The shape of the lesson
By the end of the lesson, learners can
- One curve for all: A normal distribution is fixed by its mean \(\mu\) and variance \(\sigma^2\), written \(\text{N}(\mu, \sigma^2)\).
- Using symmetry: The table only gives areas to the left of a positive \(z\).
- Working backwards: When the probability is given and a value is wanted, use the table in reverse (or the critical values table: 0.674, 0.842, 1.036, 1.282, 1.645, 1.960, 2.326, 2.576).
- Two unknowns, two equations: If neither \(\mu\) nor \(\sigma\) is known, each given probability produces one equation \(x = \mu + z\sigma\).
- Sketch: mark \(\mu\), the value(s) and the shaded area.
- Standardise: divide by \(\sigma\), not \(\sigma^2\); \(z\) to 3 d.p.
- Look up: \(\Phi(z)\) is the area to the left; use \(1 - \Phi\) and symmetry for others.
- Backwards: probability \(\to z\) from the table, with the sign from the sketch.
- Two unknowns: two equations \(x = \mu + z\sigma\), solved together.
One picture every idea returns to
If \(X \sim \text{N}(\mu, \sigma^2)\), then \(Z = \dfrac{X - \mu}{\sigma} \sim \text{N}(0, 1)\) and \(\text{P}(X < x) = \Phi\!\left(\dfrac{x - \mu}{\sigma}\right)\). Use symmetry, \(\Phi(-z) = 1 - \Phi(z)\), for negative \(z\), and always sketch the curve.
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, The bell, standardised.. Heights, masses and measurement errors are modelled by a continuous, symmetrical bell-shaped 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: One curve for all | A normal distribution is fixed by its mean \(\mu\) and variance \(\sigma^2\), written \(\text{N}(\mu, \sigma^2)\). Model the worked example Less than on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Using symmetry | The table only gives areas to the left of a positive \(z\). Model the worked example A negative z on the board, one line at a time. Your turn: \(X \sim \text{N}(120, 15^2)\). Find \(\text{P}(X > 130)\). | Pairs |
| 17 to 23 min | Teach and check: Working backwards | When the probability is given and a value is wanted, use the table in reverse (or the critical values table: 0.674, 0.842, 1.036, 1.282, 1.645, 1.960, 2.326, 2.576). Model the worked example Finding a value on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Two unknowns, two equations | If neither \(\mu\) nor \(\sigma\) is known, each given probability produces one equation \(x = \mu + z\sigma\). Model the worked example Mean and standard deviation on the board, one line at a time. | 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 Two unknowns, two equations, 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 |
|---|---|
| Dividing by the variance. \(\text{N}(50, 16)\) means \(\sigma^2 = 16\), so \(\sigma = 4\); \(\dfrac{56 - 50}{16}\) is wrong. | Return to One curve for all: ask a pair to state the correct version and show it on the lab before moving on. |
| Writing \(\Phi(0.75) - \Phi(1)\) for \(\text{P}(46 < X < 53)\). | The left end has \(z = -1\), not \(+1\); a sketch shows the area must be bigger than \(\Phi(0.75) - 0.5\). |
| Using \(z = 0.1\) or \(z = 0.8\) in the equation. | Those are probabilities; you must convert them to \(z\)-values with the table first. |
| Giving both \(z\)-values the same sign. | A value below the mean always has a negative \(z\); a quick sketch with 20 to the left of \(\mu\) and 40 to the right prevents this. |
The normal distribution: worksheet
6 questions, 23 marks. Written in the style of the papers; no past paper question is reproduced.
\(X \sim \text{N}(70, 36)\). Find (a) \(\text{P}(X < 79)\), (b) \(\text{P}(X < 64)\).
\(X \sim \text{N}(200, 12^2)\). Find \(\text{P}(185 < X < 220)\).
\(X \sim \text{N}(\mu, 8^2)\) and \(\text{P}(X > 60) = 0.25\). Find \(\mu\).
The lengths of some rods are normally distributed with mean 25 cm. 90% of the rods are shorter than 28 cm. Find the standard deviation.
