IB Maths AI HLPoisson DistributionPaper 1 & 2~6 min read
The Poisson Distribution
The Poisson distribution counts how many times something happens in a fixed window of time or space — calls to a helpline in 15 minutes, daisies per square metre. It’s discrete, written X ∼ Po(m), and has one parameter: the average rate m. Its signature feature is that the mean and variance are both equal to m. Two independent Poissons add into another Poisson, and setting one up is mostly about scaling the rate to the right time period.
📘 What you need to know
Po(m) is a discrete distribution counting occurrences in a fixed time/space, with average rate m.
Two conditions: occurrences are independent and happen at a uniform average rate.
Mean = variance = m (both in the formula booklet); SD = √m.
Values: X = 0, 1, 2, … with no upper bound; m can be any positive real.
Formula: P(X = r) = e−mmrr!, but use the GDC in the exam.
Sum of independents: X ∼ Po(m), Y ∼ Po(λ) ⇒ X+Y ∼ Po(m+λ).
Modelling: scale the rate by proportion to match the stated time period, and state your variable clearly.
Properties of the Poisson
A discrete variable X follows a Poisson if it counts independent occurrences arriving at a steady average rate. The defining quirk: its mean and variance are identical.
Poisson probability & parameters
P(X = r) = e−mmrr! E(X) = m, Var(X) = mformula, mean & variance all in the booklet ✓
🧠 Memory aid — “mean = variance = m”
The Poisson’s fingerprint is that its mean and variance are equal, both m. So the standard deviation is √m. If a real dataset has roughly equal mean and variance, a Poisson model is plausible — that equality is the quick check.
Shape of Po(m) as m grows
All Poisson graphs have a right tail; the bigger m is, the more symmetrical (bell-like) the shape.
🤔 Why is there no upper bound?
Unlike the binomial — where you can’t have more successes than trials n — a Poisson counts events with no fixed ceiling. In any window, 0, 1, 2, … events could occur, in principle without limit. The probabilities of very large counts just become vanishingly small.
WE 1
Stating Poisson assumptions (Jack’s emails)
Jack uses Po(6.25) to model the number of emails he receives during his hour lunch break. Write down two assumptions Jack has made.
the two Poisson conditions1. the emails he receives are independent of each other.2. he receives emails at a uniform average rate of 6.25 per hour during his lunch breaks.independence + uniform average rate
WE 2
Standard deviation of a Poisson
For Jack’s model Po(6.25), calculate the standard deviation of the number of emails per lunch break.
variance = m for a Poissonσ² = 6.25SD = √varianceσ = √6.25 = 2.5standard deviation = 2.5 emailsvariance equals the mean, so just square-root m.
Adding Poisson variables
Independent Poissons combine very neatly: their sum is another Poisson whose mean is just the sum of the means. No need to recompute anything from scratch.
Sum of independent PoissonsX ∼ Po(m), Y ∼ Po(λ) ⇒ X + Y ∼ Po(m + λ)
NOT in the booklet — know the “add the means” rule ✗
Extends to any number: for independent X1 ∼ Po(m1), …, Xn ∼ Po(mn), the total X1 + … + Xn ∼ Po(m1 + … + mn). Just add up all the rates.
WE 3
Combining two independent Poissons
X ∼ Po(6.25) and Y ∼ Po(4) are independent. Write down the distribution of X + Y.
sum of independent Poissons → add the meansm = 6.25 + 4 = 10.25X + Y ∼ Po(10.25)this single combined Poisson is what you’d use for P(X + Y > 7) etc.
Modelling with a Poisson
Most exam questions hand you a scenario and ask you to set up the model. The key step is scaling the rate to the time period in the question.
🧭 Recipe — setting up a Poisson model
Identify the occurrence — what single event are you counting (a car passing, a faulty item)?
Scale the rate by proportion to the stated period — e.g. 10 cars in 5 min → 120 cars in an hour.
State the variable clearly — e.g. “let X be the number of cars passing the camera in 10 minutes”.
WriteX ∼ Po(m) with the matched rate, then proceed to probabilities.
WE 4
Scaling the rate to a new period
A helpline receives calls at an average of 8 calls per hour, modelled by a Poisson. Find the distribution of the number of calls in a 15-minute period.
scale the rate by proportion15 min = ¼ hour → m = 8 × ¼ = 2state the variablelet C = number of calls in 15 minutesC ∼ Po(2)always re-scale m to match the exact time window in the question.
WE 5
Modelling over space + combining
Daisies grow at an average of 3 per m². A lawn patch is 4 m². Assuming a Poisson model, find the distribution of the number of daisies on the patch.
scale the rate to the aream = 3 × 4 = 12state the variablelet D = number of daisies on the 4 m² patchD ∼ Po(12)Poisson works for space as well as time — scale the rate the same way.
💡 Top tips
Mean = variance = m; SD = √m — the Poisson’s signature.
Check the conditions: independent occurrences at a uniform average rate.
Scale the rate by proportion to match the exact time or space window.
Sum of independent Poissons = Po(sum of means) — just add the rates.
Roughly equal mean and variance in real data hints a Poisson model fits.
Name the distribution for each variable when a question mixes several.
⚠ Common mistakes
Forgetting to scale m to the period asked about — using the hourly rate for a 15-minute window.
Using the variance as the SD — take √m, since Var = m.
Adding standard deviations when combining Poissons — you add the means, not the SDs.
Combining non-independent Poissons with the sum rule — independence is required.
Treating m as needing to be an integer — m can be any positive real (e.g. 6.25).
Assuming a fixed maximum — a Poisson has no upper bound, unlike a binomial.
Next up — Calculating Poisson Probabilities. Now that you can set up X ∼ Po(m) and combine Poissons, you’ll use the GDC’s Poisson PD and CD functions to find P(X = x) and cumulative probabilities — turning strict and one-sided inequalities into clean integer ranges, exactly as you did for the binomial.
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