058}, {"x": 9, "distribution": "Poisson(3)", "probability": 0.0027}, {"x": 9, "distribution": "Binomial(30, 0.1)", "probability": 0.0016}, {"x": 10, "distribution": "Poisson(3)", "probability": 0.0008}, {"x": 10, "distribution": "Binomial(30, 0.1)", "probability": 0.0004} ]}, "mark": "bar", "encoding": { "x": {"field": "x", "type": "ordinal", "title": "Count x", "axis": {"labelAngle": 0}}, "xOffset": {"field": "distribution"}, "y": {"field": "probability", "type": "quantitative", "title": "P(X = x)"}, "color": {"field": "distribution", "type": "nominal", "title": "Distribution"}, "tooltip": [{"field": "distribution"}, {"field": "x"}, {"field": "probability"}] } } ``` ## Continuous distributions Notation: $\mu$ and $\sigma$ are a mean and standard deviation, $\lambda$ a rate, $\alpha$ a shape, $\beta$ a rate (for the gamma) or a second shape (for the beta), $\nu$ or $k$ degrees of freedom df, and $\Gamma(\cdot)$ the gamma function, with $\Gamma(m) = (m-1)!$ for whole numbers. | Distribution | Story | Support | Parameters | Mean | Variance | Typical uses | |---|---|---|---|---|---|---| | Uniform | Every value in an interval equally likely | $[a, b]$ | $a < b$ | $(a+b)/2$ | $(b-a)^2/12$ | Rounding error, random start times, simulation input | | Normal | Sum of many small independent effects | all real numbers | $\mu, \sigma > 0$ | $\mu$ | $\sigma^2$ | Measurement error, heights, sample means (via the CLT) | | Exponential | Waiting time to the next event at a constant rate (memoryless) | $x \ge 0$ | rate $\lambda$ | $1/\lambda$ | $1/\lambda^2$ | Time between arrivals, service times | | Gamma | Waiting time to the $\alpha$-th event; flexible positive skew | $x > 0$ | shape $\alpha$, rate $\beta$ | $\alpha/\beta$ | $\alpha/\beta^2$ | Rainfall, insurance claims, total service time | | Beta | A probability or proportion that is itself uncertain | $[0, 1]$ | shapes $\alpha, \beta$ | $\frac{\alpha}{\alpha+\beta}$ | $\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$ | Proportions, Bayesian prior for $p$ | | Lognormal | Product of many small positive effects; log is normal | $x > 0$ | $\mu, \sigma$ of the log | $e^{\mu + \sigma^2/2}$ | $(e^{\sigma^2}-1)e^{2\mu+\sigma^2}$ | Incomes, reaction times, concentrations | | Weibull | Lifetime with a failure rate that rises, falls or stays flat | $x \ge 0$ | shape $k$, scale $\lambda$ | $\lambda\,\Gamma(1+1/k)$ | $\lambda^2[\Gamma(1+2/k) - \Gamma(1+1/k)^2]$ | Reliability, equipment lifetimes, wind speed | | t | Standardised mean when $\sigma$ is estimated | all real numbers | df $\nu$ | 0 (for $\nu > 1$) | $\nu/(\nu-2)$ (for $\nu > 2$) | t-tests, CIs for means, regression coefficients | | Chi-square | Sum of $k$ squared standard normals | $x \ge 0$ | df $k$ | $k$ | $2k$ | Variance tests, goodness of fit, contingency tables | | F | Ratio of two independent chi-squares, each divided by its df | $x \ge 0$ | df $d_1, d_2$ | $\frac{d_2}{d_2-2}$ (for $d_2 > 2$) | $\frac{2d_2^2(d_1+d_2-2)}{d_1(d_2-2)^2(d_2-4)}$ (for $d_2 > 4$) | ANOVA, comparing variances, regression F-test | **Probability density functions** $$ \text{Uniform: } f(x) = \frac{1}{b-a}, \quad a \le x \le b $$ $$ \text{Normal: } f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) $$ $$ \text{Exponential: } f(x) = \lambda e^{-\lambda x}, \qquad P(X > x) = e^{-\lambda x} $$ $$ \text{Gamma: } f(x) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} $$ $$ \text{Beta: } f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)}, \qquad B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} $$ $$ \text{Lognormal: } f(x) = \frac{1}{x\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right) $$ $$ \text{Weibull: } f(x) = \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} $$ $$ \text{t: } f(t) = \frac{\Gamma\!