A Poisson distribution calculator finds the chance of seeing exactly x events in a fixed interval when events happen at an average rate λ. The formula is P(X = x) = e^−λλ^x ÷ x!. With λ = 2.15 calls per minute, P(X = 3) = 0.1929 and P(X ≤ 3) = 0.8291. The mean and variance both equal λ.

Rule of thumb: use the Poisson distribution for counts of independent events at a steady rate. If the sample variance is much larger than the mean (variance ÷ mean above about 1.5), the data are overdispersed and a negative binomial model is safer.

The Poisson distribution is a probability distribution for counts. It tells you how likely it is to see 0, 1, 2, 3 or more events in a fixed interval of time, area, length or volume. Examples are phone calls per minute, cars per hour, typing errors per page, or birds per plot. It is named after the French mathematician Siméon Denis Poisson (1837). The name is pronounced “pwah-SON”.

The whole distribution is controlled by one number, lambda (λ). Lambda is the average number of events per interval. It is also the variance. So a Poisson variable with λ = 4 has a mean of 4, a variance of 4 and a standard deviation of 2.

What it measures: the probability mass function (PMF) gives P(X = x), the chance of exactly x events. The cumulative distribution function (CDF) gives P(X ≤ x), the chance of x events or fewer. The Poisson is discrete: X can only be 0, 1, 2, and so on. There is no continuous Poisson distribution; the continuous partner is the exponential distribution, which models the waiting time between Poisson events.

A worked one-liner: a help desk gets 43 calls in 20 minutes, so λ = 43 ÷ 20 = 2.15 calls per minute. The chance of exactly 3 calls in the next minute is e^−2.15 × 2.15³ ÷ 3! = 0.1164 × 9.938 ÷ 6 = 0.1929, about 19%.

The four conditions for a Poisson process:

• Events are counted in a fixed interval (one minute, one page, one square metre).
• Events happen independently. One event does not make another more or less likely.
• The average rate λ is constant across the interval.
• Two events cannot happen at exactly the same instant.

Who uses it: call centres plan staff with it, hospitals model patient arrivals, engineers count defects per unit, ecologists test whether animals or plants are randomly spread, insurers model claim counts, physicists count radioactive decays, and web teams model sign-ups or server requests per second.

4.1 What lambda (λ) tells you

Lambda is the average number of events in one interval. If λ = 2.15 calls per minute, then over many minutes the typical minute has about two calls. It does not mean every minute has 2.15 calls; counts are always whole numbers. Some minutes have 0 calls, some have 5. Lambda is the long-run average of all those minutes.

Lambda always belongs to an interval. A rate of 2.15 per minute is 129 per hour. If you change the interval, scale λ in the same way: λ_new = λ × (new length ÷ old length). Forgetting this step is the most common error in Poisson homework and in real planning work.

4.2 Reading P(X = x), the probability mass function

P(X = x) is the chance of exactly x events. In the worked example P(X = 3) = 0.1929. Say it in words: “In about 19 out of every 100 minutes, the desk gets exactly three calls.” The PMF values for all x add up to 1. The tallest bar is the mode. For a Poisson the mode is the whole-number part of λ, so for λ = 2.15 the most likely count is 2. When λ is a whole number, λ and λ − 1 tie as modes.

4.3 Reading the CDF and the tails

The cumulative distribution function P(X ≤ x) adds the PMF from 0 up to x. P(X ≤ 3) = 0.8291 means 83% of minutes have three calls or fewer. The right tail is the complement. Take care with the wording, because “at least 3” and “more than 3” are different questions:

• At most x: P(X ≤ x) = CDF(x) = 0.8291
• Fewer than x: P(X < x) = CDF(x − 1) = 0.6361
• At least x: P(X ≥ x) = 1 − CDF(x − 1) = 0.3639
• More than x: P(X > x) = 1 − CDF(x) = 0.1709

Because the Poisson is discrete, the “equal to” part matters. For a continuous distribution P(X ≤ 3) and P(X < 3) are the same; for a Poisson they differ by P(X = 3), which here is almost 20 percentage points.

