The shape of the lesson
By the end of the lesson, learners can
- Binomial to Poisson: When \(n\) is large and \(p\) is small, successes are rare events scattered among many trials, which is exactly the Poisson picture.
- Poisson to normal: As \(\lambda\) grows, the Poisson bars become symmetric and bell-shaped (Fig. 1).
- Choosing the right approximation: Check the parameters before choosing.
- Binomial, rare successes: \(n > 50\), \(np < 5\): use \(\text{Po}(np)\), no continuity correction.
- \(p\) close to 1: count the failures instead.
- Poisson, large mean: \(\lambda > 15\): use \(N(\lambda, \lambda)\), standard deviation \(\sqrt{\lambda}\).
- Continuity correction: list the integers included, then go half a unit beyond the last one.
- Justify: quote the numerical conditions in your answer.
One picture every idea returns to
If \(X \sim B(n, p)\) with \(n > 50\) and \(np < 5\), use \(X \approx \text{Po}(np)\). If \(X \sim \text{Po}(\lambda)\) with \(\lambda > 15\), use \(X \approx N(\lambda, \lambda)\): the variance equals the mean, and a continuity correction of \(0.5\) is needed because a discrete count is replaced by a continuous variable.
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, Swap a hard sum for an easy one.. 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: Binomial to Poisson | When \(n\) is large and \(p\) is small, successes are rare events scattered among many trials, which is exactly the Poisson picture. Model the worked example A rare defect on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Poisson to normal | As \(\lambda\) grows, the Poisson bars become symmetric and bell-shaped (Fig. 1). Model the worked example A lower tail on the board, one line at a time. Your turn: \(X \sim \text{Po}(22)\). Use a normal approximation to find \(\text{P}(X < 19)\). | Pairs |
| 21 to 30 min | Teach and check: Choosing the right approximation | Check the parameters before choosing. | Numbered Heads Together |
| 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 Choosing the right approximation, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| State the conditions with numbers: "\(n = 200 > 50\) and \(np = 3 < 5\)". | Writing "\(n\) is large and \(p\) is small" with no values often loses the mark, and using \(\text{Po}(np(1-p))\) is wrong: the Poisson parameter is the mean, \(np\). |
| The standard deviation of \(N(20, 20)\) is \(\sqrt{20}\), not 20. | Dividing by 20 is the commonest error in this topic, closely followed by forgetting the continuity correction altogether. |
| Do not apply a continuity correction when approximating binomial by Poisson. | Both are discrete, so \(\text{P}(X \le 2)\) stays \(\text{P}(X \le 2)\). |
Approximations: binomial to Poisson, Poisson to normal: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
State the conditions under which \(B(n, p)\) may be approximated by a Poisson distribution, and the parameter of that Poisson distribution.
\(X \sim B(400, 0.008)\). Use a suitable approximation to find (a) \(\text{P}(X = 3)\) (b) \(\text{P}(X \le 1)\).
\(X \sim \text{Po}(40)\). Use a normal approximation to find \(\text{P}(35 \le X < 45)\).
\(X \sim B(90, 0.025)\).
(a) Explain why a normal approximation is not suitable but a Poisson approximation is.
(b) Use the Poisson approximation to find \(\text{P}(X \ge 2)\).
A hospital's records show that 99% of patients do not react badly to a vaccine. 300 patients are vaccinated. Find, using a suitable approximation, the probability that more than 3 patients react badly.
Calls arrive at a call centre at random at a mean rate of 1.5 per minute.
(a) Find the probability that at most 2 calls arrive in a given minute.
(b) Use a suitable approximation to find the probability that more than 100 calls arrive in a given hour, justifying your choice.
Answer key and marking notes
- \(n\) large and \(p\) small: approximately \(n > 50\) and \(np < 5\).
- The Poisson parameter is \(\lambda = np\).
- \(n = 400 > 50\), \(np = 3.2 < 5\), so \(X \approx \text{Po}(3.2)\).
- (a) \(e^{-3.2}\dfrac{3.2^{3}}{6} = 0.223\) (3 s.f.).
- (b) \(e^{-3.2}(1 + 3.2) = 4.2e^{-3.2} = 0.171\) (3 s.f.).
- \(Y \sim N(40, 40)\). \(35 \le X < 45\) means \(35 \le X \le 44\), so \(\text{P}(34.5 < Y < 44.5)\).
