The shape of the lesson
By the end of the lesson, learners can
- Populations and random samples: The population is every item of interest; a sample is the part you actually observe.
- The sample mean is a random variable: Each new random sample gives a new value of \(\bar{X}\).
- The Central Limit Theorem: Even when the population is not normal (a Poisson count, a skewed waiting time), the mean of a LARGE random sample is approximately normal: \(\bar{X} \approx N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\).
- Random sampling: number the population, select with random numbers, ignore repeats and out-of-range values.
- Criticise in context: who is left out, and which way would that bias the result?
- Sample mean: \(\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\), sd \(\dfrac{\sigma}{\sqrt{n}}\).
- Shape: exact if \(X\) is normal; by the CLT if \(n\) is large; say which.
- Finding \(n\): set up an inequality in \(\sqrt{n}\) and round UP.
One picture every idea returns to
A random sample gives every member of the population an equal chance of selection, so it avoids bias. The sample mean \(\bar{X}\) of \(n\) independent observations has \(\text{E}(\bar{X}) = \mu\) and \(\text{Var}(\bar{X}) = \dfrac{\sigma^{2}}{n}\). If \(X\) is normal, \(\bar{X}\) is exactly normal; if \(n\) is large, the Central Limit Theorem says \(\bar{X}\) is approximately normal anyway.
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, One sample, many possible means.. 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: Populations and random samples | The population is every item of interest; a sample is the part you actually observe. Model the worked example Choosing with random numbers on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The sample mean is a random variable | Each new random sample gives a new value of \(\bar{X}\). Model the worked example Probability for a sample mean on the board, one line at a time. Your turn: \(X \sim N(120, 20^{2})\). A random sample of 16 is taken. Find \(\text{P}(115 < \bar{X} < 125)\). | Pairs |
| 21 to 30 min | Teach and check: The Central Limit Theorem | Even when the population is not normal (a Poisson count, a skewed waiting time), the mean of a LARGE random sample is approximately normal: \(\bar{X} \approx N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\). Model the worked example Mean of Poisson counts 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, 18 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 The Central Limit Theorem, 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 |
|---|---|
| "It is too small" or "it is not random" alone rarely scores. | Say WHO is excluded and HOW that would push the answer up or down. |
| \(\text{Var}(\bar{X}) = \dfrac{\sigma^{2}}{n}\), so the standard deviation is \(\dfrac{\sigma}{\sqrt{n}}\), not \(\dfrac{\sigma}{n}\). | With \(\sigma = 12\) and \(n = 9\) that is 4, not \(1.33\). |
| Use the CLT only when the population is NOT normal and \(n\) is large, and say so. | If the population is normal, \(\bar{X}\) is exactly normal and the CLT is not needed; if the population is not normal and \(n\) is small, neither applies. |
Populations, samples and the distribution of the sample mean: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
Explain the difference between a population and a sample, and state why a sample should be chosen randomly.
A sports club has 800 members. Describe how random numbers could be used to choose a random sample of 20 members.
To estimate the proportion of a city's adults who use public transport, a student questions people waiting at the main bus station. Give two reasons why this sample is unsatisfactory.
\(X \sim N(120, 20^{2})\). A random sample of 16 values is taken. Find \(\text{P}(115 < \bar{X} < 125)\).
\(X \sim N(\mu, 25)\). Find the smallest sample size \(n\) such that \(\text{P}(\bar{X} - \mu > 1) < 0.01\).
The masses of grapes in punnets have mean 252 g and standard deviation 6 g. The distribution of masses is not normal.
(a) A random sample of 40 punnets is taken. Explain why the sample mean mass is approximately normally distributed.
(b) Find the probability that the sample mean is less than 250 g.
(c) Explain why the same method could not be used for a sample of 4 punnets.
Answer key and marking notes
- The population is every member of the group being studied; a sample is a selection of members from it.
- Random selection gives every member an equal chance of being chosen, so the sample is unbiased and likely to be representative.
- List the members and number them 001 to 800.
- Generate random 3-digit numbers, ignoring 000, numbers above 800 and repeats, until 20 are found; choose those members.
- People who never use buses (or who use only trains) cannot be chosen, so the sample is not random.
- Everyone sampled is using public transport at that moment, so the proportion will be overestimated; also only people there at that time of day are included.
- \(\bar{X} \sim N\!\left(120, \dfrac{400}{16}\right) = N(120, 25)\), sd 5.
- \(z = \dfrac{\pm 5}{5} = \pm 1\).
- \(\Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.683\) (3 s.f.).
- \(\bar{X} - \mu \sim N\!\left(0, \dfrac{25}{n}\right)\).
- Need \(\dfrac{1}{5/\sqrt{n}} > 2.326\) (the \(z\) value with upper tail 0.01).
- \(\sqrt{n} > 11.63\), so \(n > 135.3\).
- Smallest \(n = 136\).
- (a) The sample is large (\(n = 40\)), so by the Central Limit Theorem \(\bar{X}\) is approximately normal, even though the population is not.
- (b) \(\bar{X} \approx N\!\left(252, \dfrac{36}{40}\right)\), sd \(0.9487\). \(z = \dfrac{250 - 252}{0.9487} = -2.108\).
- \(\text{P}(\bar{X} < 250) = 1 - 0.9825 = 0.0175\) (3 s.f.).
- (c) With \(n = 4\) the sample is small and the population is not normal, so the CLT does not apply and \(\bar{X}\) need not be normal.
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.
A sports club has 800 members. Which method gives a random sample of 40 members?