4.4 Mean, variance and standard deviation of a Poisson distribution

The expected value of a Poisson distribution is E(X) = λ, and the variance is Var(X) = λ. So the standard deviation is √λ. For λ = 2.15, the SD is 1.466 calls. This link between mean and variance is the fingerprint of the Poisson. It also means the relative spread, SD ÷ mean = 1 ÷ √λ, shrinks as λ grows. Rare events are noisy; common events are more predictable.

4.5 The estimate λ̂ and its confidence interval

When you paste data, the calculator estimates λ by the sample mean, λ̂ = Σx ÷ n. This is the maximum likelihood estimator (MLE) of the Poisson distribution, and it is also unbiased. The standard error is √(λ̂ ÷ n).

The confidence interval is the exact (Garwood) interval, built from chi-square quantiles. It is not symmetric: the upper arm is longer than the lower arm, because counts cannot go below zero. For the worked example the 95% CI is 1.556 to 2.896 calls per minute. A narrow interval needs a large total count Σx, not just many intervals. Doubling the observation time roughly shrinks the width by √2.

4.6 Is the Poisson a good fit? The dispersion index

The variance-to-mean ratio (the index of dispersion, s² ÷ x̄) should be close to 1. The calculator tests this with D = Σ(x − x̄)² ÷ x̄, which follows a chi-square distribution with n − 1 degrees of freedom if the data are Poisson.

• Ratio near 1, p ≥ 0.05: no evidence against the Poisson. Use it.
• Ratio well above 1, small p: overdispersion. Events cluster (contagion, patchy habitat, mixed groups). Poisson standard errors will be too small. Use a negative binomial or quasi-Poisson model.
• Ratio well below 1, small p: underdispersion. Events are too regular (territorial animals, booked appointments). The Poisson will overstate the spread.

4.7 The chi-square goodness-of-fit test

The goodness-of-fit test compares how many intervals had 0, 1, 2, … events with how many a perfect Poisson would give. Neighbouring bins are merged until every expected count is at least 5 (or the level you pick). The degrees of freedom are bins − 2, because one extra degree is lost for estimating λ from the same data. A large p-value means the observed pattern is consistent with a Poisson. It does not prove the data are Poisson; with small samples the test has little power, so read it together with the dispersion index and Chart 3.

4.8 Decision guide

4.9 Poisson distribution vs binomial, and vs normal

The binomial counts successes in n fixed trials; the Poisson counts events with no fixed upper limit. When n is large and p is small (a common guide is n ≥ 20 and p ≤ 0.05), Binomial(n, p) is close to Poisson(λ = np). That is the Poisson approximation to the binomial distribution. For example, 1,000 items with a 0.2% defect rate gives λ = 2.

When λ is large (about 20 or more), the Poisson looks like a normal distribution with mean λ and variance λ. Use a continuity correction: P(X ≤ x) ≈ Φ((x + 0.5 − λ) ÷ √λ). For small λ the normal approximation is poor because the Poisson is skewed.

4.10 Common mistakes this calculator prevents

• Mixing up ≥ and >. Every tail is shown separately with its own label.
• Forgetting to rescale λ. The report states the interval each λ belongs to.
• Using the Poisson on overdispersed data. The dispersion test runs automatically.
• Wrong degrees of freedom in the fit test. The calculator uses bins − 2, not bins − 1.
• Symmetric intervals for a rate. The exact Garwood interval is used instead of λ ± 1.96 SE.
• Non-integer or negative counts. These are rejected with a clear message.

When reporting Poisson distribution results in APA format, give four things: the estimated rate λ with its interval and unit of time or space, the confidence interval, a check that the Poisson fits (dispersion index and goodness of fit), and any probability you use for a decision. The five templates below fill in automatically after you run the calculator.

▶ Run the analysis above to auto-fill all five examples with your results.

These are the Poisson distribution formulas and equations used by this calculator, from the probability mass function out to the fit tests.

This flowchart shows how the Poisson distribution is calculated, step by step, from raw counts to a checked probability. Every box uses the worked example used in the Excel, R and Python sections: calls to a help desk in 20 one-minute intervals (2, 3, 1, 4, 2, 0, 3, 2, 1, 5, 2, 3, 1, 2, 4, 2, 3, 1, 0, 2).

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return these numbers for this dataset.

