| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Estimated variance confidence interval |
| Difficulty | Standard +0.3 This question involves standard calculations: finding unbiased estimates from summary statistics (routine formulas), then applying the linear combination of normal variables to find P(Y-X < 0.01). While it requires understanding of variance of differences and normal distribution properties, these are core S2 techniques with straightforward application and no novel problem-solving required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{est } \mu = 2.087\) | B1 | allow 2.09 |
| \(\text{est } \sigma^2 = \frac{100}{99}\left(\frac{435.57}{100} - 2.087^2\right)\) | M1 | \(1/99\,(435.57 - 208.7^2/100)\) |
| \(= 0.000132(3232)\) or \(131/990000\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y - X) = 2.12 - 2.087\ (= 0.033)\) | B1 | or \(2.12 - 2.087 - 0.01\) for \(Y - X - 0.01 < 0\), allow 2.09 for 2.087 |
| \(\text{Var}(Y - X) = 0.000144 + \text{'0.00013232'}\) | M1 | or \(\sqrt{(0.012^2 + \text{'0.00013232'})}\) |
| \(= 0.000276(32)\) | A1 | \(= 0.016623\) |
| \(\frac{0.01 - \text{'0.033'}}{\sqrt{\text{'0.00027632'}}} \quad (= -1.384)\) | M1 | \(\checkmark\) their \(E(Y-X)\) & \(\text{Var}(Y-X)\); var must be a combination of the two vars |
| \(\Phi(-1.384') = 1 - \Phi(1.384')\) | M1 | correct area/prob consistent with their working |
| \(= 0.0832\) | A1 | [6] |
# Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{est } \mu = 2.087$ | B1 | allow 2.09 |
| $\text{est } \sigma^2 = \frac{100}{99}\left(\frac{435.57}{100} - 2.087^2\right)$ | M1 | $1/99\,(435.57 - 208.7^2/100)$ |
| $= 0.000132(3232)$ or $131/990000$ | A1 | [3] | without $\frac{100}{99}$: 0.000131 M0A0 |
---
# Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y - X) = 2.12 - 2.087\ (= 0.033)$ | B1 | or $2.12 - 2.087 - 0.01$ for $Y - X - 0.01 < 0$, allow 2.09 for 2.087 |
| $\text{Var}(Y - X) = 0.000144 + \text{'0.00013232'}$ | M1 | or $\sqrt{(0.012^2 + \text{'0.00013232'})}$ | M1 |
| $= 0.000276(32)$ | A1 | $= 0.016623$ | A1 |
| $\frac{0.01 - \text{'0.033'}}{\sqrt{\text{'0.00027632'}}} \quad (= -1.384)$ | M1 | $\checkmark$ their $E(Y-X)$ & $\text{Var}(Y-X)$; var must be a combination of the two vars |
| $\Phi(-1.384') = 1 - \Phi(1.384')$ | M1 | correct area/prob consistent with their working |
| $= 0.0832$ | A1 | [6] | SR use of biased var (0.000131) in (i) and (ii): scores in (ii) B1M1A1 for 0.000275 and M1M1A1 for 0.0827 (6/6 available) |
**Total: [9]**
**Total for paper: [50]**7 The diameter, in cm, of pistons made in a certain factory is denoted by $X$, where $X$ is normally distributed with mean $\mu$ and variance $\sigma ^ { 2 }$. The diameters of a random sample of 100 pistons were measured, with the following results.
$$n = 100 \quad \Sigma x = 208.7 \quad \Sigma x ^ { 2 } = 435.57$$
(i) Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.
The pistons are designed to fit into cylinders. The internal diameter, in cm , of the cylinders is denoted by $Y$, where $Y$ has an independent normal distribution with mean 2.12 and variance 0.000144 . A piston will not fit into a cylinder if $Y - X < 0.01$.\\
(ii) Using your answers to part (i), find the probability that a randomly chosen piston will not fit into a randomly chosen cylinder.
\hfill \mbox{\textit{CAIE S2 2015 Q7 [9]}}