Probability Distributions Reference Sheet Template (Statistics, R and Python)

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

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.

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).

Probability mass functions

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).

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.

Probability density functions

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.

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).

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).

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).

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)

# 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")

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)

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))

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

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

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