How to use this Poisson distribution calculator, with the help-desk example as a guide:

1. Pick a mode. Choose Fit my count data if you have raw counts, or I already know λ if you only need probabilities.
2. Enter your counts. Type comma-separated whole numbers, for example 2, 3, 1, 4, 2, 0, .... Each number is the count in one interval.
3. Name each cluster. The name box above each textarea is editable, for example Weekday mornings. The name appears in tables, charts and reports.
4. Add clusters to compare groups. Press Add cluster. Each cluster gets its own λ, CI and fit tests.
5. Or upload a file. On the Upload tab, choose a CSV or Excel file and click the columns you want. Each clicked column becomes its own cluster.
6. Try a sample dataset. Ten examples include the classic horse-kick deaths and Rutherford–Geiger decay counts, plus an overdispersed dataset that fails the test.
7. Type your probability question. Set x for P(X = x) and the tails, and a to b for a range such as P(1 ≤ X ≤ 4).
8. Pick the confidence level and bin rule, then press Calculate Poisson Distribution. Nothing runs until you press it.
9. Read the verdict line and the fit checks first. If overdispersion is flagged, treat the Poisson probabilities with caution.
10. Check the four charts and export. Section 5 fills five reporting templates, and the buttons save a text report or a print-ready PDF.

Excel has a built-in Poisson distribution function, POISSON.DIST. It gives both the PMF and the CDF. What catches people out is the tails, the confidence interval for λ, and the goodness-of-fit test, which Excel does not do for you. The steps below rebuild every number on this page for the help-desk example.

9.1 The functions you need

9.2 Step by step

9.3 The awkward cases

Rescaling lambda. If your rate is per hour and you need per 15 minutes, multiply first: =POISSON.DIST(2,$E$2*15/60,FALSE).

Large lambda. For λ above about 20, a normal approximation is close: =NORM.DIST(x+0.5,λ,SQRT(λ),TRUE). POISSON.DIST itself stays accurate, so use it when you can.

Binomial to Poisson. Compare =BINOM.DIST(2,1000,0.002,FALSE) (0.2709) with =POISSON.DIST(2,2,FALSE) (0.2707). They agree to three decimals.

9.4 Why your Excel answer might differ from R or Python

9.5 Charting the Poisson distribution in Excel

1. Select the x column and the PMF column (G1:H10).
2. Insert → Charts → Clustered Column. Set the gap width to about 30%.
3. For an observed vs expected chart, add observed counts as columns and expected counts as a second series, then Change Chart Type → Combo and set expected to Line with Markers.
4. For the CDF, select x and the CDF column and choose Insert → Line → Line with Markers.
5. Label the x axis “Number of events per interval” and give λ in the title.

Charts 1 to 3 in Section 3 of this page draw these automatically.

9.6 Excel errors you will probably hit

No Analysis ToolPak is needed for any step above. It has a random number generator with a Poisson option, which must be enabled first under File → Options → Add-ins, but it is only useful for simulation.

R has four built-in Poisson functions: dpois for the PMF, ppois for the CDF, qpois for quantiles and rpois for random numbers. The script below uses base R only. It estimates λ, gives the exact CI with poisson.test, runs the dispersion and goodness-of-fit tests with the correct degrees of freedom, prints a plain-English verdict, and saves one 300 dpi figure.

10.1 The complete script

R, base only, saves poisson_fit.png

# ============================================================
# Poisson distribution: fit lambda, test the fit, get probabilities
# StatsUnlock - https://statsunlock.com/poisson-distribution-calculator/
# Base R only. No packages to install.
# ============================================================
set.seed(42)   # nothing random below, kept for reproducibility

# ---- 1. Your data ------------------------------------------
# Replace with your own counts, or read a CSV:
#   x <- read.csv("data.csv")$calls
x <- c(2, 3, 1, 4, 2, 0, 3, 2, 1, 5,
       2, 3, 1, 2, 4, 2, 3, 1, 0, 2)
x <- x[!is.na(x)]                 # drop missing values
label <- "Help desk calls per minute"
x_q <- 3; a <- 1; b <- 4          # probability question
alpha <- 0.05
min_expected <- 5

