The shape of the lesson
By the end of the lesson, learners can
- When the Poisson model fits: The Poisson distribution counts events in a fixed interval of time, length, area or volume.
- The formula and the mean rate: For \(X \sim \text{Po}(\lambda)\), \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\).
- Working backwards and two-stage problems: Some questions give a probability and ask for \(\lambda\); others use a Poisson probability as the \(p\) in a binomial over several intervals.
- Define: "Let \(X\) be the number of ... in ..." and write \(X \sim \text{Po}(\lambda)\).
- Scale: multiply the rate by the length of the interval in the question.
- Boundaries: write the event as an inequality (\(X \ge 3\), \(X \le 2\)) before calculating.
- Complements: \(\text{P}(X \ge k) = 1 - \text{P}(X \le k - 1)\).
- Context: state conditions and conclusions in terms of the situation.
- Accuracy: keep full calculator values; give answers to 3 s.f.
One picture every idea returns to
If events occur singly, independently and at random at a constant mean rate, the number \(X\) in a fixed interval satisfies \(X \sim \text{Po}(\lambda)\) with \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\) for \(r = 0, 1, 2, \ldots\). The mean and the variance are both \(\lambda\), and \(\lambda\) scales in proportion to the length of the interval.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Random events, counted.. Paper 6. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: When the Poisson model fits | The Poisson distribution counts events in a fixed interval of time, length, area or volume. Model the worked example Is a Poisson model suitable? on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The formula and the mean rate | For \(X \sim \text{Po}(\lambda)\), \(\text{P}(X = r) = e^{-\lambda}\dfrac{\lambda^{r}}{r!}\). Model the worked example Single values and cumulative probabilities on the board, one line at a time. Your turn: \(X \sim \text{Po}(2.5)\). Find \(\text{P}(X = 3)\). | Pairs |
| 21 to 30 min | Teach and check: Working backwards and two-stage problems | Some questions give a probability and ask for \(\lambda\); others use a Poisson probability as the \(p\) in a binomial over several intervals. Model the worked example Finding the mean from a probability on the board, one line at a time. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 23 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Working backwards and two-stage problems, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Do not just recite "independent and random". | Cambridge wants the condition in context: "the calls arrive independently of each other" or "the mean rate of calls is constant during the day". A condition with no reference to the situation usually earns nothing. |
| "More than 2" means \(X \ge 3\), so it is \(1 - \text{P}(X \le 2)\), not \(1 - \text{P}(X \le 3)\). | Write the inequality in symbols before you calculate; most lost marks here are a boundary slipped by one. |
| In a two-stage question, do not add the rates to \(\lambda = 6\) for the week and find \(\text{P}(X = 0)\): that is the probability of no faults all week, a different event. | Identify what is being counted at each stage. |
The Poisson distribution: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
Customers arrive at a bank. State two conditions needed for the number of customers arriving in a 5-minute period to be modelled by a Poisson distribution, in context.
\(X \sim \text{Po}(3.5)\). Find (a) \(\text{P}(X = 4)\) (b) \(\text{P}(2 \le X \le 4)\).
Emails arrive in an inbox at random at a mean rate of 9 per hour. Find the probability that fewer than 3 emails arrive in a 10-minute period.
The numbers of typing errors on 100 pages were counted.
| Errors | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Pages | 14 | 27 | 27 | 18 | 9 | 5 |
Calculate the mean and variance of the number of errors per page, and explain whether a Poisson model is suitable.
\(X \sim \text{Po}(\lambda)\) and \(\text{P}(X = 2) = \text{P}(X = 3)\).
(a) Show that \(\lambda = 3\). (b) Hence find \(\text{P}(X \ge 2)\).
Requests to hire a car from a small firm occur at random at a mean rate of 2.4 per day. The firm has 4 cars.
(a) Find the probability that, on a given day, the demand for cars cannot be met.
(b) Find the probability that the demand cannot be met on at least one day of a 5-day week.
(c) Find the least number of cars the firm should have so that the probability that demand cannot be met on a given day is less than 0.05.
Answer key and marking notes
- Customers arrive independently of each other (not in groups).
- Customers arrive at a constant mean rate during the period considered (and singly, at random).
- (a) \(e^{-3.5}\dfrac{3.5^{4}}{24} = 0.030\,197 \times 6.2526 = 0.189\).