- ANumber the members 001 to 800 and use random numbers to choose 40, ignoring repeats and numbers above 800
- BChoose the first 40 members on the alphabetical membership list
- CAsk the first 40 members who arrive at the club on a Monday evening
- DSend a message to all members and use the first 40 who volunteer
\(X \sim N(60, 10^{2})\). A random sample of 16 values of \(X\) is taken. Find the variance of the sample mean \(\bar{X}\).
- A\(6.25\)
- B\(2.5\)
- C\(100\)
- D\(0.625\)
\(X \sim N(120, 15^{2})\). A random sample of 20 values of \(X\) is taken. Find \(\text{P}(\bar{X} > 125)\).
- A\(0.0680\)
- B\(0.370\)
- C\(0.329\)
- D\(0.932\)
A random variable \(X\) has mean 4 and variance 9, but its distribution is NOT normal. For which of these is it valid to say that \(\bar{X}\) is approximately \(N\!\left(4, \dfrac{9}{n}\right)\)?
- AA random sample of size \(n = 60\), by the Central Limit Theorem
- BA random sample of size \(n = 5\), by the Central Limit Theorem
- CAny sample size, because a sample mean always has a normal distribution
- DNo sample size, because the population is not normal
The number of customers entering a shop per minute has distribution \(\text{Po}(3.6)\). A random sample of 45 one-minute periods is taken. Find, approximately, the probability that the mean number of customers per minute is less than 3.3.
- A\(0.144\)
- B\(0.437\)
- C\(0.288\)
- D\(0.856\)
\(X \sim N(\mu, 8^{2})\). Find the smallest sample size \(n\) for which \(\text{P}(|\bar{X} - \mu| < 2) > 0.99\).
- A\(107\)
- B\(106\)
- C\(62\)
- D\(87\)
A random variable, not normally distributed, has mean 1.8 and variance 0.9. Find, approximately, the probability that the total of 64 independent values exceeds 120.
- A\(0.264\)
- B\(0.468\)
- C\(0.252\)
- D\(0.736\)
\(X \sim N(\mu, 5^{2})\). For a random sample of 25 values, \(\text{P}(\bar{X} > 52) = 0.0228\). Find \(\mu\).
- A\(50\)
- B\(42\)
- C\(54\)
- D\(51.6\)
Quiz key
Numbering every member and choosing with random numbers gives each member the same chance of selection, so it is random. The tempting alphabetical list gives members with surnames late in the alphabet no chance at all. Monday-evening arrivals leave out members who never come on Mondays, and volunteers are self-selected (likely the most keen members).
\(\text{Var}(\bar{X}) = \dfrac{10^{2}}{16} = 6.25\). The tempting 2.5 is the standard deviation of \(\bar{X}\), \(\dfrac{10}{\sqrt{16}}\), not the variance. 100 is the variance of a single value, and 0.625 is \(\dfrac{\sigma}{n} = \dfrac{10}{16}\).
\(z = \dfrac{125 - 120}{15/\sqrt{20}} = 1.491\), so \(\text{P}(\bar{X} > 125) = 1 - \Phi(1.491) = 1 - 0.9320 = 0.0680\). The tempting \(0.370\) divides by \(\sigma = 15\), the probability for a single value, not a mean. \(0.329\) divides by the variance \(\dfrac{225}{20} = 11.25\) instead of its square root, and \(0.932\) is \(\text{P}(\bar{X} < 125)\).
The Central Limit Theorem says that for a LARGE sample the mean is approximately normal whatever the population's distribution, so \(n = 60\) is valid. The tempting \(n = 5\) quotes the right theorem but the sample is too small for it. A sample mean is exactly normal only when the population itself is normal, and the last option ignores the CLT altogether.
\(\text{Var}(\bar{X}) = \dfrac{3.6}{45} = 0.08\). \(z = \dfrac{3.3 - 3.6}{\sqrt{0.08}} = -1.061\), and \(\Phi(-1.061) = 1 - 0.8557 = 0.144\). The tempting \(0.437\) uses the variance 3.6 of a single minute, forgetting to divide by \(n\). \(0.288\) divides by \(\dfrac{3.6}{\sqrt{45}}\), treating 3.6 as a standard deviation, and \(0.856\) is the wrong tail.
\(\dfrac{2\sqrt{n}}{8} > 2.576\) gives \(\sqrt{n} > 10.304\), so \(n > 106.17\) and the smallest sample size is 107. The tempting 106 rounds down, which gives a probability just below 0.99. 62 uses 1.96 (a 95% middle), and 87 uses the one-tailed 1% value 2.326 instead of the value for 0.005 in each tail.
\(T \approx N(115.2, 57.6)\). \(z = \dfrac{120 - 115.2}{\sqrt{57.6}} = 0.632\), so \(\text{P}(T > 120) = 1 - 0.7363 = 0.264\). The tempting \(0.468\) uses \(\text{Var}(64X) = 64^{2} \times 0.9\), one value multiplied by 64 rather than 64 separate values. \(0.252\) treats 0.9 as the standard deviation (variance \(64 \times 0.81\)), and \(0.736\) is the wrong tail.
\(\Phi(z) = 1 - 0.0228 = 0.9772\) gives \(z = 2.000\). The standard deviation of \(\bar{X}\) is \(\dfrac{5}{\sqrt{25}} = 1\), so \(\dfrac{52 - \mu}{1} = 2\) and \(\mu = 50\). The tempting 42 uses \(\sigma = 5\) instead of \(\dfrac{\sigma}{\sqrt{n}}\). 54 has the sign wrong (52 must be ABOVE the mean for a small upper-tail probability), and 51.6 uses \(\dfrac{\sigma}{n} = 0.2\).
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.