# ---- 2. Estimate lambda ------------------------------------
n   <- length(x)
tot <- sum(x)
lam <- tot / n                    # MLE = sample mean
s2  <- var(x)                     # var() uses n - 1
se  <- sqrt(lam / n)
ci  <- poisson.test(tot, n, conf.level = 1 - alpha)$conf.int   # exact Garwood CI

# ---- 3. Dispersion test ------------------------------------
D    <- sum((x - lam)^2) / lam
df_d <- n - 1
p_d  <- 2 * min(pchisq(D, df_d), pchisq(D, df_d, lower.tail = FALSE))

# ---- 4. Goodness of fit, bins merged until E >= 5 ----------
poisson_gof <- function(x, lam, min_e) {
  n <- length(x); mx <- max(x)
  lo <- c(); hi <- c(); O <- c(); E <- c()
  start <- 0; e <- 0; o <- 0
  for (k in 0:mx) {
    e <- e + dpois(k, lam) * n
    o <- o + sum(x == k)
    if (e >= min_e && k < mx) {
      lo <- c(lo, start); hi <- c(hi, k); O <- c(O, o); E <- c(E, e)
      start <- k + 1; e <- 0; o <- 0
    }
  }
  lo <- c(lo, start); hi <- c(hi, NA)
  O  <- c(O, sum(x >= start))
  E  <- c(E, ppois(start - 1, lam, lower.tail = FALSE) * n)
  m <- length(E)
  if (E[m] < min_e && m > 1) {
    O[m - 1] <- O[m - 1] + O[m]; E[m - 1] <- E[m - 1] + E[m]; hi[m - 1] <- NA
    lo <- lo[-m]; hi <- hi[-m]; O <- O[-m]; E <- E[-m]
  }
  chi <- sum((O - E)^2 / E)
  df  <- length(E) - 2            # minus 1 for estimating lambda
  p   <- if (df > 0) pchisq(chi, df, lower.tail = FALSE) else NA
  list(table = data.frame(from = lo, to = hi, observed = O, expected = round(E, 3)),
       chi = chi, df = df, p = p)
}
g <- poisson_gof(x, lam, min_expected)

# ---- 5. Print everything -----------------------------------
cat(label, "\n")
cat(sprintf("n = %d, total = %d, lambda = %.4f, variance = %.4f\n", n, tot, lam, s2))
cat(sprintf("SD of Poisson = %.4f, SE of lambda = %.4f\n", sqrt(lam), se))
cat(sprintf("%d%% exact CI for lambda = %.4f to %.4f\n", round((1 - alpha) * 100), ci[1], ci[2]))
cat(sprintf("variance / mean = %.4f, D = %.4f, df = %d, p = %.4f\n", s2 / lam, D, df_d, p_d))
print(g$table)
cat(sprintf("goodness of fit chi-square(%d) = %.4f, p = %.4f\n", g$df, g$chi, g$p))
cat(sprintf("P(X = %d)  = %.4f\n", x_q, dpois(x_q, lam)))
cat(sprintf("P(X <= %d) = %.4f\n", x_q, ppois(x_q, lam)))
cat(sprintf("P(X < %d)  = %.4f\n", x_q, ppois(x_q - 1, lam)))
cat(sprintf("P(X >= %d) = %.4f\n", x_q, ppois(x_q - 1, lam, lower.tail = FALSE)))
cat(sprintf("P(X > %d)  = %.4f\n", x_q, ppois(x_q, lam, lower.tail = FALSE)))
cat(sprintf("P(%d <= X <= %d) = %.4f\n", a, b, ppois(b, lam) - ppois(a - 1, lam)))
ok <- p_d >= 0.05 && (is.na(g$p) || g$p >= 0.05)
cat("Verdict:", if (ok) "the Poisson distribution fits these counts.\n"
    else "the Poisson does not fit well; check for overdispersion.\n")

