Topic 4.17 · teacher page · Higher Level

Lambda belongs to the interval

The scaling step that costs most of the marks, and the identity that tells you the model is wrong.

The one thing to do with the animation

Stretch the window to two hours and ask for λ.

Many will say 3, because 3 is the number in the question. It is 6, and the readout says so while the quoted rate has not changed at all.

Note what the mean and variance readouts do: they move together, always, because they are the same number. That is the identity examiners use when they ask whether Poisson is appropriate, and it arrives here for free.

Telling it apart from the binomial

A binomial has a fixed n and stops there. A Poisson has no n and no largest value. The language in the question is the tell: “out of 50” is binomial, “per hour” or “per page” is Poisson.

Give them five one-line scenarios and have them sort rather than compute. It takes four minutes and it prevents the wrong distribution being chosen under exam pressure, which is unrecoverable.

The answers

1. λ for 250 m2 × (250/100) = 5.
2. sd of Po(9)Variance is λ = 9, so sd = 3.
3. P(X = 0) for Po(1)e−1 = 0.368.
4. Mean 4.1, variance 11.8B, Poisson fits poorly. The variance is nearly three times the mean, so the events are clumping.

Where the marks go

1 markScaling λ to the interval the question asks about, before any calculator work.

1 markReading the inequality correctly, with “at least 2” as 1 − P(X ≤ 1).

1 markJustifying the model: constant average rate, independent events, mean roughly equal to variance.

What each wrong answer tells you

They giveWhat it means
2 (Q1)Used the quoted rate unchanged. The single most common Poisson error and the reason for the widget.
500 (Q1)Multiplied by 250 instead of by 250/100.
2.5 (Q1)Found the number of 100 m lengths and forgot to multiply by the rate.
9 (Q2)Gave the variance. It equals λ, which is the trap: the number looks like an answer.
0.632 (Q3)1 minus the answer, so they found P(X ≥ 1).
0 (Q3)Thought zero arrivals impossible. It is the single most likely count when λ = 1.
A or C (Q4)Either missed that the test is mean against variance, or fitted a model they had just disproved. Both are worth naming out loud.

Other things they will say

"Can X be bigger than λ?" Far bigger. There is no upper limit at all, which is the structural difference from the binomial and worth repeating.

"What if the rate changes during the day?" Then Poisson is the wrong model for the whole day, and noticing that is the mark on a modelling question. A single peak hour can still be Poisson on its own.

"Why e?" It falls out of the limit of a binomial with n large and p small. Worth one sentence, not a derivation, unless they ask twice.

A possible order

 What is happening
1Stretch the window. Ask for λ at two hours. Let the wrong answer be said out loud first.
2Sorting five scenarios into Poisson or binomial.
3The formula, the calculator, and the four questions.
4A question needing two intervals added, which is where λ adding becomes useful rather than abstract.
5The mean-against-variance check on a real data set, with a written verdict.

Two things not to say

Do not write λ = 3 on the board and leave it there for the whole lesson. It is the thing that gets copied into a two-hour question.

Do not present the mean-equals-variance identity as a curiosity. It is the diagnostic, and questions award marks for applying it.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Scale lambdaMotorbike taxis arrive at 3 an hour. State λ for a two hour period.
    λ = 6. It scales with the interval.
  2. A single probabilityFor λ = 6 find P(X = 4).
    0.134

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. At least twoPotholes occur at 2 per 100 m. For a 75 m stretch, λ = 1.5. Find P(at least 2).
    1 − P(0) − P(1) = 1 − 0.2231 − 0.3347 = 0.442.
  2. Combine two sourcesTwo independent stands have λ = 2 and λ = 3 an hour. Find λ for the pair and P(X ≤ 1).
    λ = 5, and P(X ≤ 1) = 0.0404. Independent Poisson counts add.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. The mistake the page exists forA student takes λ = 3 for a two hour window because the rate is 3 an hour. Explain the error and its effect.
    λ is a count for the interval, not a rate, so it must be 6. Using 3 models half the time and makes large counts look far less likely than they are.
  2. Test the assumptionsGive two reasons a Poisson model could fail for taxi arrivals across a whole day.
    The rate is not constant, since arrivals peak at 8am, so one λ cannot describe the day. And arrivals clump rather than occurring independently, as a single arriving train delivers several passengers at once.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.