| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Challenging +1.2 This question requires applying the CLT to a uniform distribution (needing to recall variance formula for uniform), then constructing a confidence interval for a parameter. It's more challenging than routine CLT applications because students must work backwards from the sample mean to find bounds on 'a', requiring algebraic manipulation beyond standard textbook exercises. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{Var}(X) = \frac{(a+5-a+1)^2}{12} = [=3]\) | M1 | |
| \(\bar{X} \sim N(a+2, \frac{3}{50})\) | A1, A1ft | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(17.2 - 1.96 \times \sqrt{\frac{3}{50}} < \mu < 17.2 + 1.96 \times \sqrt{\frac{3}{50}}\) | B1 M1 | |
| \(17.2 - 1.96 \times \sqrt{\frac{3}{50}} < a+2 < 17.2 + 1.96 \times \sqrt{\frac{3}{50}}\) | B1 | |
| \(14.7 < a < 15.7\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) 1st B1 for correct use of \(z = 1.96\) in an attempt e.g. \(\bar{x} \pm z\sigma\) or \(\bar{x} \pm z\sigma^2\). Where \( | z | > 1.5\) accept just + or just –. Answer of (16.7, 17.7) scores B1M1B0A0 |
| M1 for \(17.2 \pm z \times \sqrt{\frac{"3"}}{50}}\) where \( | z | > 1.5\) accept just + or just –. Answer of (16.7, 17.7) scores B1M1B0A0 |
| 2nd B1 for either of the inequalities with \(a + 2\) and any \(z\) (\( | z | > 1.5\)) or \(a = 15.2 \pm z \times \sqrt{\frac{"3"}{50}}\) |
**(a)** $\text{Var}(X) = \frac{(a+5-a+1)^2}{12} = [=3]$ | M1 |
$\bar{X} \sim N(a+2, \frac{3}{50})$ | A1, A1ft | (3)
**Notes:**
M1 for a correct expression for Var(X) in terms of $a$ or Var(X) = 3
1st A1 for normal and correct mean must be $a + 2$
NB N(17.2, ...) is A0 and N(17.2, $\frac{3}{50}$) is M1A0A1
2nd A1ft for correct Var($\bar{X}$), i.e. (their "3")/50
**(b)** $17.2 - 1.96 \times \sqrt{\frac{3}{50}} < \mu < 17.2 + 1.96 \times \sqrt{\frac{3}{50}}$ | B1 M1 |
$17.2 - 1.96 \times \sqrt{\frac{3}{50}} < a+2 < 17.2 + 1.96 \times \sqrt{\frac{3}{50}}$ | B1 |
$14.7 < a < 15.7$ | A1 | (4)
**[Total 7]**
**Notes:**
**(b)** 1st B1 for correct use of $z = 1.96$ in an attempt e.g. $\bar{x} \pm z\sigma$ or $\bar{x} \pm z\sigma^2$. Where $|z| > 1.5$ accept just + or just –. Answer of (16.7, 17.7) scores B1M1B0A0
M1 for $17.2 \pm z \times \sqrt{\frac{"3"}}{50}}$ where $|z| > 1.5$ accept just + or just –. Answer of (16.7, 17.7) scores B1M1B0A0
2nd B1 for either of the inequalities with $a + 2$ and any $z$ ($|z| > 1.5$) or $a = 15.2 \pm z \times \sqrt{\frac{"3"}{50}}$
A1 for awrt 14.7 and 15.7
---6. The continuous random variable $X$ is uniformly distributed over the interval
$$[ a - 1 , a + 5 ]$$
where $a$ is a constant.\\
Fifty observations of $X$ are taken, giving a sample mean of 17.2
\begin{enumerate}[label=(\alph*)]
\item Use the Central Limit Theorem to find an approximate distribution for $\bar { X }$.
\item Hence find a 95\% confidence interval for $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2013 Q6 [7]}}