The shape of the lesson
By the end of the lesson, learners can
- Scaling one variable: aX + b: Multiplying by \(a\) stretches the spread by \(|a|\), so the variance is multiplied by \(a^{2}\).
- Sums and differences: variances add: For independent \(X\) and \(Y\), \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)\): taking away an uncertain amount adds uncertainty.
- Sums of Poisson variables: If \(X \sim \text{Po}(\lambda)\) and \(Y \sim \text{Po}(\mu)\) are independent, then \(X + Y \sim \text{Po}(\lambda + \mu)\).
- Define the combination, e.g. \(D = X_1 + X_2 + X_3 - 2Y\), so the question becomes \(\text{P}(D > 0)\).
- Mean: apply the coefficients directly.
- Variance: square every coefficient and ADD, for sums and differences alike.
- Separate items or one item scaled? \(n\) items: \(n\sigma^{2}\); one item times \(n\): \(n^{2}\sigma^{2}\).
- Shape: normal stays normal; Poisson sums stay Poisson.
- Standardise with the square root of the variance.
One picture every idea returns to
\(\text{E}(aX + bY) = a\text{E}(X) + b\text{E}(Y)\) always. For independent \(X\) and \(Y\), \(\text{Var}(aX + bY) = a^{2}\text{Var}(X) + b^{2}\text{Var}(Y)\), so variances ADD even for a difference. Constants shift the mean but not the variance. Combinations of independent normals are normal; sums of independent Poissons are Poisson.
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, Means add, variances add too.. 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: Scaling one variable: aX + b | Multiplying by \(a\) stretches the spread by \(|a|\), so the variance is multiplied by \(a^{2}\). Model the worked example Mean and variance of a transformed variable on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Sums and differences: variances add | For independent \(X\) and \(Y\), \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)\): taking away an uncertain amount adds uncertainty. Model the worked example A total of several items on the board, one line at a time. Your turn: \(X \sim N(10, 4)\). Find \(\text{P}(X_1 + X_2 > 23)\) for two independent values, and say why this differs from \(\text{P}(2X > 23)\). | Pairs |
| 21 to 30 min | Teach and check: Sums of Poisson variables | If \(X \sim \text{Po}(\lambda)\) and \(Y \sim \text{Po}(\mu)\) are independent, then \(X + Y \sim \text{Po}(\lambda + \mu)\). Model the worked example Combining two sources 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, 22 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 Sums of Poisson variables, 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 |
|---|---|
| \(\text{Var}(3X - 5)\) is not \(3 \times 4 - 5\), and not \(3 \times 4\). | The factor is squared and the constant disappears: \(9 \times 4 = 36\). |
| The commonest Paper 6 error: \(\text{Var}(A - B) = 9 - 16\). | A variance can never be negative; for independent variables they always add. |
| Before adding Poisson parameters, put them on the same interval. | A rate of 2 per hour and 0.5 per 10 minutes give \(\text{Po}(2 + 3) = \text{Po}(5)\) for an hour, not \(\text{Po}(2.5)\). |
Linear combinations of random variables: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
\(\text{E}(X) = 6\) and \(\text{Var}(X) = 2.5\). Find \(\text{E}(4X - 3)\) and \(\text{Var}(4X - 3)\).
\(X \sim N(30, 16)\) and \(Y \sim N(20, 9)\) are independent. Find \(\text{P}(X + Y > 55)\).
\(X \sim N(10, 4)\). \(X_1\) and \(X_2\) are two independent observations of \(X\).
(a) State the distributions of \(2X\) and of \(X_1 + X_2\).
(b) Find \(\text{P}(X_1 + X_2 > 23)\).
Machine A breaks down at random at a mean rate of 0.4 per week and, independently, machine B at a mean rate of 0.6 per week. Find the probability of at least 3 breakdowns in total in a 2-week period.
Men's masses are \(N(78, 10^{2})\) kg and women's masses are \(N(64, 8^{2})\) kg. A lift carries 4 men and 4 women, chosen at random. Find the probability that their total mass exceeds 600 kg.
A café pours cups of coffee with volumes \(C \sim N(250, 9)\) ml. A jug holds a volume \(J \sim N(2040, 400)\) ml. All volumes are independent.
(a) Find the probability that a cup contains more than 255 ml.
(b) Find the probability that one jug is enough to fill 8 cups.
Answer key and marking notes
- \(\text{E}(4X - 3) = 4 \times 6 - 3 = 21\).
- \(\text{Var}(4X - 3) = 16 \times 2.5 = 40\).