\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\,\Gamma\!\left(\frac{\nu}{2}\right)} \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}} $$ $$ \text{Chi-square: } f(x) = \frac{x^{k/2 - 1} e^{-x/2}}{2^{k/2}\,\Gamma(k/2)} $$ $$ \text{F: } f(x) = \frac{1}{B\!\left(\frac{d_1}{2}, \frac{d_2}{2}\right)} \left(\frac{d_1}{d_2}\right)^{d_1/2} x^{d_1/2 - 1} \left(1 + \frac{d_1}{d_2}x\right)^{-\frac{d_1+d_2}{2}} $$ You rarely compute these densities by hand. What matters is recognising the shape (symmetric or skewed, bounded or not), knowing the mean and variance, and asking software for areas. ## Relationships between distributions The distributions are one family, not a list. Knowing the links lets you rebuild a forgotten formula and explains why tests use the t, chi-square and F distributions at all. ```mermaid flowchart TD BE["Bernoulli(p)"] -->|"sum of n"| BI["Binomial(n, p)"] HY["Hypergeometric"] -->|"population large"| BI BI -->|"n large, p small, np = lambda"| PO["Poisson(lambda)"] BI -->|"n large (limit)"| NO["Normal"] PO -->|"lambda large (limit)"| NO BE -->|"trials until first success"| GE["Geometric(p)"] GE -->|"sum of r"| NB["Negative binomial"] PO -->|"gaps between events"| EX["Exponential(rate)"] EX -->|"sum of alpha"| GA["Gamma(alpha, rate)"] GA -->|"shape k/2, rate 1/2"| CH["Chi-square(k)"] NO -->|"square of a standard normal"| CH NO -->|"Z over root of chi-square/df"| TT["t(df)"] CH -->|"ratio of two, each over df"| FF["F(d1, d2)"] TT -->|"t squared"| FF TT -->|"df large (limit)"| NO NO -->|"exponentiate"| LN["Lognormal"] WE["Weibull"] -->|"shape 1"| EX BT["Beta(1, 1)"] -->|"equals"| UN["Uniform(0, 1)"] ``` The most useful links in practice: - **Binomial to Poisson**: when $n$ is large and $p$ small, $\text{Binomial}(n, p) \approx \text{Poisson}(np)$ (see the chart above). - **Binomial and Poisson to normal**: a common rule of thumb (not a theorem) is that with $np \ge 10$ and $n(1-p) \ge 10$, or $\lambda \ge 10$ or so, a normal curve with the same mean and variance is a reasonable approximation (some textbooks use 5 instead of 10; the closer $p$ is to 0 or 1, the larger $n$ must be); this is the basis of the z-test for proportions (see academic/stats-proportion-tests). - **Normal to t, chi-square and F**: the t distribution appears when $\sigma$ is replaced by the sample SD, the chi-square when squared standardised values are summed, and the F as a ratio of two variance estimates. This is why t-tests, chi-square tests and ANOVA use them (see academic/stats-sampling-distributions). - **Exponential and Poisson**: if events arrive as a Poisson process with rate $\lambda$, the waiting time between events is exponential with the same rate. Counts and waits are two views of one process. ## R and Python functions R uses a prefix system: `d` density or mass, `p` cumulative probability $P(X \le x)$, `q` quantile, `r` random draws. scipy.stats uses methods on a distribution object: `.pmf` or `.pdf`, `.cdf`, `.ppf` (quantile), `.sf` (survival function $P(X > x) = 1 -$ cdf), `.isf` (inverse survival) and `.rvs` (draws; prefer `numpy.random.default_rng(seed)` for reproducible simulation). | Distribution | R (d, p, q, r stem) | scipy.stats | Parameterisation trap | |---|---|---|---| | Bernoulli | `dbinom(x, 1, p)` | `bernoulli(p)` | None | | Binomial | `dbinom(x, size, prob)` | `binom(n, p)` | None | | Poisson | `dpois(x, lambda)` | `poisson(mu)` | scipy calls the mean `mu` | | Geometric | `dgeom(x, prob)` | `geom(p)` | R counts **failures** before the first success (starts at 0); scipy counts **trials** (starts at 1). `dgeom(2, p)` equals `geom.pmf(3, p)`. | | Negative binomial | `dnbinom(x, size, prob)` or `dnbinom(x, size, mu = m)` | `nbinom(n, p)` | Both count failures. R allows the mean form `size, mu`; scipy needs `n = size` and `p = size / (size + mu)`. | | Hypergeometric | `dhyper(x, m, n, k)`: m successes, n failures, k draws | `hypergeom(M, n, N)`: M total, n successes, N draws | The letters mean different things. R's `n` is the number of failures; scipy's `n` is the number of successes. | | Uniform | `dunif(x, min, max)` | `uniform(loc, scale)` | scipy: `loc = a`, `scale = b - a` (not `b`) | | Normal | `dnorm(x, mean, sd)` | `norm(loc, scale)` | Both use the SD, not the variance | | Exponential | `dexp(x, rate)` | `expon(scale)` | scipy `scale = 1 / rate` | | Gamma | `dgamma(x, shape, rate)` (or `scale =`) | `gamma(a, scale)` | scipy `scale = 1 / rate`; R's third positional argument is the rate | | Beta | `dbeta(x, shape1, shape2)` | `beta(a, b)` | None | | Lognormal | `dlnorm(x, meanlog, sdlog)` | `lognorm(s, scale)` | scipy `s = sdlog`, `scale = exp(meanlog)` | | Weibull | `dweibull(x, shape, scale)` | `weibull_min(c, scale)` | scipy calls the shape `c` | | t | `dt(x, df)` | `t(df)` | None | | Chi-square | `dchisq(x, df)` | `chi2(df)` | None | | F | `df(x, df1, df2)` | `f(dfn, dfd)` | In R, `df` is also a common variable name; `df()` is the F density | Upper tails: in R add `lower.tail = FALSE` to a `p` or `q` function; in scipy use `.sf` or `.isf`. For a discrete variable, "at least 6" is "more than 5", so the upper-tail call uses 5, not 6. ## Worked example A (fully solved): late returns at Harbour City Library **Question.** Harbour City Library (fictional) receives late book returns at an average of 3 per hour, and returns arrive independently. What is the probability of 6 or more late returns in a given hour? **Step 1: choose the distribution.** Counts of independent events in a fixed window with a known average rate: Poisson with $\lambda = 3$. Check: the support (0, 1, 2, ...) matches a count, and there is no fixed upper limit. **Step 2: write the event.** $P(X \ge 6) = 1 - P(X \le 5)$. **Step 3: by hand (optional check).** $$ P(X \le 5) = e^{-3}\left(1 + 3 + \frac{9}{2} + \frac{27}{6} + \frac{81}{24} + \frac{243}{120}\right) = 0.04979 \times 18.4 \approx 0.9161 $$ so $P(X \ge 6) \approx 0.0839$. **Step 4: software.** R `ppois(5, 3, lower.tail = FALSE)` and scipy `poisson.sf(5, 3)` both give 0.0839 (see the verification output below). Both use 5, because $X \ge 6$ is the same event as $X > 5$. **Step 5: interpret.** About 8% of hours, or roughly one hour in twelve, will see 6 or more late returns. If the library adds a second person to the returns desk only to cope with hours of 6 or more late returns, it should expect to need that help in about one hour in twelve, assuming the rate really is constant across the day (it often is not: check the assumption against hourly data). ## Worked example B (half solved): underbaked loaves at Brightwell Bakery **Question.