# ---- 6. One figure: observed vs expected -------------------
k   <- 0:max(max(x), ceiling(lam + 4 * sqrt(lam)))
obs <- sapply(k, function(v) sum(x == v))
ex  <- dpois(k, lam) * n
png("poisson_fit.png", width = 7, height = 4.2, units = "in", res = 300)
par(mar = c(4.5, 4.5, 3, 1))
mids <- barplot(obs, names.arg = k, col = "#60a5fa", border = "#1d4ed8",
                ylim = c(0, max(obs, ex) * 1.2),
                xlab = "Number of events per interval (x)",
                ylab = "Number of intervals", main = label)
lines(mids, ex, type = "b", pch = 19, col = "#c2410c", lwd = 2)
legend("topright", bty = "n", fill = c("#60a5fa", NA), border = c("#1d4ed8", NA),
       lty = c(NA, 1), pch = c(NA, 19), col = c(NA, "#c2410c"),
       legend = c("Observed intervals", sprintf("Poisson expected, lambda = %.2f", lam)))
dev.off()
cat("Saved poisson_fit.png\n")

10.2 What each part does

10.3 What the figure shows

Blue bars are how many minutes had 0, 1, 2, … calls. The orange line is how many a perfect Poisson with λ = 2.15 would give. The bars follow the line closely, with slightly more 2s and fewer 0s and 1s than expected, which matches the non-significant fit test (p = 0.390).

10.4 Problems and fixes

10.5 One-liners

In Python the Poisson distribution lives in scipy.stats.poisson. The script below matches the R script number for number: λ, the exact CI, the dispersion test, the goodness-of-fit test with df = bins − 2, every tail probability and a 300 dpi figure.

11.1 The complete script

Python 3, saves poisson_fit.png

# ============================================================
# Poisson distribution: fit lambda, test the fit, get probabilities
# StatsUnlock - https://statsunlock.com/poisson-distribution-calculator/
# pip install numpy scipy matplotlib pandas
# ============================================================
import numpy as np
from scipy import stats
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

np.random.seed(42)  # nothing random below, kept for reproducibility

# ---- 1. Your data ------------------------------------------
# Replace with your own counts, or read a CSV:
#   import pandas as pd
#   x = pd.read_csv("data.csv")["calls"].dropna().astype(int).to_numpy()
x = np.array([2, 3, 1, 4, 2, 0, 3, 2, 1, 5,
              2, 3, 1, 2, 4, 2, 3, 1, 0, 2])
label = "Help desk calls per minute"
x_q, a, b = 3, 1, 4          # probability question
alpha = 0.05
min_expected = 5

x = x[~np.isnan(x)] if x.dtype.kind == "f" else x   # drop missing values

# ---- 2. Estimate lambda ------------------------------------
n = len(x)
T = x.sum()
lam = T / n                                   # MLE = sample mean
s2 = x.var(ddof=1)                            # sample variance, n - 1
se = np.sqrt(lam / n)
lo = stats.chi2.ppf(alpha / 2, 2 * T) / (2 * n) if T > 0 else 0.0
hi = stats.chi2.ppf(1 - alpha / 2, 2 * T + 2) / (2 * n)

# ---- 3. Dispersion test ------------------------------------
D = ((x - lam) ** 2).sum() / lam
df_d = n - 1
p_d = 2 * min(stats.chi2.cdf(D, df_d), stats.chi2.sf(D, df_d))

# ---- 4. Goodness of fit, bins merged until E >= 5 ----------
def poisson_gof(x, lam, min_e):
    n, mx = len(x), x.max()
    bins, start, E, O = [], 0, 0.0, 0
    for k in range(mx + 1):
        E += stats.poisson.pmf(k, lam) * n
        O += (x == k).sum()
        if E >= min_e and k < mx:
            bins.append([start, k, O, E]); start, E, O = k + 1, 0.0, 0
    bins.append([start, None, (x >= start).sum(), stats.poisson.sf(start - 1, lam) * n])
    if bins[-1][3] < min_e and len(bins) > 1:
        last = bins.pop()
        bins[-1][1] = None; bins[-1][2] += last[2]; bins[-1][3] += last[3]
    chi = sum((o - e) ** 2 / e for _, _, o, e in bins)
    df = len(bins) - 2                        # minus 1 for estimating lambda
    p = stats.chi2.sf(chi, df) if df > 0 else float("nan")
    return bins, chi, df, p

bins, chi, df_g, p_g = poisson_gof(x, lam, min_expected)