- \(X + Y \sim N(50, 25)\).
- \(z = \dfrac{55 - 50}{5} = 1\).
- \(\text{P} = 1 - \Phi(1) = 1 - 0.8413 = 0.159\) (3 s.f.).
- (a) \(2X \sim N(20, 16)\); \(X_1 + X_2 \sim N(20, 8)\).
- (b) \(z = \dfrac{23 - 20}{\sqrt{8}} = 1.061\).
- \(\text{P} = 1 - 0.8556 = 0.144\) (3 s.f.). (Using \(2X\) wrongly would give \(z = 0.75\) and \(0.227\).)
- Total in 2 weeks: \(\text{Po}(2 \times 0.4 + 2 \times 0.6) = \text{Po}(2)\).
- \(\text{P}(T \ge 3) = 1 - e^{-2}(1 + 2 + 2) = 1 - 5e^{-2}\).
- \(= 1 - 0.6767 = 0.323\) (3 s.f.).
- \(T = M_1 + \cdots + M_4 + W_1 + \cdots + W_4\).
- \(\text{E}(T) = 4 \times 78 + 4 \times 64 = 568\); \(\text{Var}(T) = 4 \times 100 + 4 \times 64 = 656\).
- \(z = \dfrac{600 - 568}{\sqrt{656}} = 1.249\).
- \(\text{P}(T > 600) = 1 - 0.8942 = 0.106\) (3 s.f.).
- (a) \(z = \dfrac{255 - 250}{3} = 1.667\), so \(\text{P} = 1 - 0.9522 = 0.0478\) (3 s.f.).
- (b) Let \(D = J - (C_1 + \cdots + C_8)\). We need \(\text{P}(D > 0)\).
- \(\text{E}(D) = 2040 - 8 \times 250 = 40\); \(\text{Var}(D) = 400 + 8 \times 9 = 472\).
- \(z = \dfrac{0 - 40}{\sqrt{472}} = -1.841\), so \(\text{P}(D > 0) = \Phi(1.841) = 0.967\) (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.
\(\text{E}(X) = 7\) and \(\text{Var}(X) = 3\). Find \(\text{E}(5 - 2X)\) and \(\text{Var}(5 - 2X)\).
- A\(\text{E} = -9\), \(\text{Var} = 12\)
- B\(\text{E} = -9\), \(\text{Var} = -1\)
- C\(\text{E} = -9\), \(\text{Var} = 6\)
- D\(\text{E} = -9\), \(\text{Var} = -12\)
\(X\) and \(Y\) are independent random variables with \(\text{Var}(X) = 5\) and \(\text{Var}(Y) = 4\). Find \(\text{Var}(3X - 2Y)\).
- A\(61\)
- B\(29\)
- C\(7\)
- D\(23\)
\(X \sim N(10, 4)\). A single value of \(X\) is observed and multiplied by 3. What is the distribution of \(3X\)?
- A\(N(30, 36)\)
- B\(N(30, 12)\)
- C\(N(30, 4)\)
- D\(N(10, 36)\)
In a hospital, emergency admissions by ambulance arrive at random at a mean rate of 2.1 per hour, and walk-in emergencies arrive independently at random at a mean rate of 1.4 per hour. Find the probability that there are at least 2 emergency arrivals in total in a given hour.
- A\(0.864\)
- B\(0.185\)
- C\(0.970\)
- D\(0.253\)
Chocolates have masses \(N(25, 1.2^{2})\) grams and empty boxes have masses \(N(40, 3^{2})\) grams, all independent. A full box contains 20 chocolates. Find the probability that a full box has mass more than 545 g.
- A\(0.208\)
- B\(0.418\)
- C\(0.168\)
- D\(0.176\)
The times taken by Amira and Bryn to solve a puzzle are \(A \sim N(52, 16)\) and \(B \sim N(48, 9)\) seconds, independently. Find the probability that Amira takes longer than Bryn.
- A\(0.788\)
- B\(0.935\)
- C\(0.716\)
- D\(0.212\)
\(X \sim N(30, 4)\) and \(Y \sim N(18, 1)\), independently. Find the probability that the sum of two independent values of \(X\) is greater than three times a single value of \(Y\).
- A\(0.927\)
- B\(0.885\)
- C\(0.965\)
- D\(0.738\)
\(X \sim \text{Po}(2.5)\) and \(Y \sim \text{Po}(1.5)\) are independent, and \(T = X + 2Y\). Which statement is correct?