The masses of bags of flour are normally distributed. 10% of bags weigh less than 495 g and 5% weigh more than 520 g. Find the mean and standard deviation of the masses.
The heights of adults in a town are modelled by \(\text{N}(165, 6^2)\) cm.
(a) Find the probability that an adult is taller than 175 cm. (b) In a sample of 400 adults, how many would you expect to be taller than 175 cm? (c) Three adults are chosen at random. Find the probability that exactly one is taller than 175 cm. (d) Find the height exceeded by 20% of adults.
Answer key and marking notes
- \(\sigma = 6\). (a) \(z = \dfrac{79 - 70}{6} = 1.5\): \(\Phi(1.5) = 0.9332\).
- (b) \(z = \dfrac{64 - 70}{6} = -1\): \(1 - \Phi(1) = 1 - 0.8413 = 0.1587\).
- \(z_1 = \dfrac{185 - 200}{12} = -1.25\), \(z_2 = \dfrac{220 - 200}{12} = 1.667\).
- \(\Phi(1.667) - [1 - \Phi(1.25)] = 0.9522 - 0.1056 = 0.8466\).
- 0.847 (3 s.f.).
- 60 is above the mean, with \(\Phi(z) = 0.75\), so \(z = 0.674\).
- \(\dfrac{60 - \mu}{8} = 0.674\).
- \(\mu = 60 - 5.392 = 54.6\) (3 s.f.).
- \(\Phi(z) = 0.9\) gives \(z = 1.282\).
- \(\dfrac{28 - 25}{\sigma} = 1.282\).
- \(\sigma = \dfrac{3}{1.282} = 2.34\) cm (3 s.f.).
- Lower tail 10%: \(z = -1.282\), so \(\mu - 1.282\sigma = 495\).
- Upper tail 5%: \(z = 1.645\), so \(\mu + 1.645\sigma = 520\).
- Subtract: \(2.927\sigma = 25\), so \(\sigma = 8.54\) g (3 s.f.).
- \(\mu = 495 + 1.282 \times 8.541 = 506\) g (3 s.f.).
- (a) \(z = \dfrac{10}{6} = 1.667\); \(1 - \Phi(1.667) = 1 - 0.9522 = 0.0478\).
- (b) \(400 \times 0.0478 = 19.1\), so about 19 adults.
- (c) Binomial with \(n = 3\), \(p = 0.0478\): \(3 \times 0.0478 \times 0.9522^2 = 0.130\) (3 s.f.).
- (d) \(\Phi(z) = 0.8\) gives \(z = 0.842\); height \(= 165 + 0.842 \times 6 = 170\) cm (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.
\(X \sim \text{N}(72, 36)\). Find \(\text{P}(X < 80)\).
- A0.909
- B0.0913
- C0.588
- D0.409
\(X \sim \text{N}(200, 15^2)\). Find \(\text{P}(X > 185)\).
- A0.841
- B0.159
- C0.527
- D0.341
\(X \sim \text{N}(60, 8^2)\). Find \(\text{P}(52 < X < 70)\).
- A0.736
- B0.0531
- C0.264
- D0.394
\(X \sim \text{N}(40, 5^2)\). Find the value of \(k\) such that \(\text{P}(X < k) = 0.05\).
- A31.8
- B48.2
- C30.2
- D39.8
\(X \sim \text{N}(50, \sigma^2)\) and \(\text{P}(X > 58) = 0.2\). Find \(\sigma\).
- A9.50
- B6.74
- C15.3
- D90.3
\(X\) is normally distributed with \(\text{P}(X < 35) = 0.3\) and \(\text{P}(X > 60) = 0.1\). Find \(\mu\) and \(\sigma\).
- A\(\mu = 42.3\), \(\sigma = 13.8\)
- B\(\mu = 17.7\), \(\sigma = 33.0\)
- C\(\mu = 41.0\), \(\sigma = 11.5\)
- D\(\mu = 44.9\), \(\sigma = 11.8\)
The lengths of 500 bolts are normally distributed with mean 30 mm and standard deviation 0.4 mm. A bolt is rejected if it is shorter than 29.5 mm or longer than 30.6 mm. How many bolts would you expect to be rejected?