** At Brightwell Bakery (fictional), each loaf is underbaked with probability 0.05, independently. A quality check takes 10 loaves. Find (a) the probability that exactly 1 is underbaked and (b) the probability that 2 or more are underbaked. **Step 1: choose the distribution.** A fixed number of independent trials with the same success probability: [ ] with $n = [\ ]$ and $p = [\ ]$. **Step 2: write the events.** (a) $P(X = 1)$. (b) $P(X \ge 2) = 1 - P(X \le [\ ])$. **Step 3: by hand for (a).** $$ P(X = 1) = \binom{10}{1}(0.05)^1(0.95)^{9} = 10 \times 0.05 \times [\ \ ] \approx [\ \ ] $$ **Step 4: software.** R: `dbinom(1, 10, 0.05)` gives 0.3151 and `pbinom(1, 10, 0.05)` gives 0.9139. So (b) is [ ]. The scipy equivalents are `binom.pmf([ ], 10, 0.05)` and `binom.sf([ ], 10, 0.05)`. **Step 5: interpret.** If the bakery's process is in control, a batch of 10 with 2 or more underbaked loaves will occur in about [ ] of checks, so seeing it [ ]. ## Exercise C (unsolved): exam scores and waiting times 1. Scores on a Lakemoor University (fictional) statistics exam are approximately normal with mean 65 and SD 10. What proportion of students score above 80? What score marks the 90th percentile? 2. Waiting time in the Brightwell Bakery queue is exponential with a mean of 4 minutes. What is the probability a customer waits more than 6 minutes? Write the scipy call carefully. 3. A cart at Harbour City Library holds 20 returned books, 5 of them damaged. A volunteer picks 4 at random without replacement. What is the probability that exactly 1 is damaged? Write both the R and scipy calls. ## Lookup verification code **R (base R only)** ```r # Worked lookups (fictional settings) # 1 Binomial: 10 loaves, P(defect) = 0.05; P(X = 1), P(X <= 1) cat(sprintf("dbinom(1, 10, 0.05) = %.4f\n", dbinom(1, 10, 0.05))) cat(sprintf("pbinom(1, 10, 0.05) = %.4f\n", pbinom(1, 10, 0.05))) # 2 Poisson: library late returns, mean 3 per hour; P(X >= 6) cat(sprintf("1 - ppois(5, 3) = %.4f\n", 1 - ppois(5, 3))) cat(sprintf("ppois(5, 3, lower.tail = FALSE) = %.4f\n", ppois(5, 3, lower.tail = FALSE))) # 3 Normal: exam scores N(65, 10^2); P(X > 80) and 90th percentile cat(sprintf("pnorm(80, 65, 10, lower.tail = FALSE) = %.4f\n", pnorm(80, 65, 10, lower.tail = FALSE))) cat(sprintf("qnorm(0.90, 65, 10) = %.4f\n", qnorm(0.90, 65, 10))) cat(sprintf("qnorm(0.975) = %.4f\n", qnorm(0.975))) # 4 Exponential: waiting time, rate 0.25 per minute (mean 4); P(T > 6) cat(sprintf("pexp(6, rate = 0.25, lower.tail = FALSE) = %.4f\n", pexp(6, rate = 0.25, lower.tail = FALSE))) # 5 t, chi-square, F critical values cat(sprintf("qt(0.975, df = 9) = %.4f\n", qt(0.975, df = 9))) cat(sprintf("qchisq(0.95, df = 3) = %.4f\n", qchisq(0.95, df = 3))) cat(sprintf("qf(0.95, df1 = 2, df2 = 27) = %.4f\n", qf(0.95, 2, 27))) # 6 Geometric: P(first success on trial 3), p = 0.2 (R counts failures) cat(sprintf("dgeom(2, 0.2) = %.4f\n", dgeom(2, 0.2))) # 7 Negative binomial: mean 4, size 2 (overdispersed counts); P(X = 0) cat(sprintf("dnbinom(0, size = 2, mu = 4) = %.4f\n", dnbinom(0, size = 2, mu = 4))) # 8 Hypergeometric: 20 books, 5 damaged, draw 4; P(exactly 1 damaged) cat(sprintf("dhyper(1, m = 5, n = 15, k = 4) = %.4f\n", dhyper(1, 5, 15, 4))) # 9 Gamma shape 2 rate 0.5; Beta(2, 5); Lognormal meanlog 1 sdlog 0.5 cat(sprintf("pgamma(6, shape = 2, rate = 0.5) = %.4f\n", pgamma(6, shape = 2, rate = 0.5))) cat(sprintf("pbeta(0.3, 2, 5) = %.4f\n", pbeta(0.3, 2, 5))) cat(sprintf("plnorm(4, meanlog = 1, sdlog = 0.5) = %.4f\n", plnorm(4, 1, 0.5))) cat(sprintf("punif(0.3, 0, 2) = %.4f\n", punif(0.3, 