# ---- 5. Probabilities --------------------------------------
P = stats.poisson
print(f"{label}")
print(f"n = {n}, total = {T}, lambda = {lam:.4f}, variance = {s2:.4f}")
print(f"SD of Poisson = {np.sqrt(lam):.4f}, SE of lambda = {se:.4f}")
print(f"{int((1-alpha)*100)}% exact CI for lambda = {lo:.4f} to {hi:.4f}")
print(f"variance / mean = {s2/lam:.4f}, D = {D:.4f}, df = {df_d}, p = {p_d:.4f}")
for lo_k, hi_k, o, e in bins:
    tag = f"{lo_k}" if hi_k == lo_k else (f">={lo_k}" if hi_k is None else f"{lo_k}-{hi_k}")
    print(f"  bin {tag:>5}: observed {o:3d}, expected {e:7.3f}")
print(f"goodness of fit chi-square({df_g}) = {chi:.4f}, p = {p_g:.4f}")
print(f"P(X = {x_q})  = {P.pmf(x_q, lam):.4f}")
print(f"P(X <= {x_q}) = {P.cdf(x_q, lam):.4f}")
print(f"P(X < {x_q})  = {P.cdf(x_q - 1, lam):.4f}")
print(f"P(X >= {x_q}) = {P.sf(x_q - 1, lam):.4f}")
print(f"P(X > {x_q})  = {P.sf(x_q, lam):.4f}   # scipy sf(k) is P(X > k)")
print(f"P({a} <= X <= {b}) = {P.cdf(b, lam) - P.cdf(a - 1, lam):.4f}")
ok = p_d >= 0.05 and (np.isnan(p_g) or p_g >= 0.05)
print("Verdict:", "the Poisson distribution fits these counts." if ok
      else "the Poisson does not fit well; check for overdispersion.")

# ---- 6. One figure: observed vs expected -------------------
k = np.arange(0, max(x.max(), int(lam + 4 * np.sqrt(lam))) + 1)
obs = np.array([(x == v).sum() for v in k])
exp_ = P.pmf(k, lam) * n
fig, ax = plt.subplots(figsize=(7, 4.2))
ax.bar(k, obs, color="#60a5fa", edgecolor="#1d4ed8", label="Observed intervals")
ax.plot(k, exp_, "o-", color="#c2410c", lw=2, label=f"Poisson expected, lambda = {lam:.2f}")
ax.set_xlabel("Number of events per interval (x)")
ax.set_ylabel("Number of intervals")
ax.set_title(label)
ax.set_xticks(k)
ax.legend(frameon=False)
fig.tight_layout()
fig.savefig("poisson_fit.png", dpi=300)
print("Saved poisson_fit.png")

Output for the worked example: n = 20, total = 43, lambda = 2.1500, 95% CI 1.5560 to 2.8960, D = 15.1395 (df 19, p = 0.5726), goodness of fit χ²(1) = 0.7394, p = 0.3899, P(X = 3) = 0.1929, P(X ≤ 3) = 0.8291, P(X ≥ 3) = 0.3639, P(1 ≤ X ≤ 4) = 0.8163.

11.2 What each part does

11.3 What the figure shows

The figure is the same as the R version: observed minutes as blue bars, expected Poisson counts as an orange line. Use it as the main figure when you report a Poisson fit.

11.4 Problems and fixes

11.5 One-liners

12.1 Poisson distribution table, P(X = x)

Conclusion: read down a column for your λ. The largest value in each column sits at x = λ (or λ − 1 and λ tied when λ is a whole number).

12.2 Cumulative Poisson probability table, P(X ≤ x)

Conclusion: P(X ≥ x) = 1 minus the value one row above. For λ = 2, P(X ≥ 4) = 1 − 0.8571 = 0.1429.

12.3 Mean, variance, standard deviation and shape by λ

Conclusion: skewness falls as λ rises. Above λ ≈ 20 the Poisson is nearly symmetric and a normal approximation works well.

12.4 Poisson approximation to the binomial

Conclusion: with λ held at 2, the approximation improves as n grows and p shrinks. Use it when n ≥ 20 and p ≤ 0.05.

