The shape of the lesson
By the end of the lesson, learners can
- When the bars look normal: A binomial distribution is roughly bell-shaped when \(n\) is large and \(p\) is not too close to 0 or 1.
- The continuity correction: A binomial variable takes whole numbers only; a normal variable is continuous.
- Between, and exactly: Strict and non-strict inequalities at each end are corrected separately.
- In context: In a word problem, define \(X\) and its binomial distribution first, check \(np > 5\) and \(nq > 5\), then state the normal approximation and its continuity-corrected inequality before standardising.
- Define: \(X \sim \text{B}(n, p)\) in context.
- Check: \(np > 5\) and \(nq > 5\).
- Approximate: \(Y \sim \text{N}(np, npq)\); \(\sigma = \sqrt{npq}\).
- Correct: each whole number \(r\) covers \(r - 0.5\) to \(r + 0.5\); decide whether each boundary is included.
- Standardise: \(z = \dfrac{y - np}{\sqrt{npq}}\), then use \(\Phi\) and symmetry.
One picture every idea returns to
If \(X \sim \text{B}(n, p)\) with \(np > 5\) and \(nq > 5\) (where \(q = 1 - p\)), then approximately \(X \sim \text{N}(np, npq)\). Each whole number \(r\) is represented by the interval from \(r - 0.5\) to \(r + 0.5\): this is the continuity correction.
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, Bars become a curve.. When \(n\) is large, adding dozens of binomial terms by hand is hopeless, but the bars of the distribution trace out a normal curve. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: When the bars look normal | A binomial distribution is roughly bell-shaped when \(n\) is large and \(p\) is not too close to 0 or 1. Model the worked example Setting up the approximation on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: The continuity correction | A binomial variable takes whole numbers only; a normal variable is continuous. Model the worked example At most on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Between, and exactly | Strict and non-strict inequalities at each end are corrected separately. Model the worked example Between on the board, one line at a time. Your turn: \(X \sim \text{B}(80, 0.35)\). Use a normal approximation to find \(\text{P}(25 \le X \le 30)\). | Pairs |
| 23 to 30 min | Teach and check: In context | In a word problem, define \(X\) and its binomial distribution first, check \(np > 5\) and \(nq > 5\), then state the normal approximation and its continuity-corrected inequality before standardising. Model the worked example Late flights on the board, one line at a time. | Round Table |
| 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 In context, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Using \(\sqrt{np}\) or \(npq\) as the standard deviation. | The variance is \(npq = 14.4\), so \(\sigma = \sqrt{14.4} = 3.795\). |
| Correcting in the wrong direction. | Ask 'is the boundary value included?' For \(X \ge 30\) it is, so you go DOWN to 29.5 to take in its whole bar. |
| Answering 0 for \(\text{P}(X = 25)\). | That is true for a continuous variable, but \(X\) is binomial; the continuity correction turns 25 into the interval 24.5 to 25.5. |
| Treating 'more than 25' as \(\text{P}(Y > 24.5)\). 25 itself is not included, so the boundary moves up to 25.5. | Return to In context: ask a pair to state the correct version and show it on the lab before moving on. |
The normal approximation to the binomial: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
\(X \sim \text{B}(80, 0.35)\). Show that a normal approximation is appropriate and state the approximating distribution.
\(X \sim \text{B}(80, 0.35)\). Use a normal approximation to find \(\text{P}(X > 32)\).
\(X \sim \text{B}(80, 0.35)\). Use a normal approximation to find \(\text{P}(25 \le X \le 30)\).
A fair die is thrown 120 times. Use a suitable approximation to find the probability that at least 25 sixes are obtained.
30% of customers prefer brand A. A random sample of 200 customers is taken. Use a suitable approximation to find the probability that between 50 and 70 customers inclusive prefer brand A.
An airline sells 110 tickets for a flight with 104 seats. Each ticket holder turns up independently with probability 0.9.
(a) Use a suitable approximation to find the probability that more passengers turn up than there are seats. (b) The airline does this on 5 separate flights. Find the probability that at least one of the flights has more passengers than seats.
Answer key and marking notes
- \(np = 28 > 5\), \(nq = 52 > 5\).