0, 2))) cat(sprintf("pweibull(3, shape = 1.5, scale = 4) = %.4f\n", pweibull(3, 1.5, 4))) set.seed(2026); x <- rpois(5, 3); cat("rpois(5, 3) with seed 2026:", x, "\n") ``` ```text dbinom(1, 10, 0.05) = 0.3151 pbinom(1, 10, 0.05) = 0.9139 1 - ppois(5, 3) = 0.0839 ppois(5, 3, lower.tail = FALSE) = 0.0839 pnorm(80, 65, 10, lower.tail = FALSE) = 0.0668 qnorm(0.90, 65, 10) = 77.8155 qnorm(0.975) = 1.9600 pexp(6, rate = 0.25, lower.tail = FALSE) = 0.2231 qt(0.975, df = 9) = 2.2622 qchisq(0.95, df = 3) = 7.8147 qf(0.95, df1 = 2, df2 = 27) = 3.3541 dgeom(2, 0.2) = 0.1280 dnbinom(0, size = 2, mu = 4) = 0.1111 dhyper(1, m = 5, n = 15, k = 4) = 0.4696 pgamma(6, shape = 2, rate = 0.5) = 0.8009 pbeta(0.3, 2, 5) = 0.5798 plnorm(4, meanlog = 1, sdlog = 0.5) = 0.7801 punif(0.3, 0, 2) = 0.1500 pweibull(3, shape = 1.5, scale = 4) = 0.4777 rpois(5, 3) with seed 2026: 4 3 1 2 3 ``` **Python (scipy.stats)** ```python from scipy import stats import numpy as np print(f"binom.pmf(1, 10, 0.05) = {stats.binom.pmf(1, 10, 0.05):.4f}") print(f"binom.cdf(1, 10, 0.05) = {stats.binom.cdf(1, 10, 0.05):.4f}") print(f"poisson.sf(5, 3) = {stats.poisson.sf(5, 3):.4f}") print(f"norm.sf(80, 65, 10) = {stats.norm.sf(80, loc=65, scale=10):.4f}") print(f"norm.ppf(0.90, 65, 10) = {stats.norm.ppf(0.90, loc=65, scale=10):.4f}") print(f"norm.ppf(0.975) = {stats.norm.ppf(0.975):.4f}") print(f"expon.sf(6, scale=4) = {stats.expon.sf(6, scale=1/0.25):.4f}") print(f"t.ppf(0.975, 9) = {stats.t.ppf(0.975, 9):.4f}") print(f"chi2.ppf(0.95, 3) = {stats.chi2.ppf(0.95, 3):.4f}") print(f"f.ppf(0.95, 2, 27) = {stats.f.ppf(0.95, 2, 27):.4f}") print(f"geom.pmf(3, 0.2) = {stats.geom.pmf(3, 0.2):.4f}") size, mu = 2, 4 p = size / (size + mu) print(f"nbinom.pmf(0, n=2, p={p:.4f}) = {stats.nbinom.pmf(0, size, p):.4f}") print(f"hypergeom.pmf(1, M=20, n=5, N=4) = {stats.hypergeom.pmf(1, 20, 5, 4):.4f}") print(f"gamma.cdf(6, a=2, scale=2) = {stats.gamma.cdf(6, a=2, scale=1/0.5):.4f}") print(f"beta.cdf(0.3, 2, 5) = {stats.beta.cdf(0.3, 2, 5):.4f}") print(f"lognorm.cdf(4, s=0.5, scale=e^1) = {stats.lognorm.cdf(4, s=0.5, scale=np.exp(1)):.4f}") print(f"uniform.cdf(0.3, loc=0, scale=2) = {stats.uniform.cdf(0.3, loc=0, scale=2):.4f}") print(f"weibull_min.cdf(3, c=1.5, scale=4) = {stats.weibull_min.cdf(3, 1.5, scale=4):.4f}") rng = np.random.default_rng(2026); print("rng.poisson(3, 5) with seed 2026:", rng.poisson(3, 5)) ``` ```text binom.pmf(1, 10, 0.05) = 0.3151 binom.cdf(1, 10, 0.05) = 0.9139 poisson.sf(5, 3) = 0.0839 norm.sf(80, 65, 10) = 0.0668 norm.ppf(0.90, 65, 10) = 77.8155 norm.ppf(0.975) = 1.9600 expon.sf(6, scale=4) = 0.2231 t.ppf(0.975, 9) = 2.2622 chi2.ppf(0.95, 3) = 7.8147 f.ppf(0.95, 2, 27) = 3.3541 geom.pmf(3, 0.2) = 0.1280 nbinom.pmf(0, n=2, p=0.3333) = 0.1111 hypergeom.pmf(1, M=20, n=5, N=4) = 0.4696 gamma.cdf(6, a=2, scale=2) = 0.8009 beta.cdf(0.3, 2, 5) = 0.5798 lognorm.cdf(4, s=0.5, scale=e^1) = 0.7801 uniform.cdf(0.3, loc=0, scale=2) = 0.1500 weibull_min.cdf(3, c=1.5, scale=4) = 0.4777 rng.poisson(3, 5) with seed 2026: [3 3 6 3 3] ``` **What the output teaches.