- A\(\text{E}(T) = 5.5\), \(\text{Var}(T) = 8.5\), and \(T\) does not have a Poisson distribution
- B\(\text{E}(T) = 5.5\), \(\text{Var}(T) = 8.5\), and \(T \sim \text{Po}(5.5)\)
- C\(\text{E}(T) = 5.5\), \(\text{Var}(T) = 5.5\), and \(T \sim \text{Po}(5.5)\)
- D\(\text{E}(T) = 4\), \(\text{Var}(T) = 4\), and \(T \sim \text{Po}(4)\)
Quiz key
\(\text{E}(5 - 2X) = 5 - 2 \times 7 = -9\). \(\text{Var}(5 - 2X) = (-2)^{2} \times 3 = 12\). The tempting \(-1\) applies the whole expression to the variance (\(5 - 2 \times 3\)); a variance can never be negative. 6 multiplies by 2 instead of \(2^{2}\), and \(-12\) forgets that \((-2)^{2}\) is positive.
\(\text{Var}(3X - 2Y) = 3^{2} \times 5 + (-2)^{2} \times 4 = 45 + 16 = 61\). The tempting 29 subtracts the variances (\(45 - 16\)); for independent variables variances always add, even for a difference. 7 is \(3 \times 5 - 2 \times 4\) (no squaring and subtracting), and 23 is \(3 \times 5 + 2 \times 4\) (no squaring).
\(\text{E}(3X) = 30\) and \(\text{Var}(3X) = 3^{2} \times 4 = 36\), so \(3X \sim N(30, 36)\). The tempting \(N(30, 12)\) is the distribution of \(X_1 + X_2 + X_3\), the sum of THREE independent values, which is less spread out because highs and lows partly cancel. \(N(30, 4)\) leaves the variance unchanged, and \(N(10, 36)\) forgets to scale the mean.
Total \(T \sim \text{Po}(2.1 + 1.4) = \text{Po}(3.5)\). \(\text{P}(T \ge 2) = 1 - e^{-3.5}(1 + 3.5) = 1 - 4.5e^{-3.5} = 0.864\). The tempting \(0.253\) finds at least 2 of EACH type and multiplies, which is a different event. \(0.185\) is \(\text{P}(T = 2)\) only, and \(0.970\) is \(\text{P}(T \ge 1)\).
Mean \(= 20 \times 25 + 40 = 540\). Variance \(= 20 \times 1.44 + 9 = 37.8\). \(z = \dfrac{545 - 540}{\sqrt{37.8}} = 0.813\), so the probability is \(1 - \Phi(0.813) = 1 - 0.7919 = 0.208\). The tempting \(0.418\) uses \(\text{Var}(20C) = 400 \times 1.44\), treating the box as one chocolate multiplied by 20. \(0.168\) adds standard deviations (\(20 \times 1.2 + 3\)) instead of variances, and \(0.176\) leaves out the box's variance.
\(D = A - B \sim N(52 - 48, 16 + 9) = N(4, 25)\). \(\text{P}(D > 0) = \text{P}\!\left(Z > \dfrac{-4}{5}\right) = \Phi(0.8) = 0.788\). The tempting \(0.935\) uses \(\text{Var}(D) = 16 - 9 = 7\); variances add even for a difference. \(0.716\) adds the standard deviations (\(4 + 3 = 7\)) instead of the variances, and \(0.212\) is the probability that Bryn takes longer.
\(\text{E}(X_1 + X_2 - 3Y) = 60 - 54 = 6\). \(\text{Var} = 4 + 4 + 3^{2} \times 1 = 17\). \(\text{P}(> 0) = \Phi\!\left(\dfrac{6}{\sqrt{17}}\right) = \Phi(1.455) = 0.927\). The tempting \(0.885\) uses \(2X\) (variance \(16\)) for the two separate values. \(0.965\) uses \(\text{Var}(3Y) = 3\) instead of 9, and \(0.738\) squares the wrong number, using \(\text{Var}(3Y) = 9^{2} \times 1\).
\(\text{E}(T) = 2.5 + 2 \times 1.5 = 5.5\) and \(\text{Var}(T) = 2.5 + 2^{2} \times 1.5 = 8.5\). A Poisson variable has variance equal to its mean, and \(8.5 \ne 5.5\), so \(T\) is not Poisson. The tempting second option has the right numbers but the wrong conclusion: only plain sums of Poisson counts are Poisson. The third uses \(\text{Var}(2Y) = 2 \times 1.5\), and the fourth ignores the factor 2 altogether.
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.