| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution properties with standard bookwork calculations. Part (a) uses the empirical rule (±3σ covers ~all data), part (b) requires calculating z-scores and probabilities to check plausibility, and part (c) verifies consistency of summary statistics. All techniques are routine S1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sigma \approx \dfrac{10}{a}\) or \(\dfrac{20}{b}\) or \(\dfrac{\text{range}}{b}\) or \(10c\) or \(20d\) | M1 | OE; with \(2 \leq a \leq 4\), \(4 \leq b \leq 8\), or with \(c\) or \(d\) in equivalent percentages; Cannot be implied from correct answer (justification required) |
| \(\mathbf{2.5}\) or \(\mathbf{3.3(OE)}\) or \(\mathbf{5}\) | A1 | |
| Total | 2 | |
| SC: Award B1 for only \(2.5\) or \(3.3\text{(OE)}\) or \(5\) with no justification; Award B0 for any other answer with no justification or with incorrect justification (eg \(\sqrt{10} = 3.16\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Valid statement involving: 391 and 405 OR 401 and 415 OR 24 and 10 OR 391 and 415 and \(10/24\) with linking statement | B1 | Allow 'set weight' to imply 415 and/or 'mean' to imply 391; B0 for 10 linked to \(\sigma\) |
| \(95.5 >\) (value of \(\sigma\) of \(2.5\) or \(3.3\text{(OE)}\) or \(5\)) | B1 | Accept \(\neq\) rather than \(>\); clear correct numerical comparison |
| Neither (likely to be) correct | Bdep1 | Dependent on B1 B1 |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean or \(\bar{y} = \dfrac{8210.0}{10} = \mathbf{821}\) OR \(\sum y = \mathbf{8200}\) | B1 | CAO |
| Variance \(= \dfrac{110.00}{9} = \mathbf{12.2}\) or \(\dfrac{110.00}{10} = \mathbf{11}\) OR SD \(= \mathbf{3.5}\) or \(\mathbf{3.3}\) | B1 | AWRT; CAO; award on value, ignore notation; AWRT |
| \(821\) is similar to/within 10 of \(820\) OR \(8210\) is within \(100\) of \(8200\) | B1 | OE; clear correct numerical comparison of 821 with 820; allow 'set weight' to imply 820; or clear correct numerical comparison of 8210 with 8200 but do not accept 'within 10' here |
| \(3.5\) or \(3.3\) is similar to a value of \(\sigma\) of \(3.3\text{(OE)}\) or \(2.5\) | B1 | Clear correct numerical comparison |
| Total | 4 |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sigma \approx \dfrac{10}{a}$ or $\dfrac{20}{b}$ or $\dfrac{\text{range}}{b}$ or $10c$ or $20d$ | M1 | OE; with $2 \leq a \leq 4$, $4 \leq b \leq 8$, or with $c$ or $d$ in equivalent percentages; **Cannot** be implied from correct answer (justification required) |
| $\mathbf{2.5}$ **or** $\mathbf{3.3(OE)}$ **or** $\mathbf{5}$ | A1 | |
| **Total** | **2** | |
| SC: Award B1 for only $2.5$ or $3.3\text{(OE)}$ or $5$ with no justification; Award B0 for any other answer with no justification or with incorrect justification (eg $\sqrt{10} = 3.16$) | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Valid statement involving: 391 and 405 **OR** 401 and 415 **OR** 24 and 10 **OR** 391 and 415 and $10/24$ with linking statement | B1 | Allow 'set weight' to imply 415 and/or 'mean' to imply 391; B0 for 10 linked to $\sigma$ |
| $95.5 >$ (value of $\sigma$ of $2.5$ or $3.3\text{(OE)}$ or $5$) | B1 | Accept $\neq$ rather than $>$; clear correct numerical comparison |
| **Neither** (likely to be) **correct** | Bdep1 | Dependent on B1 B1 |
| **Total** | **3** | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean or $\bar{y} = \dfrac{8210.0}{10} = \mathbf{821}$ **OR** $\sum y = \mathbf{8200}$ | B1 | CAO |
| Variance $= \dfrac{110.00}{9} = \mathbf{12.2}$ **or** $\dfrac{110.00}{10} = \mathbf{11}$ **OR** SD $= \mathbf{3.5}$ **or** $\mathbf{3.3}$ | B1 | AWRT; CAO; award on **value**, ignore notation; AWRT |
| $821$ is similar to/within 10 of $820$ **OR** $8210$ is within $100$ of $8200$ | B1 | OE; clear correct numerical comparison of 821 with 820; allow 'set weight' to imply 820; **or** clear correct numerical comparison of 8210 with 8200 but do **not** accept 'within 10' here |
| $3.5$ or $3.3$ is similar to a value of $\sigma$ of $3.3\text{(OE)}$ or $2.5$ | B1 | Clear correct numerical comparison |
| **Total** | **4** | |7 A machine, which cuts bread dough for loaves, can be adjusted to cut dough to any specified set weight. For any set weight, $\mu$ grams, the actual weights of cut dough are known to be approximately normally distributed with a mean of $\mu$ grams and a fixed standard deviation of $\sigma$ grams.
It is also known that the machine cuts dough to within 10 grams of any set weight.
\begin{enumerate}[label=(\alph*)]
\item Estimate, with justification, a value for $\sigma$.
\item The machine is set to cut dough to a weight of 415 grams.
As a training exercise, Sunita, the quality control manager, asked Dev, a recently employed trainee, to record the weight of each of a random sample of 15 such pieces of dough selected from the machine's output. She then asked him to calculate the mean and the standard deviation of his 15 recorded weights.
Dev subsequently reported to Sunita that, for his sample, the mean was 391 grams and the standard deviation was 95.5 grams.
Advise Sunita on whether or not each of Dev's values is likely to be correct. Give numerical support for your answers.
\item Maria, an experienced quality control officer, recorded the weight, $y$ grams, of each of a random sample of 10 pieces of dough selected from the machine's output when it was set to cut dough to a weight of 820 grams. Her summarised results were as follows.
$$\sum y = 8210.0 \quad \text { and } \quad \sum ( y - \bar { y } ) ^ { 2 } = 110.00$$
Explain, with numerical justifications, why both of these values are likely to be correct.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q7 [9]}}