| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Forward transformation: find new statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for sample mean/variance and linear transformations of random variables. Part (a) requires direct substitution into well-known formulas, while part (b) uses the basic rules that E(aX+b) = aE(X)+b and Var(aX+b) = a²Var(X). No problem-solving or conceptual insight is needed beyond recalling these standard results. |
| Spec | 5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{20.5}{40} = 0.5125\) | B1 | Accept \(0.513\) or \(\frac{41}{80}\). Condone \(\frac{20.5}{40}\) |
| \(\frac{40}{39}\left(\frac{10.728}{40} - (0.5125^2)\right)\) or \(\frac{1}{39}\left(10.728 - \frac{20.50^2}{40}\right)\) | M1 | Biased variance (\(0.005544\) or \(\frac{887}{160000}\)) scores M0 A0 |
| \(0.0056859\) or \(0.00569\) (3 sf) or \(\frac{887}{156000}\) | A1 | CAO |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([11 \times 0.5125 + 0.5] = 6.1375\) or \(\frac{491}{80}\) or \(6.14\) (3sf) | B1 FT | FT *their* \(0.5125\) |
| \(11^2 \times 0.0056859\) | M1 | With nothing added. Using *their* variance in (a) (no sd/var confusion) |
| \(0.688\) (3sf) | A1 | CAO |
| 3 |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{20.5}{40} = 0.5125$ | B1 | Accept $0.513$ or $\frac{41}{80}$. Condone $\frac{20.5}{40}$ |
| $\frac{40}{39}\left(\frac{10.728}{40} - (0.5125^2)\right)$ or $\frac{1}{39}\left(10.728 - \frac{20.50^2}{40}\right)$ | M1 | Biased variance ($0.005544$ or $\frac{887}{160000}$) scores M0 A0 |
| $0.0056859$ or $0.00569$ (3 sf) or $\frac{887}{156000}$ | A1 | CAO |
| | **3** | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[11 \times 0.5125 + 0.5] = 6.1375$ or $\frac{491}{80}$ or $6.14$ (3sf) | B1 FT | FT *their* $0.5125$ |
| $11^2 \times 0.0056859$ | M1 | With nothing added. Using *their* variance in (a) (no sd/var confusion) |
| $0.688$ (3sf) | A1 | CAO |
| | **3** | |
---1 The mass, in kilograms, of a block of cheese sold in a supermarket is denoted by the random variable $M$. The masses of a random sample of 40 blocks are summarised as follows.
$$n = 40 \quad \Sigma m = 20.50 \quad \Sigma m ^ { 2 } = 10.7280$$
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the population mean and variance of $M$.
\item The price, $\$ P$, of a block of cheese of mass $M \mathrm {~kg}$ is found using the formula $P = 11 M + 0.50$. Find estimates of the population mean and variance of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q1 [6]}}