| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Known variance confidence intervals |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring standard calculations: computing sample mean and unbiased variance from summary statistics, then constructing a confidence interval with known variance. Both are routine procedures with no conceptual challenges or problem-solving required, making it easier than average for A-level. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \left(\frac{6046}{36}\right) = 167.94...\) | B1 | awrt 168 |
| \(s^2 = \frac{1016338 - 36 \times \bar{x}^2}{35}\) | M1 | for a correct expression for \(s^2\), follow through their mean. Beware it is very "sensitive" |
| \(s^2 = 27.0253....\) | A1 | awrt 27.0 (Accept 27) |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(99\%\) Confidence Interval is: \(\bar{x} \pm 2.5758 \times \frac{5.1}{\sqrt{36}}\) | M1A1ft | |
| \(= (165.755..., 170.133...)\) | A1 A1 | awrt (166,170) |
| (5 marks) |
**Part (a)**
$\bar{x} = \left(\frac{6046}{36}\right) = 167.94...$ | B1 | awrt 168
$s^2 = \frac{1016338 - 36 \times \bar{x}^2}{35}$ | M1 | for a correct expression for $s^2$, follow through their mean. Beware it is very "sensitive"
$s^2 = 27.0253....$ | A1 | awrt 27.0 (Accept 27)
| | | (3 marks)
**Part (b)**
$99\%$ Confidence Interval is: $\bar{x} \pm 2.5758 \times \frac{5.1}{\sqrt{36}}$ | M1A1ft |
$= (165.755..., 170.133...)$ | A1 A1 | awrt (166,170)
| | | (5 marks)
**Guidance Notes:**
M1 for substituting their values in $\bar{x} \pm z \times \frac{s}{\sqrt{n}}$ or $\bar{x} \pm z \times \frac{5.1}{\sqrt{36}}$ where $z$ is a recognizable value from tables
1st A1 follow through their mean and their $z$ (to 2dp) in $\bar{x} \pm z \times \frac{5.1}{\sqrt{36}}$
Beware: $167.94 \pm 2.5758 \times \frac{5.1^2}{36} \to (166.07...,169.8...)$ but scores B1M0A0A0
Correct answer only in (b) scores 0/5
2nd & 3rd A marks depend upon 2.5758 and M mark.
---\begin{enumerate}
\item Some biologists were studying a large group of wading birds. A random sample of 36 were measured and the wing length, $x \mathrm {~mm}$ of each wading bird was recorded. The results are summarised as follows
\end{enumerate}
$$\sum x = 6046 \quad \sum x ^ { 2 } = 1016338$$
(a) Calculate unbiased estimates of the mean and the variance of the wing lengths of these birds.
Given that the standard deviation of the wing lengths of this particular type of bird is actually 5.1 mm ,\\
(b) find a $99 \%$ confidence interval for the mean wing length of the birds from this group.\\
\hfill \mbox{\textit{Edexcel S3 2008 Q1 [8]}}