- A86
- B414
- C33
- D53
\(X \sim \text{N}(100, 12^2)\). Find the value of \(a\) such that \(\text{P}(100 - a < X < 100 + a) = 0.9\).
- A19.7
- B15.4
- C23.5
- D10.8
Quiz key
\(\sigma = 6\), so \(z = \dfrac{80 - 72}{6} = 1.333\) and \(\text{P}(X < 80) = \Phi(1.333) = 0.9087 = 0.909\) (3 s.f.). The tempting 0.588 divides by the variance 36, giving \(z = 0.222\). 0.0913 is the right-hand tail \(\text{P}(X > 80)\), and 0.409 is the area between the mean and 80 only.
\(z = \dfrac{185 - 200}{15} = -1\). The area to the right of \(-1\) equals, by symmetry, the area to the left of \(+1\): \(\Phi(1) = 0.8413 = 0.841\) (3 s.f.). The tempting 0.159 is \(\Phi(-1)\), the area to the LEFT of 185. 0.527 divides by \(15^2\), and 0.341 is only the area between 185 and the mean.
\(z_1 = \dfrac{52 - 60}{8} = -1\), \(z_2 = \dfrac{70 - 60}{8} = 1.25\). \(\text{P} = \Phi(1.25) - [1 - \Phi(1)] = 0.8944 - 0.1587 = 0.7357 = 0.736\) (3 s.f.). The tempting 0.0531 is \(\Phi(1.25) - \Phi(1)\), which treats the left end as \(z = +1\). 0.264 is the probability of lying outside the interval, and 0.394 stops at the mean.
\(\Phi(z) = 0.05\) gives \(z = -1.645\) (negative, since \(k\) is below the mean). \(\dfrac{k - 40}{5} = -1.645\), so \(k = 40 - 8.225 = 31.775 = 31.8\) (3 s.f.). The tempting 48.2 uses \(+1.645\), which puts \(k\) above the mean. 30.2 uses 1.960, the value for 2.5% in each tail, and 39.8 uses the probability 0.05 as if it were a \(z\)-value.
\(\Phi(z) = 0.8\) gives \(z = 0.842\). \(\dfrac{8}{\sigma} = 0.842\), so \(\sigma = \dfrac{8}{0.842} = 9.50\) (3 s.f.). The tempting 6.74 multiplies \(8 \times 0.842\) instead of dividing. 15.3 uses \(z = 0.524\), the value for 0.7, and 90.3 is \(\sigma^2\).
\(\Phi(z) = 0.7\) gives 0.524, so 35 has \(z = -0.524\): \(\mu - 0.524\sigma = 35\). \(\Phi(z) = 0.9\) gives \(z = 1.282\): \(\mu + 1.282\sigma = 60\). Subtract: \(1.806\sigma = 25\), \(\sigma = 13.84\), and \(\mu = 35 + 0.524 \times 13.84 = 42.25\). So \(\mu = 42.3\), \(\sigma = 13.8\) (3 s.f.). The tempting \(\mu = 17.7\) gives both \(z\)-values the same sign. \(\mu = 41.0\) uses 1.645 for the 10% tail, and \(\mu = 44.9\) uses 0.842 (the value for 0.8) at 35.
\(z_1 = \dfrac{29.5 - 30}{0.4} = -1.25\), \(z_2 = \dfrac{30.6 - 30}{0.4} = 1.5\). P(accepted) \(= \Phi(1.5) - [1 - \Phi(1.25)] = 0.9332 - 0.1056 = 0.8276\). P(rejected) \(= 0.1724\), so expect \(500 \times 0.1724 = 86.2\), about 86 bolts. The tempting 414 is the number accepted. 33 counts only the bolts that are too long, and 53 only those that are too short.
Each tail holds 0.05, so \(\Phi\!\left(\dfrac{a}{12}\right) = 0.95\), giving \(\dfrac{a}{12} = 1.645\) and \(a = 19.74 = 19.7\) (3 s.f.). The tempting 15.4 uses 1.282, putting all 10% in one tail. 23.5 uses 1.960, the value for 95% in the middle, and 10.8 is \(0.9 \times 12\).
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.