12.5 Chi-square critical values for the fit and dispersion tests

Conclusion: the worked goodness-of-fit χ² of 0.739 on 1 df is far below 3.841, so the Poisson is not rejected.

12.6 Variance-to-mean ratio guide

Conclusion: these bands are guides only. Always confirm with the dispersion test, whose power depends on n.

Eight worked examples of the Poisson distribution, from the textbook case to a clear failure. Each shows the observed counts against a perfect Poisson.

Study design: count the number of events in many equal, non-overlapping intervals of time or space, with the same conditions in every interval. Each interval gives one whole-number count.

1. Define the event exactly. Write down what counts as one event (one call answered, one defect over 1 mm, one nest with eggs). Vague rules add noise that looks like overdispersion.
2. Fix the interval length before you start. One minute, one page, one 10 m × 10 m quadrat. Every interval must be the same size, or you need a rate model with an offset.
3. Keep conditions steady. Do not mix busy and quiet hours in one sample. If the rate changes with time of day, record it as a separate cluster.
4. Choose intervals at random or systematically. Picking only “interesting” intervals biases λ upward.
5. Make the intervals independent. Leave gaps between intervals if events can trigger each other, such as repeat calls.
6. Record zeros. An interval with no events is a real observation with the value 0. Leaving zeros out inflates λ and wrecks the fit test.
7. Record missing intervals as blank, never as 0. A broken counter is not a zero count.
8. Collect enough intervals. Aim for at least 30 intervals, and a total count T of at least 30, so the CI is useful and the fit test has several bins.
9. Log the conditions. Date, time, observer and any unusual event, so you can explain outliers later.
10. Plan the model check. Decide now that you will use the dispersion test and switch to a negative binomial model if p < 0.05.

Minimum sample: 30 or more intervals and a total count of 30 or more. With fewer, report λ and its exact CI but treat the fit tests as weak.

Pre-registration line: “Counts per fixed interval will be modelled as Poisson; if the index of dispersion test gives p < .05, a negative binomial model will be used instead.”

Common mistakes: unequal intervals, dropping zero counts, typing 0 for missing data, pooling different time periods, and counting the same event twice across neighbouring intervals.

This free Poisson distribution calculator is designed for students, teachers, analysts and researchers who work with counts of random events.

Four examples of the Poisson distribution:

• Operations: a call centre averages 2.15 calls per minute; what is the chance of 5 or more in one minute?
• Health care: an emergency room averages 6 arrivals per hour; how many nurses cover 95% of hours?
• Quality control: 0.8 defects per circuit board; what share of boards will be defect-free? P(X = 0) = e^−0.8 = 0.449.
• Ecology: are nests spread at random across quadrats, or clumped? The dispersion test answers this.

Design guidance: choose an interval long enough that most intervals have at least one event (λ around 1 to 10 is easiest to work with), but short enough that the rate stays steady within it.

17.1 Assumptions of the Poisson distribution

1. Counts in fixed intervals. Check: all intervals are equal. If broken: λ mixes different exposures and is biased toward the longer intervals.
2. Independence. Check: plot counts in time order and look for runs. If broken (contagion): variance is inflated, SEs are too small, p-values too small.
3. Constant rate. Check: compare λ across clusters or halves of the data. If broken: overdispersion appears, and tail probabilities are underestimated.
4. No simultaneous events. Check: events are recorded one by one. If broken (batches): counts jump in groups and variance is too large.
5. Mean equals variance. Check: the dispersion test. If broken: overdispersion makes the Poisson CI too narrow; underdispersion makes it too wide.

17.2 Limitations of this calculator

• It fits one λ per cluster. To model λ as a function of predictors, use Poisson regression (a GLM).
• It does not fit a negative binomial or zero-inflated model. It flags the need for one.
• The goodness-of-fit test has low power with fewer than about 30 intervals. Use a likelihood-based test or simulation for small samples.
• It does not compare two rates formally. Use an exact conditional test or Poisson regression.
• It cannot fix biased sampling. A precise λ from unrepresentative intervals is still wrong.

These peer-reviewed journal articles support the methods in this Poisson distribution calculator: the classic Poisson datasets, exact confidence intervals for λ, the Poisson approximation to the binomial, dispersion and goodness-of-fit tests, and modern models for overdispersed and zero-inflated counts.

