| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Unknown variance confidence intervals |
| Difficulty | Standard +0.3 This is a straightforward application of the t-distribution confidence interval formula with unknown variance. Students must calculate sample mean and standard deviation from summary statistics, then apply the standard formula with t₇ critical value. Part (b) requires simple interpretation of whether 440g lies in the interval. While it involves multiple steps, each is routine for FS2 students with no conceptual challenges or novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = 444\) | M1 | For finding mean and attempting \(s^2\) |
| \(s_x^2 = \frac{1577314 - 8 \times 444^2}{7} = \frac{226}{7} = 32.2857...\) | A1 | For correct mean and \(s^2\) (accept awrt 3sf) |
| \(t_7(5\%)\) 2-tail cv \(= 2.365\) | B1 | For correct cv of 2.365 or better |
| 95% CI for \(\mu\): \(444 \pm 2.365 \times \sqrt{\frac{32.2857...}{8}}\) | M1 | For use of correct formula, ft their mean, \(s_x\) and cv for \(t\) (use of 1.96 is M0) |
| \(= (439.248...,\ 448.75...)\) = awrt (439, 449) | A1 | For awrt (439, 449) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 440 is in CI so the average contents statement is OK | B1 | For correct statement about 440 and interval and conclusion |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 444$ | M1 | For finding mean and attempting $s^2$ |
| $s_x^2 = \frac{1577314 - 8 \times 444^2}{7} = \frac{226}{7} = 32.2857...$ | A1 | For correct mean and $s^2$ (accept awrt 3sf) |
| $t_7(5\%)$ 2-tail cv $= 2.365$ | B1 | For correct cv of 2.365 or better |
| 95% CI for $\mu$: $444 \pm 2.365 \times \sqrt{\frac{32.2857...}{8}}$ | M1 | For use of correct formula, ft their mean, $s_x$ and cv for $t$ (use of 1.96 is M0) |
| $= (439.248...,\ 448.75...)$ = awrt **(439, 449)** | A1 | For awrt (439, 449) |
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## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 440 is in CI so the average contents statement is OK | B1 | For correct statement about 440 and interval and conclusion |2 Jemima makes jam to sell in a local shop. The jam is sold in jars and the weight of jam in a jar is normally distributed.
Jemima takes a random sample of 8 of her jars of jam and weighs the contents of each jar, $x$ grams. Her results are summarised as follows
$$\sum x = 3552 \quad \sum x ^ { 2 } = 1577314$$
\begin{enumerate}[label=(\alph*)]
\item Calculate a 95\% confidence interval for the mean weight of jam in a jar.
The labels on the jars state that the average contents weigh 440 grams.
\item State, giving a reason, whether or not Jemima should be concerned about the labels on her jars of jam.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2020 Q2 [6]}}