| Exam Board | WJEC |
|---|---|
| Module | Further Unit 5 (Further Unit 5) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | CI from raw data list |
| Difficulty | Standard +0.3 This is a straightforward confidence interval calculation with a small sample from a normal distribution. Part (a) requires computing sample mean and standard deviation, then applying the t-distribution formula (7 marks suggests showing all steps). Part (b) tests conceptual understanding that CLT is not needed since the population is already stated to be normal. Standard Further Maths statistics question with no novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma x = 249.6\), \(\bar{x} = 31.2\) | B1 | — |
| \(\Sigma x^2 = 7792.26\) | B1 | — |
| \(s^2 = \frac{1}{n-1}\left(\Sigma x^2 - n\bar{x}^2\right) = \frac{237}{350} = 0.677...\) | B1 | — |
| \(\text{DF} = 7\) | B1 | — |
| \(t \text{ value} = 2.365\) | B1 | FT their DOF |
| \(\text{Standard error} = \sqrt{\frac{0.677...}{8}}\) | B1 | si |
| \(\text{CL} = 31.2 \pm 2.365 \times \sqrt{\frac{0.677...}{8}}\) | M1 | FT their \(\bar{x}\), \(t\) value and s.e. |
| \(95\%\text{CI} = [30.5, 31.9]\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Appropriate explanation. e.g. The Central Limit Theorem is not required because the underlying distribution is normal. e.g. The Central Limit Theorem is not used because \(n\) is small. | E1 | — |
## (a)
$\Sigma x = 249.6$, $\bar{x} = 31.2$ | B1 | —
$\Sigma x^2 = 7792.26$ | B1 | —
$s^2 = \frac{1}{n-1}\left(\Sigma x^2 - n\bar{x}^2\right) = \frac{237}{350} = 0.677...$ | B1 | —
$\text{DF} = 7$ | B1 | —
$t \text{ value} = 2.365$ | B1 | FT their DOF
$\text{Standard error} = \sqrt{\frac{0.677...}{8}}$ | B1 | si
$\text{CL} = 31.2 \pm 2.365 \times \sqrt{\frac{0.677...}{8}}$ | M1 | FT their $\bar{x}$, $t$ value and s.e.
$95\%\text{CI} = [30.5, 31.9]$ | A1 | cao
**Total: [8]**
## (b)
Appropriate explanation. e.g. The Central Limit Theorem is not required because the underlying distribution is normal. e.g. The Central Limit Theorem is not used because $n$ is small. | E1 | —
**Total: [8]**
---A coffee shop produces biscuits to sell. The masses, in grams, of the biscuits follow a normal distribution with mean $\mu$. Eight biscuits are chosen at random and their masses, in grams, are recorded. The results are given below.
32.1 \quad 29.9 \quad 31.0 \quad 31.1 \quad 32.5 \quad 30.8 \quad 30.7 \quad 31.5
\begin{enumerate}[label=(\alph*)]
\item Calculate a 95\% confidence interval for $\mu$ based on this sample. [7]
\item Explain the relevance or otherwise of the Central Limit Theorem in your calculations. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 5 2019 Q1 [8]}}