1. Rutherford, E., & Geiger, H. (1910). The probability variations in the distribution of α particles. Philosophical Magazine, 20(118), 698–707. https://doi.org/10.1080/14786441008636955
2. Garwood, F. (1936). Fiducial limits for the Poisson distribution. Biometrika, 28(3/4), 437–442. https://doi.org/10.1093/biomet/28.3-4.437
3. Clarke, R. D. (1946). An application of the Poisson distribution. Journal of the Institute of Actuaries, 72(3), 481. https://doi.org/10.1017/S0020268100035435
4. Fisher, R. A. (1950). The significance of deviations from expectation in a Poisson series. Biometrics, 6(1), 17–24. https://doi.org/10.2307/3001420
5. Cochran, W. G. (1954). Some methods for strengthening the common χ² tests. Biometrics, 10(4), 417–451. https://doi.org/10.2307/3001616
6. Hodges, J. L., & Le Cam, L. (1960). The Poisson approximation to the Poisson binomial distribution. Annals of Mathematical Statistics, 31(3), 737–740. https://doi.org/10.1214/aoms/1177705799
7. Le Cam, L. (1960). An approximation theorem for the Poisson binomial distribution. Pacific Journal of Mathematics, 10(4), 1181–1197. https://doi.org/10.2140/pjm.1960.10.1181
8. Chen, L. H. Y. (1975). Poisson approximation for dependent trials. Annals of Probability, 3(3), 534–545. https://doi.org/10.1214/aop/1176996359
9. Ahrens, J. H., & Dieter, U. (1982). Computer generation of Poisson deviates from modified normal distributions. ACM Transactions on Mathematical Software, 8(2), 163–179. https://doi.org/10.1145/355993.355997
10. Good, I. J. (1986). Some statistical applications of Poisson's work. Statistical Science, 1(2), 157–170. https://doi.org/10.1214/ss/1177013690
11. Dean, C., & Lawless, J. F. (1989). Tests for detecting overdispersion in Poisson regression models. Journal of the American Statistical Association, 84(406), 467–472. https://doi.org/10.1080/01621459.1989.10478792
12. Cameron, A. C., & Trivedi, P. K. (1990). Regression-based tests for overdispersion in the Poisson model. Journal of Econometrics, 46(3), 347–364. https://doi.org/10.1016/0304-4076(90)90014-K
13. Ulm, K. (1990). A simple method to calculate the confidence interval of a standardized mortality ratio (SMR). American Journal of Epidemiology, 131(2), 373–375. https://doi.org/10.1093/oxfordjournals.aje.a115507
14. Lambert, D. (1992). Zero-inflated Poisson regression, with an application to defects in manufacturing. Technometrics, 34(1), 1–14. https://doi.org/10.2307/1269547
15. Böhning, D. (1994). A note on a test for Poisson overdispersion. Biometrika, 81(2), 418–419. https://doi.org/10.1093/biomet/81.2.418
16. Gardner, W., Mulvey, E. P., & Shaw, E. C. (1995). Regression analyses of counts and rates: Poisson, overdispersed Poisson, and negative binomial models. Psychological Bulletin, 118(3), 392–404. https://doi.org/10.1037/0033-2909.118.3.392
17. Barker, L. (2002). A comparison of nine confidence intervals for a Poisson parameter when the expected number of events is ≤ 5. The American Statistician, 56(2), 85–89. https://doi.org/10.1198/000313002317572736
18. Ver Hoef, J. M., & Boveng, P. L. (2007). Quasi-Poisson vs. negative binomial regression: How should we model overdispersed count data? Ecology, 88(11), 2766–2772. https://doi.org/10.1890/07-0043.1
19. Coxe, S., West, S. G., & Aiken, L. S. (2009). The analysis of count data: A gentle introduction to Poisson regression and its alternatives. Journal of Personality Assessment, 91(2), 121–136. https://doi.org/10.1080/00223890802634175
20. O'Hara, R. B., & Kotze, D. J. (2010). Do not log-transform count data. Methods in Ecology and Evolution, 1(2), 118–122. https://doi.org/10.1111/j.2041-210X.2010.00021.x