Inverse Poisson Calculator

The inverse Poisson distribution finds the smallest count k with P(X ≤ k) ≥ p, for a Poisson variable X with mean λ. It answers "how many events should I plan for?" where the Poisson CDF answers "how likely is k events or fewer?".

• X counts events in a fixed interval: calls per hour, defects per metre, arrivals per day.
• For λ = 4 and p = 0.95, k = 8: P(X ≤ 7) = 0.9489 falls short and P(X ≤ 8) = 0.9786 reaches 0.95.
• The answer is a whole number, so the probability at k usually lands above p, not on it.

Other names: the Poisson quantile function, qpois in R and poisson.ppf in Python.

To calculate the inverse Poisson, add the probabilities P(X = 0), P(X = 1), P(X = 2), … until the running total reaches p; the last count added is k. For λ = 4 and p = 0.95:

1. P(X = 0) = e^(−4) = 0.0183. Total 0.0183.
2. Each next term is the last one times λ/i: 0.0733, 0.1465, 0.1954, 0.1954, 0.1563, 0.1042, 0.0595.
3. The total after k = 7 is 0.9489, still below 0.95.
4. Add P(X = 8) = 0.0298. The total is 0.9786, so k = 8.

The rule "times λ/i" saves work: P(X = i) = P(X = i − 1) × λ / i, so no factorials are needed.

Pick a mode, type the numbers, and the calculator returns k, λ or an interval as you type. The steps:

1. Choose Find k for a count, Find λ for a rate, or Interval for λ for a confidence interval from an observed count.
2. For k: type the mean λ and the probability p, then pick the rule. "At most k" gives P(X ≤ k) ≥ p. "More than k" gives P(X > k) ≤ p.
3. Type p as a decimal (0.95), a percent (95%) or a fraction (19/20).
4. Read the answer, the probabilities on each side of k, the normal approximation and the most likely count.
5. Open the probability bars or the cumulative graph, and the worked steps.

The result can be copied, shared as a link, printed or kept in the history list.

The inverse Poisson has no closed formula: k is the smallest whole number where the Poisson CDF reaches p.

• Cumulative probability: P(X ≤ k) = e^(−λ) Σᵢ₌₀ᵏ λⁱ / i!.
• Probability of exactly i events: P(X = i) = e^(−λ) λⁱ / i!.
• Recurrence: P(X = i) = P(X = i − 1) · λ / i, starting from e^(−λ).
• Right tail: the smallest k with P(X > k) ≤ p is Q(1 − p).
• Gamma link: P(X ≤ k) = Q(k + 1, λ), the regularized upper incomplete gamma function.
• Mean and variance: both equal λ, so the standard deviation is √λ.

Software jumps close with the normal guess λ + z√λ, then steps up or down one count at a time.

The inverse Poisson stacks P(X = 0), P(X = 1), … end to end; k is the block that pushes the total past p.

Read the table backwards: p = 0.5 lands at k = 4, p = 0.9 at k = 7 and p = 0.95 at k = 8. Any p from 0.9489 up to 0.9786 gives the same k = 8.

To find λ, solve e^(−λ) Σᵢ₌₀ᵏ λⁱ / i! = p for λ; the left side falls as λ grows, so one λ fits and bisection finds it.

• P(X = 0) = 0.05: e^(−λ) = 0.05 gives λ = −ln 0.05 = 2.9957.
• P(X ≤ 3) = 0.5: λ = 3.6721, so 3.6721 is the mean whose median count is 3.
• P(X ≤ 10) = 0.95: λ = 6.1690.

The same λ comes from the chi-square distribution: λ = χ²(1 − p; 2k + 2) / 2. For k = 3 and p = 0.05, λ = 15.5073 / 2 = 7.7537.

The exact (Garwood) confidence interval for λ after k events runs from λ₁ with P(X ≥ k) = α/2 to λ₂ with P(X ≤ k) = α/2. After 12 events, the 95% interval is 6.2006 to 20.9616.

Divide both bounds by the exposure to get a rate. 30 events in 10 years gives a 90% interval of 2.16 to 4.07 events a year.

Zero events still gives an upper bound: 3.689 at 95% two-sided, or 2.996 one-sided. The "rule of 3" uses this: no events in n trials puts the rate below about 3/n.

For large λ, k ≈ ⌈λ + z√λ − 0.5⌉, where z = Φ⁻¹(p) and the 0.5 is a continuity correction. The approximation improves as λ grows, since the skewness 1/√λ shrinks.

• λ = 4, p = 0.95: the approximation gives 7, the exact answer is 8.
• λ = 10, p = 0.95: both give 15.
• λ = 100, p = 0.95: the approximation gives 116, the exact answer is 117.
• λ = 120, p = 0.99: the approximation gives 145, the exact answer is 146.

The Poisson distribution leans right, so the normal guess tends to fall 1 short in the upper tail. Use the exact sum when the count sets stock, staff or capacity.

The graph of the inverse Poisson is a staircase: k stays flat while p moves inside one step of the CDF, then jumps by 1.

• The Poisson CDF F(k) is a step graph that climbs from e^(−λ) towards 1.
• Read it sideways: draw a level at p, move right to the first step at or above it, and read k.
• The steps are tallest near k = λ, where the probabilities are largest.
• For λ = 4, every p from 0.4335 to 0.6288 returns k = 4.

R has qpois(p, λ), Python has scipy.stats.poisson.ppf(p, λ) and MATLAB has poissinv(p, λ); Excel and the TI-84 have no inverse Poisson function.

• R: qpois(0.95, 4) returns 8. qpois(0.05, 4, lower.tail = FALSE) gives the right-tail answer, 8.
• Python: from scipy.stats import poisson; poisson.ppf(0.95, 4) returns 8.0.
• MATLAB: poissinv(0.95, 4) returns 8.
• Excel and Google Sheets: list k in one column and =POISSON.DIST(k, 4, TRUE) next to it, then take the first row at or above p.
• TI-84: run poissoncdf(4, k) for k = 0, 1, 2, … until the result reaches p.

The inverse Poisson table lists the smallest k with P(X ≤ k) ≥ p for means from 0.5 to 100. Click a row to load that λ with p = 0.95 into the calculator.

The 0.5 column is the median. For whole-number λ the median equals λ; the 99% column sits about 2.3√λ above the mean.

The Poisson CDF turns a count into a probability; the inverse Poisson turns a probability back into a count. P(X ≤ 8) = 0.9786 for λ = 4, and qpois(0.9786, 4) = 8.

The round trip Q(F(k)) = k always holds. F(Q(p)) = p holds only when p lands on a step; otherwise F(Q(p)) is a little above p.

The inverse Poisson sets capacity: how many staff, parts, servers or beds cover random demand a chosen share of the time.

• Call centre: 120 calls an hour on average. Planning for 146 calls covers 99% of hours.
• Spare parts: 3 failures a month on average. A stock of 8 parts runs out in fewer than 1 month in 250.
• Web server: 40 requests a second on average. Capacity for 55 covers 99% of seconds.
• Hospital beds: 18 admissions a day on average. 25 free beds cover 95% of days.
• Quality control: with 2 defects per roll expected, a roll with more than qpois(0.99, 2) = 6 defects flags a problem.