- (b) \(\text{P}(2) + \text{P}(3) + \text{P}(4) = e^{-3.5}\left(\dfrac{3.5^{2}}{2} + \dfrac{3.5^{3}}{6} + \dfrac{3.5^{4}}{24}\right)\)
- \(= 0.030\,197 \times (6.125 + 7.1458 + 6.2526) = 0.030\,197 \times 19.523 = 0.590\).
- \(\lambda = 9 \times \dfrac{10}{60} = 1.5\), so \(X \sim \text{Po}(1.5)\).
- \(\text{P}(X < 3) = \text{P}(X \le 2) = e^{-1.5}\left(1 + 1.5 + \dfrac{1.5^{2}}{2}\right) = 3.625e^{-1.5}\).
- \(= 0.809\) (3 s.f.).
- \(\sum fx = 0 + 27 + 54 + 54 + 36 + 25 = 196\), so the mean is \(\dfrac{196}{100} = 1.96\).
- \(\sum fx^{2} = 0 + 27 + 108 + 162 + 144 + 125 = 566\), so the variance is \(\dfrac{566}{100} - 1.96^{2} = 5.66 - 3.8416 = 1.82\).
- The mean and variance are close, which supports a Poisson model with \(\lambda \approx 1.96\), provided errors occur independently and at random.
- (a) \(e^{-\lambda}\dfrac{\lambda^{2}}{2} = e^{-\lambda}\dfrac{\lambda^{3}}{6}\); dividing by \(e^{-\lambda}\lambda^{2}\) gives \(\dfrac{1}{2} = \dfrac{\lambda}{6}\), so \(\lambda = 3\).
- (b) \(\text{P}(X \ge 2) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} = 1 - 0.1991 = 0.801\).
- (a) \(X \sim \text{Po}(2.4)\). Demand is not met when \(X > 4\): \(\text{P}(X \ge 5) = 1 - \text{P}(X \le 4)\).
- \(\text{P}(X \le 4) = e^{-2.4}\left(1 + 2.4 + 2.88 + 2.304 + 1.3824\right) = 9.9664e^{-2.4} = 0.9041\), so \(\text{P}(X > 4) = 0.0959\).
- (b) \(1 - (1 - 0.0959)^{5} = 1 - 0.9041^{5} = 1 - 0.6042 = 0.396\).
- (c) With \(n\) cars we need \(\text{P}(X > n) < 0.05\). \(n = 4\): \(0.0959\) (too big). \(n = 5\): \(\text{P}(X > 5) = 0.0357 < 0.05\).
- The firm needs at least 5 cars.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
For which of these would a Poisson distribution be LEAST suitable as a model?
- AThe number of buses arriving at a stop in an hour, where the buses run to a fixed timetable
- BThe number of misprints on a page of a long novel
- CThe number of radioactive particles reaching a detector in one second
- DThe number of calls to a helpline in 5 minutes during a steady mid-morning period
\(X \sim \text{Po}(4.2)\). Find \(\text{P}(X = 3)\), correct to 3 significant figures.
- A\(0.185\)
- B\(0.395\)
- C\(0.194\)
- D\(0.790\)
Customers enter a shop at random at a mean rate of 1.5 per hour. Find the probability that at least one customer enters in a 2-hour period.
- A\(0.950\)
- B\(0.777\)
- C\(0.0498\)
- D\(0.149\)
Messages arrive on a phone at random at a mean rate of 8 per hour. Find the probability that fewer than 3 messages arrive in a 15-minute period.
- A\(0.677\)
- B\(0.857\)
- C\(0.271\)
- D\(0.0138\)
\(X\) has a Poisson distribution with standard deviation 1.8. Find \(\text{P}(X \le 1)\).
- A\(0.166\)
- B\(0.463\)
- C\(0.612\)
- D\(0.127\)
\(X \sim \text{Po}(\lambda)\) and \(\text{P}(X = 4) = 2\,\text{P}(X = 3)\). Find \(\text{P}(X = 2)\).
- A\(0.0107\)
- B\(0.0446\)
- C\(0.271\)
- D\(0.0138\)
Emails arrive in an inbox at random at a mean rate of 0.9 per 10 minutes. An hour is split into six 10-minute periods. Find the probability that at least 5 of the six periods contain at least one email.
