The Poisson distribution

• Write the MODEL line first — “\(X\) = number of calls in 15 minutes, \(X \sim \text{Po}(1)\)” — because the interval sets \(\lambda\), and \(\lambda\) sets every number after it.
• \(\text{P}(X = r) = \dfrac{e^{-\lambda}\lambda^r}{r!}\) with \(\lambda\) for this interval; \(n > 50\) and \(np < 5\) for \(\text{Po}(np)\); \(\lambda > 15\) for \(\text{N}(\lambda, \lambda)\), with the correction.

• 6.1.1 I can use — use formulae to calculate probabilities for the distribution Po(λ)
• 6.1.2 I can use — use the fact that if X ~ Po(λ) then the mean and variance of X are each equal to λ
• 6.1.3 I can understand — understand the relevance of the Poisson distribution to the distribution of random events, and use the Poisson distribution as a model
• 6.1.4 I can use — use the Poisson distribution as an approximation to the binomial distribution where appropriate
• 6.1.5 I can use — use the normal distribution, with continuity correction, as an approximation to the Poisson distribution where appropriate

Accuracy for this chapter. Keep \(e^{-\lambda}\) to at least five significant figures through the working (0.082085, 0.13534, 0.049787) and round once, to 3 significant figures, at the end. Keep \(\sigma = \sqrt{\lambda}\) unrounded (4.4721, 5.4772), write \(z\) to 3 decimal places, and read Φ to 4 decimal places with the ADD column for the third decimal of \(z\) as written: Φ(1.006) = 0.8413 + 0.0014 = 0.8427 (row 1.0, column 0, ADD column 6). Show every term of a cumulative sum; a total from the calculator with no terms is an unsupported answer.