- \(z_1 = \dfrac{34.5 - 40}{\sqrt{40}} = -0.870\), \(z_2 = \dfrac{44.5 - 40}{\sqrt{40}} = 0.712\).
- \(\Phi(0.712) - \Phi(-0.870) = 0.7618 - (1 - 0.8078) = 0.570\) (3 s.f.).
- (a) \(np = 2.25 < 5\), so the binomial is too skewed for a normal approximation; \(n = 90 > 50\) and \(np < 5\), so \(\text{Po}(2.25)\) is suitable.
- (b) \(\text{P}(X \ge 2) = 1 - e^{-2.25}(1 + 2.25) = 1 - 3.25 \times 0.1054 = 0.657\) (3 s.f.).
- \(R \sim B(300, 0.01)\); \(n > 50\), \(np = 3 < 5\), so \(R \approx \text{Po}(3)\).
- \(\text{P}(R > 3) = 1 - e^{-3}\left(1 + 3 + 4.5 + 4.5\right) = 1 - 13e^{-3}\).
- \(= 1 - 0.6472 = 0.353\) (3 s.f.).
- (a) \(X \sim \text{Po}(1.5)\): \(e^{-1.5}(1 + 1.5 + 1.125) = 3.625e^{-1.5} = 0.809\) (3 s.f.).
- (b) In an hour \(T \sim \text{Po}(90)\). Since \(\lambda = 90 > 15\), use \(Y \sim N(90, 90)\).
- More than 100 means 101 or more: \(\text{P}(Y > 100.5)\).
- \(z = \dfrac{100.5 - 90}{\sqrt{90}} = 1.107\), so \(1 - \Phi(1.107) = 1 - 0.8659 = 0.134\) (3 s.f.).
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.
\(X\) is the number of successes in 60 independent trials, each with probability of success 0.95. Which of these is the most suitable approximation to use?
- AThe number of failures, \(60 - X\), is approximately \(\text{Po}(3)\)
- B\(X\) is approximately \(\text{Po}(57)\)
- C\(X\) is approximately \(N(57, 2.85)\)
- DThe number of failures, \(60 - X\), is approximately \(\text{Po}(2.85)\)
0.8% of the light bulbs made in a factory are faulty. A random sample of 400 bulbs is taken. Use a suitable approximation to find the probability that at most 1 bulb in the sample is faulty.
- A\(0.171\)
- B\(0.130\)
- C\(0.380\)
- D\(0.829\)
\(X \sim \text{Po}(25)\). Use a suitable approximation to find \(\text{P}(X < 20)\).
- A\(0.136\)
- B\(0.159\)
- C\(0.184\)
- D\(0.413\)
For which of these random variables is a normal distribution whose variance equals its mean the most suitable approximation?
- A\(X \sim \text{Po}(22)\)
- B\(X \sim \text{Po}(6.5)\)
- C\(X \sim B(30, 0.5)\)
- D\(X \sim B(200, 0.01)\)
\(X \sim \text{Po}(40)\). Use a suitable approximation to find \(\text{P}(35 \le X < 45)\).
- A\(0.570\)
- B\(0.616\)
- C\(0.524\)
- D\(0.0999\)
Each day a machine makes 600 items, and 0.5% of items are defective, independently. Use a suitable approximation to find the probability that there are at least 2 defective items on each of two given days.
- A\(0.641\)
- B\(0.801\)
- C\(0.983\)
- D\(0.333\)
Requests reach a website at random at a mean rate of 3 per minute. Use a suitable approximation to find the probability of more than 200 requests in an hour.
- A\(0.0633\)
- B\(0.0680\)
- C\(0.0731\)
- D\(0.937\)
\(X \sim \text{Po}(\lambda)\), where \(\lambda\) is large. Using a normal approximation, \(\text{P}(X \le 30) = 0.0665\). Find \(\lambda\).
- A\(40.0\)
- B\(39.4\)
- C\(38.9\)
- D\(23.3\)
Quiz key
\(X \sim B(60, 0.95)\) has \(np = 57\) but \(nq = 3 < 5\), so a normal approximation is not valid, and \(\text{Po}(57)\) is wrong because the variance of \(X\) (2.85) is nowhere near its mean. Count failures instead: \(60 - X \sim B(60, 0.05)\) with \(n = 60 > 50\) and \(np = 3 < 5\), so \(60 - X \approx \text{Po}(3)\). The tempting \(\text{Po}(2.85)\) uses the variance \(npq\) as the Poisson parameter, but the parameter is the mean, \(np = 3\).