- \(\text{N}(28, 18.2)\), since \(npq = 80 \times 0.35 \times 0.65 = 18.2\).
- \(X > 32\) means 33 or more: \(\text{P}(Y > 32.5)\), \(Y \sim \text{N}(28, 18.2)\).
- \(z = \dfrac{32.5 - 28}{\sqrt{18.2}} = \dfrac{4.5}{4.266} = 1.055\).
- \(1 - \Phi(1.055) = 1 - 0.8543 = 0.146\) (3 s.f.).
- \(\text{P}(24.5 < Y < 30.5)\).
- \(z_1 = \dfrac{24.5 - 28}{4.266} = -0.820\), \(z_2 = \dfrac{30.5 - 28}{4.266} = 0.586\).
- \(\Phi(0.586) - [1 - \Phi(0.820)] = 0.7211 - 0.2061 = 0.515\) (3 s.f.).
- \(X \sim \text{B}(120, \tfrac{1}{6})\): \(np = 20\), \(nq = 100\), both over 5.
- \(Y \sim \text{N}\left(20, \tfrac{50}{3}\right)\), \(\sigma = 4.082\).
- \(\text{P}(Y > 24.5)\): \(z = \dfrac{4.5}{4.082} = 1.102\).
- \(1 - \Phi(1.102) = 1 - 0.8648 = 0.135\) (3 s.f.).
- \(X \sim \text{B}(200, 0.3)\): \(np = 60\), \(nq = 140\). \(Y \sim \text{N}(60, 42)\), \(\sigma = 6.481\).
- \(\text{P}(49.5 < Y < 70.5)\): \(z = \pm\dfrac{10.5}{6.481} = \pm 1.620\).
- \(2\Phi(1.620) - 1 = 2(0.9474) - 1 = 0.895\) (3 s.f.).
- (a) \(X \sim \text{B}(110, 0.9)\): \(np = 99\), \(nq = 11\), both over 5. \(Y \sim \text{N}(99, 9.9)\), \(\sigma = 3.146\).
- More than 104 means 105 or more: \(\text{P}(Y > 104.5)\), \(z = \dfrac{5.5}{3.146} = 1.748\).
- \(1 - \Phi(1.748) = 1 - 0.9598 = 0.0402\) (3 s.f.).
- (b) \(1 - (1 - 0.0402)^5 = 1 - 0.9598^5 = 0.186\) (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.
Which of these binomial distributions can suitably be approximated by a normal distribution?
- A\(\text{B}(40, 0.15)\)
- B\(\text{B}(100, 0.04)\)
- C\(\text{B}(60, 0.93)\)
- D\(\text{B}(9, 0.5)\)
\(X \sim \text{B}(75, 0.36)\). Which normal distribution should be used to approximate \(X\)?
- A\(\text{N}(27, 17.28)\)
- B\(\text{N}(27, 4.157)\)
- C\(\text{N}(27, 27)\)
- D\(\text{N}(48, 17.28)\)
\(X \sim \text{B}(n, p)\) is approximated by \(Y \sim \text{N}(np, npq)\). With a continuity correction, \(\text{P}(15 \le X < 22)\) becomes:
- A\(\text{P}(14.5 < Y < 21.5)\)
- B\(\text{P}(14.5 < Y < 22.5)\)
- C\(\text{P}(15.5 < Y < 21.5)\)
- D\(\text{P}(15 < Y < 22)\)
\(X \sim \text{B}(50, 0.4)\). Use a suitable approximation to find \(\text{P}(X \le 16)\).
- A0.156
- B0.124
- C0.0970
- D0.385
\(X \sim \text{B}(120, 0.25)\). Use a suitable approximation to find \(\text{P}(X \ge 35)\).
- A0.171
- B0.146
- C0.123
- D0.829
\(X \sim \text{B}(90, 0.3)\). Use a suitable approximation to find \(\text{P}(24 < X \le 30)\).
- A0.507
- B0.579
- C0.435
- D0.510
\(X \sim \text{B}(100, 0.45)\). Use a normal approximation to estimate \(\text{P}(X = 45)\).
