| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a straightforward application of standard formulas for unbiased estimates and linear combinations of normal variables. Part (a) requires direct substitution into memorized formulas, while part (b) involves forming T-S and using normal distribution tables—both routine procedures for S2 with no conceptual challenges or novel problem-solving required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{est}\,\mu = 14\), accept \(\frac{560}{40}\) | B1 | |
| \(\text{est}\,\sigma^2 = \frac{40}{39}\left(\frac{7850}{40} - 14^2\right)\) or \(\frac{1}{39}\left(7850 - \frac{560^2}{40}\right)\) | M1 | |
| \(0.25641\) or \(0.256\) (3sf) | A1 | Accept \(\frac{10}{39}\). Without \(\frac{40}{39}\) i.e. biased: est \(\sigma^2 = 0.25\) M0 A0. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(S - T) = 14.2 - \text{'14'} [= 0.2]\) | B1FT | FT *their* 14. |
| \(\text{Var}(S - T) = 0.3 + \text{'0.256'} [= 0.55641]\) | B1FT | Accept \(\frac{217}{390}\). FT *their* 0.256 including FT biased. \(\text{Var}(S-T) = 0.55\). |
| \(\frac{0.1 - \text{'0.2'}}{\sqrt{\text{'0.55641'}}} [= -0.134]\) | M1 | Standardising with *their* values (note biased gives \(-0.135\)). FT *their* E & Var. |
| \(P(S - T > 0.1) = 1 - \Phi(\text{'-0.134'}) = \Phi(\text{'0.134'})\) | M1 | Finding correct area consistent with *their* values. |
| \(0.553\) (3sf) | A1 | Use of biased gives \(0.554\) (3sf) can score the A1. Similar scheme for \(P(T - S) < -0.1\). Similar scheme for \(S - T - 0.1 > 0\). And \(T - S + 0.1 < 0\). |
| 5 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{est}\,\mu = 14$, accept $\frac{560}{40}$ | B1 | |
| $\text{est}\,\sigma^2 = \frac{40}{39}\left(\frac{7850}{40} - 14^2\right)$ or $\frac{1}{39}\left(7850 - \frac{560^2}{40}\right)$ | M1 | |
| $0.25641$ or $0.256$ (3sf) | A1 | Accept $\frac{10}{39}$. Without $\frac{40}{39}$ i.e. biased: est $\sigma^2 = 0.25$ **M0 A0**. |
| | **3** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(S - T) = 14.2 - \text{'14'} [= 0.2]$ | B1FT | FT *their* 14. |
| $\text{Var}(S - T) = 0.3 + \text{'0.256'} [= 0.55641]$ | B1FT | Accept $\frac{217}{390}$. FT *their* 0.256 including FT biased. $\text{Var}(S-T) = 0.55$. |
| $\frac{0.1 - \text{'0.2'}}{\sqrt{\text{'0.55641'}}} [= -0.134]$ | M1 | Standardising with *their* values (note biased gives $-0.135$). FT *their* E & Var. |
| $P(S - T > 0.1) = 1 - \Phi(\text{'-0.134'}) = \Phi(\text{'0.134'})$ | M1 | Finding correct area consistent with *their* values. |
| $0.553$ (3sf) | A1 | Use of biased gives $0.554$ (3sf) can score the **A1**. Similar scheme for $P(T - S) < -0.1$. Similar scheme for $S - T - 0.1 > 0$. And $T - S + 0.1 < 0$. |
| | **5** | |
---6 The random variable $T$ denotes the time, in seconds, for 100 m races run by Tania. $T$ is normally distributed with mean $\mu$ and variance $\sigma ^ { 2 }$. A random sample of 40 races run by Tania gave the following results.
$$n = 40 \quad \Sigma t = 560 \quad \Sigma t ^ { 2 } = 7850$$
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.\\
The random variable $S$ denotes the time, in seconds, for 100 m races run by Suki. $S$ has the independent distribution $\mathrm { N } ( 14.2,0.3 )$.
\item Using your answers to part (a), find the probability that, in a randomly chosen 100 m race, Suki's time will be at least 0.1 s more than Tania's time.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q6 [8]}}