- A\(0.223\)
- B\(0.0736\)
- C\(0.180\)
- D\(0.0441\)
Calls reach a call centre at random at a mean rate of 0.6 per minute. Find the least whole number of minutes \(t\) for which the probability of at least one call in \(t\) minutes is greater than 0.99.
- A\(8\)
- B\(7\)
- C\(4\)
- D\(1\)
Quiz key
Timetabled buses arrive at planned, regularly spaced times, not at random, so the counts would vary far less than a Poisson model predicts (the variance would be well below the mean). Misprints, radioactive particles and calls in a steady period all occur at random and independently at a roughly constant rate, so they are classic Poisson situations. The tempting helpline option would only fail if the rate changed during the period, but the stem says the period is steady.
\(\text{P}(X = 3) = e^{-4.2}\dfrac{4.2^{3}}{3!} = 0.014\,996 \times 12.348 = 0.185\). The tempting \(0.395\) is \(\text{P}(X \le 3)\), a cumulative probability, not a single value. \(0.194\) is \(\text{P}(X = 4)\) and \(0.790\) is \(\text{P}(X \ge 3)\).
\(\lambda = 1.5 \times 2 = 3\), so \(X \sim \text{Po}(3)\) and \(\text{P}(X \ge 1) = 1 - e^{-3} = 1 - 0.0498 = 0.950\). The tempting \(0.777\) is \(1 - e^{-1.5}\), which forgets to scale the rate to the 2-hour period. \(0.0498\) is \(\text{P}(X = 0)\), the complement, and \(0.149\) is \(\text{P}(X = 1)\) only.
\(\lambda = 8 \times \dfrac{15}{60} = 2\). \(\text{P}(X < 3) = \text{P}(X \le 2) = e^{-2}\left(1 + 2 + \dfrac{2^{2}}{2}\right) = 5e^{-2} = 0.677\). The tempting \(0.857\) is \(\text{P}(X \le 3)\), which includes 3 when the question says fewer than 3. \(0.271\) is \(\text{P}(X = 2)\) only, and \(0.0138\) uses \(\lambda = 8\) without scaling to 15 minutes.
Variance \(= 1.8^{2} = 3.24\), so \(\lambda = 3.24\). \(\text{P}(X \le 1) = e^{-3.24}(1 + 3.24) = 4.24e^{-3.24} = 0.166\). The tempting \(0.463\) takes \(\lambda = 1.8\), treating the standard deviation as the mean. \(0.612\) uses \(\lambda = \sqrt{1.8}\), and \(0.127\) is \(\text{P}(X = 1)\) only.
\(e^{-\lambda}\dfrac{\lambda^{4}}{24} = 2e^{-\lambda}\dfrac{\lambda^{3}}{6}\) gives \(\dfrac{\lambda}{24} = \dfrac{2}{6}\), so \(\lambda = 8\). Then \(\text{P}(X = 2) = e^{-8}\dfrac{8^{2}}{2} = 32e^{-8} = 0.0107\). The tempting \(0.0446\) comes from \(\dfrac{\lambda}{3} = 2\), i.e. \(\lambda = 6\), a slip in the factorials (\(\dfrac{4!}{3!} = 4\), not 3). \(0.271\) comes from turning the ratio upside down (\(\dfrac{4}{\lambda} = 2\), \(\lambda = 2\)), and \(0.0138\) is \(\text{P}(X \le 2)\) with \(\lambda = 8\).
\(p = 1 - e^{-0.9} = 0.5934\). The number of periods with an email is \(B(6, 0.5934)\). \(\text{P}(\ge 5) = 6p^{5}(1 - p) + p^{6} = 0.1798 + 0.0437 = 0.223\). The tempting \(0.0736\) leaves out the factor \(\binom{6}{5} = 6\). \(0.180\) is exactly 5 periods only, and \(0.0441\) uses \(e^{-0.9}\) (no emails) as the success probability.
\(1 - e^{-0.6t} > 0.99\) gives \(e^{-0.6t} < 0.01\), so \(0.6t > \ln 100 = 4.605\) and \(t > 7.68\). The least whole number is 8. The tempting 7 rounds 7.68 down, but at \(t = 7\) the probability is \(1 - e^{-4.2} = 0.985\), not enough. 4 comes from using \(\log_{10} 100 = 2\) instead of \(\ln 100\), and 1 comes from solving \(e^{-0.6t} < 0.99\), mixing up the event and its complement.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.