OCR S1 2013 June — Question 4 6 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeStandard combined mean and SD
DifficultyModerate -0.8 This is a straightforward S1 question testing basic understanding of linear transformations of data (part i), weighted means (part ii), and interpretation of standard deviation (part iii). All parts require direct application of standard formulas with minimal problem-solving. The conceptual demand is low—recognizing that linear transformations affect mean but not standard deviation, and that consistency (low SD) doesn't guarantee accuracy (closeness to true value). This is easier than average A-level maths questions.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance

At a stall in a fair, contestants have to estimate the mass of a cake. A group of 10 people made estimates, \(m\) kg, and for each person the value of \((m - 5)\) was recorded. The mean and standard deviation of \((m - 5)\) were found to be 0.74 and 0.13 respectively.
  1. Write down the mean and standard deviation of \(m\). [2]
The mean and standard deviation of the estimates made by another group of 15 people were found to be 5.6 kg and 0.19 kg respectively.
  1. Calculate the mean of all 25 estimates. [2]
  2. Fiona claims that if a group's estimates are more consistent, they are likely to be more accurate. Given that the true mass of the cake is 5.65 kg, comment on this claim. [2]

(i)
Answer: \(5.74\) or \(0.13\) or 'the same'
Marks: B1, B1[2]
Guidance:
- NB 0.13 seen within working; B0 for \(0.13\) scores B0 for \(\frac{s_x^2}{10} - (\text{their mean})^2 = 0.13^2\), scores B0
- eg \(\frac{s_x^2}{10} - (\text{their mean})^2 = 0.13^2\) scores B0 for 0.13
(ii)
Answer:
- \((10\times 5.74' + 15\times 5.6) + 25oe = 5.656 = 5.66\) (3 sf)
Marks: M1, A1ft, [2]
Guidance:
- eg \(5.74\times \frac{2}{5} + 5.6\times \frac{3}{5}\)
- ft their 5.74
- NB 5.7 with no wking: M0A0 even if already penalised elsewhere for over-rounding
(iii)
Answer:
- 1st gp (or one gp) is more consistent (or less spread oe) but less accurate (or mean further from true mean oe)
Marks: B1ft, B1ft, [2]
Guidance:
- 2nd gp (or one gp) more accurate or etc but less consistent or etc
- If neither B1 scored, but state 'consistency does not imply accuracy' or similar: SC B1
- Ignore all other, eg ignore 'Claim false' or 'Claim true' etc even if contradicts other statements
- Reference to mean of all 25 does not score
- Follow through their values for 1st gp: eg if 1st gp sd = 5.13: 1st gp less accurate and less consistent oe B1B1
- Similar for other it.
### (i)
**Answer:** $5.74$ or $0.13$ or 'the same'

**Marks:** B1, B1[2]

**Guidance:**
- NB 0.13 seen within working; B0 for $0.13$ scores B0 for $\frac{s_x^2}{10} - (\text{their mean})^2 = 0.13^2$, scores B0
- eg $\frac{s_x^2}{10} - (\text{their mean})^2 = 0.13^2$ scores B0 for 0.13

### (ii)
**Answer:**
- $(10\times 5.74' + 15\times 5.6) + 25oe = 5.656 = 5.66$ (3 sf)

**Marks:** M1, A1ft, [2]

**Guidance:**
- eg $5.74\times \frac{2}{5} + 5.6\times \frac{3}{5}$
- ft their 5.74
- NB 5.7 with no wking: M0A0 even if already penalised elsewhere for over-rounding

### (iii)
**Answer:**
- 1st gp (or one gp) is more consistent (or less spread oe) but less accurate (or mean further from true mean oe)

**Marks:** B1ft, B1ft, [2]

**Guidance:**
- 2nd gp (or one gp) more accurate or etc but less consistent or etc
- If neither B1 scored, but state 'consistency does not imply accuracy' or similar: SC B1
- Ignore all other, eg ignore 'Claim false' or 'Claim true' etc even if contradicts other statements
- Reference to mean of all 25 does not score
- Follow through their values for 1st gp: eg if 1st gp sd = 5.13: 1st gp less accurate and less consistent oe B1B1
- Similar for other it.

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At a stall in a fair, contestants have to estimate the mass of a cake. A group of 10 people made estimates, $m$ kg, and for each person the value of $(m - 5)$ was recorded. The mean and standard deviation of $(m - 5)$ were found to be 0.74 and 0.13 respectively.

\begin{enumerate}[label=(\roman*)]
\item Write down the mean and standard deviation of $m$. [2]
\end{enumerate}

The mean and standard deviation of the estimates made by another group of 15 people were found to be 5.6 kg and 0.19 kg respectively.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the mean of all 25 estimates. [2]

\item Fiona claims that if a group's estimates are more consistent, they are likely to be more accurate. Given that the true mass of the cake is 5.65 kg, comment on this claim. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR S1 2013 Q4 [6]}}