- A0.0804
- B0
- C0.0402
- D0.540
An airline finds that 8% of passengers who buy tickets do not turn up, independently. It sells 220 tickets for a flight with 205 seats. Use a suitable approximation to find the probability that there is a seat for every passenger who turns up.
- A0.779
- B0.699
- C0.221
- D0.741
Quiz key
\(\text{B}(40, 0.15)\): \(np = 6\) and \(nq = 34\), both greater than 5. \(\text{B}(100, 0.04)\) has \(np = 4\), so the tempting idea that a large \(n\) is enough fails. \(\text{B}(60, 0.93)\) has \(nq = 4.2\), and \(\text{B}(9, 0.5)\) has \(np = nq = 4.5\).
Mean \(np = 75 \times 0.36 = 27\); variance \(npq = 75 \times 0.36 \times 0.64 = 17.28\). So \(\text{N}(27, 17.28)\). The tempting \(\text{N}(27, 4.157)\) puts the standard deviation \(\sqrt{17.28}\) where the variance belongs. \(\text{N}(27, 27)\) uses \(np\) as the variance, and 48 is \(nq\), the mean number of failures.
The whole numbers included are 15 to 21. The bar for 15 starts at 14.5 and the bar for 21 ends at 21.5, so \(\text{P}(14.5 < Y < 21.5)\). The tempting 22.5 includes the bar for 22, although \(X < 22\) excludes it. 15.5 leaves out 15, and the last option makes no continuity correction.
\(np = 20\), \(nq = 30\), both over 5. \(Y \sim \text{N}(20, 12)\), \(\sigma = 3.464\). \(\text{P}(Y < 16.5)\): \(z = \dfrac{16.5 - 20}{3.464} = -1.010\), so \(1 - \Phi(1.010) = 1 - 0.8438 = 0.156\) (3 s.f.). The tempting 0.124 has no continuity correction (16). 0.0970 corrects to 15.5, which leaves out \(X = 16\), and 0.385 divides by the variance 12.
\(Y \sim \text{N}(30, 22.5)\), \(\sigma = 4.743\). \(\text{P}(Y > 34.5)\): \(z = \dfrac{34.5 - 30}{4.743} = 0.949\), so \(1 - \Phi(0.949) = 1 - 0.8287 = 0.171\) (3 s.f.). The tempting 0.146 has no correction. 0.123 corrects to 35.5, which leaves out \(X = 35\), and 0.829 is the left-hand area.
\(Y \sim \text{N}(27, 18.9)\), \(\sigma = 4.347\). \(\text{P}(24.5 < Y < 30.5)\): \(z_1 = -0.575\), \(z_2 = 0.805\). \(\Phi(0.805) - [1 - \Phi(0.575)] = 0.7896 - 0.2826 = 0.507\) (3 s.f.). The tempting 0.579 starts at 23.5 and so wrongly includes 24. 0.435 stops at 29.5 and leaves out 30, and 0.510 makes no correction at either end.
\(Y \sim \text{N}(45, 24.75)\), \(\sigma = 4.975\). \(\text{P}(44.5 < Y < 45.5)\): \(z = \pm 0.101\), so \(2\Phi(0.101) - 1 = 2(0.5402) - 1 = 0.0804\) (3 s.f.), close to the exact binomial value 0.0800. The tempting 0 treats \(X\) as continuous, but \(X\) is binomial. 0.0402 uses only half the bar, from 45 to 45.5, and 0.540 is \(\text{P}(Y < 45.5)\).
\(X \sim \text{B}(220, 0.08)\): \(np = 17.6\), \(nq = 202.4\), both over 5. \(Y \sim \text{N}(17.6, 16.192)\), \(\sigma = 4.024\). At most 205 turn up means \(X \ge 15\), so \(\text{P}(Y > 14.5)\): \(z = -0.770\), giving \(\Phi(0.770) = 0.7794 = 0.779\) (3 s.f.). The tempting 0.699 corrects to 15.5, which leaves out \(X = 15\) (exactly 205 passengers, who all get seats). 0.221 is the probability that someone is left without a seat, and 0.741 makes no continuity correction.
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.