** - Every exact lookup agrees to four decimals between R and Python once the parameterisation is translated: `pexp(6, rate = 0.25)` matches `expon.sf(6, scale=4)`, `dgeom(2, 0.2)` matches `geom.pmf(3, 0.2)`, and `dnbinom(0, size = 2, mu = 4)` matches `nbinom.pmf(0, 2, 1/3)`. - The random draws differ (4 3 1 2 3 in R, 3 3 6 3 3 in Python) even with the same seed, because R and numpy use different generators. Seeds make each language reproducible, not the two languages identical. - The critical values you will meet in tests are here too: 1.96 for a two-sided 95% z interval, 2.2622 for a t interval with df = 9, 7.8147 for a chi-square test with df = 3, and 3.3541 for an F-test with df = 2 and 27. ## Common mistakes and misconceptions | Wrong statement | Correct statement | |---|---| | "The normal pdf at 80 is 0.0130, so the probability of scoring exactly 80 is 0.0130." | A density is not a probability. For a continuous variable, $P(X = 80) = 0$; ask for an area such as $P(79.5 < X < 80.5)$. | | "P(X ≥ 6) for a Poisson is `1 - ppois(6, 3)`." | That gives $P(X > 6) = P(X \ge 7)$. For discrete variables use `1 - ppois(5, 3)` or `ppois(5, 3, lower.tail = FALSE)`. | | "`expon(6)` in scipy is the exponential with rate 6." | After any shape parameters, the next positional argument of a scipy distribution is `loc`; `expon` has no shape parameter, so `expon(6)` shifts the start to 6 (its mean becomes 7). Use `expon(scale=1/6)` for rate 6. | | "R's `dgeom` and scipy's `geom.pmf` give the same answer for the same x." | R counts failures (from 0), scipy counts trials (from 1). Shift x by one. | | "The normal is the default for any data." | Choose by the data-generating story and the support. Counts, proportions, waiting times and incomes usually need another distribution or a transformation. | | "Poisson data must have variance equal to the mean." | That is the model's assumption, not a guarantee. Real counts are often overdispersed; then a negative binomial fits better. | | "The t distribution is for small samples, the normal for large ones." | The t arises whenever $\sigma$ is estimated from the data, whatever the sample size. It approaches the normal as df grows, so the choice matters less for large n. | | "`rnorm(10, 5, 4)` draws from a normal with variance 4." | R's third argument is the SD, so the variance is 16. The same holds for scipy's `scale`. | ## Self-test: which distribution? 1. Each of 25 students passes a quiz independently with probability 0.8. X = number who pass. Name the distribution, and give its mean and variance. 2. X = number of help-desk emails the library receives in one hour, averaging 2 per hour. Name the distribution and give the R and scipy calls for $P(X \ge 4)$, with the value. 3. X = time in minutes until the next customer walks into the bakery, arrivals at a constant average rate. 4. A cart holds 30 books, 8 of them overdue. Six books are picked without replacement. X = number overdue. 5. A student retakes a driving test until passing, with a 0.6 chance of passing each attempt. X = number of attempts. Which R function would you use, and what value of x do you pass for "passes on the 2nd attempt"? 6. A test statistic is the ratio of two independent variance estimates from normal samples. Which distribution does it follow under the null hypothesis? ## My notes and next steps | Topic | Confidence 1-5 | What still confuses me | Next action and date | |---|---|---|---| | *Example:* Discrete upper tails | 2 | Which number to pass for "at least" | Redo Example A in both languages, YYYY-MM-DD | | Choosing a distribution from a story | [ ] | [ ] | [ ] | | Continuous distributions and areas | [ ] | [ ] | [ ] | | t, chi-square, F and their df | [ ] | [ ] | [ ] | | R vs scipy parameterisation | [ ] | [ ] | [ ] | Distributions my course uses that are not on this sheet: [List them and add rows above]. Mistakes to copy into my error