• “Calls arrive at 4 per hour, so the number in 15 minutes is \(\text{Po}(4)\).” Correct Scale \(\lambda\) to the interval. \(\lambda\) is the rate multiplied by the length of the interval: 15 minutes is a quarter of an hour, so \(\lambda = 1\). A different part of the same question may need a different \(\lambda\) — 2 for 30 minutes, 12 for 3 hours. Write the interval in the MODEL line every time.
• “\(\text{P}(X \ge 2) = 1 - \text{P}(X \le 2)\).” Correct The complement of \(X \ge 2\) is \(X \le 1\). \(\text{P}(X \ge 2) = 1 - \text{P}(X = 0) - \text{P}(X = 1)\). List the values on each side before you subtract: \(\{0, 1\}\) against \(\{2, 3, 4, \dots\}\).
• “\(X \sim \text{Po}(30)\), so \(X \approx \text{N}(30, \sqrt{30})\).” Correct The second parameter is the variance, \(\lambda\), not \(\sqrt{\lambda}\). \(Y \sim \text{N}(30, 30)\), and it is when you standardise that you divide by \(\sigma = \sqrt{30} = 5.4772\).
• “\(n = 200\), so \(\text{B}(200, p)\) can be approximated by a Poisson.” Correct Two conditions, both stated: \(n > 50\) and \(np < 5\), approximately. \(\text{B}(200, 0.1)\) has \(np = 20\), so the Poisson does not apply (the normal of chapter 27 does). Neither condition is in MF19.
• “\(\text{P}(X = 3) = \dfrac{e^{-\lambda}\lambda^3}{3}\).” Correct The denominator is \(3! = 6\). And keep \(e^{-\lambda}\) to at least five significant figures: rounding \(e^{-2.5} = 0.082085\) to 0.08 moves the answer in the second figure.
• “4 calls per hour, so in 15 minutes \(\lambda = 4\).” Repair \(\lambda\) scales with the interval. 15 minutes is a quarter of an hour, so \(\lambda = 4 \times \tfrac{1}{4} = 1\).
• “12 calls per 3 hours, so for the 3-hour period \(\lambda = \tfrac{12}{3} = 4\).” Repair For the 3-hour interval \(\lambda\) is 12. The rate 4 per hour belongs to a one-hour interval; dividing gave the wrong interval.
• “\(\text{P}(X \ge 2) = 1 - \text{P}(X \le 2)\).” Repair The complement of \(X \ge 2\) is \(X \le 1\): \(\text{P}(X \ge 2) = 1 - \text{P}(0) - \text{P}(1)\).
• “\(\text{P}(X = 3) = \dfrac{e^{-\lambda}\lambda^3}{3}\).” Repair The denominator is \(3! = 6\), not 3.
• “\(e^{-2.5} \approx 0.08\), so \(\text{P}(X = 3) = \dfrac{0.08 \times 15.625}{6} = 0.208\).” Repair Keep \(e^{-2.5} = 0.082085\) through the working: the answer is 0.214. Early rounding moved it in the second figure.
• “\(X \sim \text{Po}(4.2)\), so \(\text{Var}(X) = \sqrt{4.2}\).” Repair The variance is \(\lambda = 4.2\); the standard deviation is \(\sqrt{4.2} = 2.05\).
• “The mean is 5 and the variance is 12, so \(X \sim \text{Po}(5)\).” Repair A Poisson has mean = variance. With the variance more than twice the mean the model is not appropriate — say so, and suggest why (events clustering, so not independent or not singly).
• “Events happen at 3 per hour, therefore they are Poisson.” Repair A rate alone is not enough. The events must occur randomly, independently, singly and at a constant average rate; state the conditions in the context of the question.
• “\(\text{B}(30, 0.1) \approx \text{Po}(3)\).” Repair \(n = 30\) is not more than 50. The condition is \(n > 50\) and \(np < 5\); use the binomial formula.
• “\(\text{B}(1000, 0.02) \approx \text{Po}(20)\).” Repair \(np = 20\) is not less than 5. Use the normal approximation \(\text{N}(20, 19.6)\) of chapter 27 instead.
• “\(\text{Po}(30) \approx \text{N}(30, \sqrt{30})\).” Repair The second parameter is the variance: \(\text{N}(30, 30)\), and \(\sigma = \sqrt{30}\) is used when standardising.
• “For \(\text{Po}(30)\), \(\text{P}(X > 35) \approx \text{P}(Y > 34.5)\).” Repair \(X > 35\) means \(X \ge 36\), so the boundary is 35.5: \(\text{P}(Y > 35.5)\). Only \(X \ge 35\) gives 34.5.
• “\(X \sim \text{Po}(8)\); use \(\text{N}(8, 8)\) to find \(\text{P}(X \le 5)\).” Repair \(\lambda = 8\) is not more than 15. Add the Poisson terms directly.
• “\(e^{-\lambda} = 0.2\), so \(\lambda = 0.2\).” Repair Take logarithms: \(\lambda = -\ln 0.2 = \ln 5 = 1.61\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 29 adds two independent Poisson variables: when you reach it, redo worked example 2 and notice that its MODEL lines already scale \(\lambda\). Chapter 32 tests a hypothesis about \(\lambda\): when you reach it, redo worked example 4, whose complement \(1 - \text{P}(X \le 2)\) is a tail probability of exactly the kind a test uses. Recalling a method inside a new problem is worth more than another pass over this chapter on its own, and Paper 6 assumes everything before it, so nothing here is ever finished with.

• Chapter 28 · Probability & Statistics 2 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 2 content, examined in Paper 6. Paper 6 (Probability & Statistics 2) is offered only as part of the A Level, where it is 20%. It assumes the whole of the Paper 5 content and the calculus of Paper 3. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question on this section can set up a context (calls, flaws, orders) and ask for the conditions of the model, then probabilities over two or three different intervals; give a frequency table and ask whether a Poisson model is plausible; name an unknown mean through a given probability and ask you to find it; or ask for “a suitable approximation”, where the choice and its justification are part of the answer.
• MF19 prints the Poisson probability, the mean and the variance, and the normal table. It does not print the four conditions, the scaling of the mean to the interval, the approximation conditions (n more than 50 and np less than 5; mean more than 15) or the continuity correction. Those are in the MF19 card.
• An unscaled mean; a complement taken on the wrong side; a cumulative total with no terms shown (an unsupported calculator answer earns nothing); a variance written where a standard deviation belongs; an approximation used without its conditions; a missing or backwards continuity correction; and early rounding of e−λ. Answers to 3 significant figures, with the terms kept to five.