\(X \sim B(400, 0.008)\) with \(n > 50\) and \(np = 3.2 < 5\), so \(X \approx \text{Po}(3.2)\). \(\text{P}(X \le 1) = e^{-3.2}(1 + 3.2) = 4.2e^{-3.2} = 0.171\). The tempting \(0.130\) is \(\text{P}(X = 1)\) only, forgetting \(X = 0\). \(0.380\) is \(\text{P}(X \le 2)\), and \(0.829\) is \(\text{P}(X \ge 2)\), the complement.
\(Y \sim N(25, 25)\), standard deviation 5. \(X < 20\) means \(X \le 19\), so use \(\text{P}(Y < 19.5)\): \(z = \dfrac{19.5 - 25}{5} = -1.1\) and \(\Phi(-1.1) = 1 - 0.8643 = 0.1357 = 0.136\). The tempting \(0.159\) uses 20 with no continuity correction (\(z = -1\)); \(0.184\) uses 20.5, which wrongly includes 20; \(0.413\) divides by the variance 25 instead of the standard deviation 5.
A Poisson variable has variance equal to its mean, and \(\lambda = 22 > 15\), so \(\text{Po}(22) \approx N(22, 22)\). \(\text{Po}(6.5)\) has \(\lambda\) too small, so use the Poisson formula directly. The tempting \(B(30, 0.5)\) can be approximated by a normal, but by \(N(15, 7.5)\), whose variance is not its mean. \(B(200, 0.01)\) has \(np = 2 < 5\), so it is approximated by \(\text{Po}(2)\), not a normal.
\(Y \sim N(40, 40)\). \(35 \le X \le 44\) becomes \(\text{P}(34.5 < Y < 44.5)\). \(z_1 = \dfrac{34.5 - 40}{\sqrt{40}} = -0.870\), \(z_2 = \dfrac{44.5 - 40}{\sqrt{40}} = 0.712\). \(\Phi(0.712) - \Phi(-0.870) = 0.7618 - 0.1922 = 0.570\). The tempting \(0.616\) uses 45.5, wrongly including 45; \(0.524\) uses 35.5, wrongly excluding 35; \(0.0999\) divides by 40 instead of \(\sqrt{40}\).
\(n = 600 > 50\), \(np = 3 < 5\), so one day's count is approximately \(\text{Po}(3)\). \(\text{P}(X \ge 2) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} = 0.8009\). Both days: \(0.8009^{2} = 0.641\). The tempting \(0.801\) is for one day only. \(0.983\) combines the days into \(\text{Po}(6)\) and finds \(\text{P}(\ge 2)\) in total, which is a different event (it allows 4 on one day and 0 on the other). \(0.333\) is \(\text{P}(X \ge 3)\) on both days, a boundary slipped by one.
\(X \sim \text{Po}(180) \approx Y \sim N(180, 180)\). \(X > 200\) means \(X \ge 201\), so \(\text{P}(Y > 200.5)\): \(z = \dfrac{20.5}{\sqrt{180}} = 1.528\), and \(1 - \Phi(1.528) = 1 - 0.9367 = 0.0633\). The tempting \(0.0680\) uses 200 with no continuity correction; \(0.0731\) uses 199.5, which wrongly includes 200; \(0.937\) is the probability of at most 200.
\(\Phi(z) = 0.0665\) gives \(z = -1.502\). So \(\dfrac{30.5 - \lambda}{\sqrt{\lambda}} = -1.502\), i.e. \(\lambda - 1.502\sqrt{\lambda} - 30.5 = 0\). With \(u = \sqrt{\lambda}\): \(u = \dfrac{1.502 + \sqrt{1.502^{2} + 122}}{2} = 6.325\), so \(\lambda = 40.0\). The tempting \(39.4\) leaves out the continuity correction (uses 30); \(38.9\) corrects the wrong way (29.5); \(23.3\) uses \(z = +1.502\), but 30 must be below the mean when the probability is so small.
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.