log (academic/stats-error-log): [Mistake, why it happened, the fix]. ## Sources and further reading Based on standard distribution theory and the NIST/SEMATECH e-Handbook, with function names and parameterisations checked against the R and SciPy documentation and by running the code above. - NIST/SEMATECH e-Handbook of Statistical Methods, section 1.3.6 Probability Distributions, https://www.itl.nist.gov/div898/handbook/ (accessed 2026-10-25). Public domain. - R Core Team (2025). *R: A Language and Environment for Statistical Computing*, version 4.5.1, package `stats` (help pages for `Distributions`, `NegBinomial`, `Geometric`, `Hypergeometric`). - Virtanen, P., et al. (2020). SciPy 1.0: fundamental algorithms for scientific computing in Python. *Nature Methods*, 17(3), 261–272. (Documentation for `scipy.stats` consulted for version 1.18.1.) - Student (1908). The probable error of a mean. *Biometrika*, 6(1), 1–25. (Origin of the t distribution; Student was W. S. Gosset.) - Weibull, W. (1951). A statistical distribution function of wide applicability. *Journal of Applied Mechanics*, 18(3), 293–297. - Dunlosky, J., Rawson, K. A., Marsh, E. J., Nathan, M. J., and Willingham, D. T. (2013). Improving students' learning with effective learning techniques. *Psychological Science in the Public Interest*, 14(1), 4–58. (Interleaved practice in the self-test.) - Further reading (by name only): de Smith, M. J. (2024). *Statistical Analysis Handbook* (2024 edition, StatsRef); Diez, D. M., Çetinkaya-Rundel, M., and Barr, C. D., *OpenIntro Statistics* (CC BY-SA). ## Answer key **Example B (gaps).** Binomial with $n = 10$, $p = 0.05$. (b) $P(X \ge 2) = 1 - P(X \le 1)$. By hand: $(0.95)^9 \approx 0.6302$, so $P(X = 1) = 10 \times 0.05 \times 0.6302 \approx 0.3151$. (b) is $1 - 0.9139 = 0.0861$. scipy: `binom.pmf(1, 10, 0.05)` and `binom.sf(1, 10, 0.05)`. About 9% of checks would show 2 or more underbaked loaves even when the process is in control, so one such batch is not strong evidence of a problem; a run of them would be. **Exercise C.** 1. Normal(65, 10): $P(X > 80) = 0.0668$ (`pnorm(80, 65, 10, lower.tail = FALSE)`, `norm.sf(80, 65, 10)`), so about 6.7% of students. 90th percentile: 77.8 (`qnorm(0.90, 65, 10)` gives 77.8155). 2. Exponential with mean 4, so rate 0.25: $P(T > 6) = e^{-0.25 \times 6} = e^{-1.5} \approx 0.2231$. scipy: `expon.sf(6, scale=4)`. Writing `expon.sf(6, 0.25)` would set `loc = 0.25`, a different distribution. 3. Hypergeometric: $\binom{5}{1}\binom{15}{3}/\binom{20}{4} = 5 \times 455 / 4845 \approx 0.4696$. R `dhyper(1, 5, 15, 4)`; scipy `hypergeom.pmf(1, 20, 5, 4)`. **Self-test.** 1. Binomial(25, 0.8): mean $25 \times 0.8 = 20$, variance $25 \times 0.8 \times 0.2 = 4$. 2. Poisson(2). R `ppois(3, 2, lower.tail = FALSE)`, scipy `poisson.sf(3, 2)`; both give 0.1429 (checked by running both). 3. Exponential (a constant rate of arrivals means memoryless waiting times). 4. Hypergeometric with $N = 30$, $K = 8$, $n = 6$ (without replacement from a small population). 5. Geometric with $p = 0.6$. R counts failures, so "passes on the 2nd attempt" is one failure: `dgeom(1, 0.6)` $= 0.4 \times 0.6 = 0.24$. In scipy it would be `geom.pmf(2, 0.6)`. 6. F distribution with df equal to the two variance estimates' df. ## Revision history | Version | Date | Author | Change | |---|---|---|---| | 0.1 | YYYY-MM-DD | [Your name] | First version from template | | [0.2] | YYYY-MM-DD | [Your name] | [Added distributions used in my course] |