| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from continuous grouped data |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing standard procedures: drawing a histogram with unequal class widths (requiring frequency density calculation), computing mean and standard deviation from grouped data using midpoints, applying the outlier rule (mean ± 2 or 3 standard deviations), and comparing two distributions. All techniques are routine textbook exercises with no problem-solving or novel insight required, making it easier than average. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Weight | \(30 \leqslant w < 50\) | \(50 \leqslant w < 60\) | \(60 \leqslant w < 70\) | \(70 \leqslant w < 80\) | \(80 \leqslant w < 90\) |
| Frequency | 11 | 10 | 18 | 14 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Frequency densities: \(30 \le w < 50\): fd \(= 0.55\); \(50 \le w < 60\): fd \(= 1\); \(60 \le w < 70\): fd \(= 1.8\); \(70 \le w < 80\): fd \(= 1.4\); \(80 \le w < 90\): fd \(= 0.7\) | M1 | For fd's — at least 3 correct. Accept any suitable unit for fd such as freq per 10g. M1 can also be gained from freq per: 5.5, 10, 18, 14, 7 (at least 3 correct) or similar. If fd not explicitly given, M1 and A1 can be gained from all heights correct (within half a square) on histogram |
| All fd's correct | A1 | |
| Linear scales on both axes and labels | G1 | Linear scale and label on vertical axis in relation to first M1 mark, i.e. fd or frequency density, or if relevant freq/10, etc (NOT e.g. fd/10). Allow scale given as fd\(\times\)10 or similar. Accept f/w or f/cw. Ignore horizontal label |
| Width of bars correct (no gaps, drawn at 30, 50, etc.) | G1 | Must be drawn at 30, 50, 60 etc. NOT 29.5 or 30.5 etc. No gaps allowed. No inequality labels on their own such as \(30 \le W < 50\) etc but allow if 30, 50, 60 etc occur at correct boundary position |
| Height of bars correct | G1 | Height of bars — must be linear vertical scale. FT of heights dep on at least 3 heights correct and all must agree with their fds. If fds not given and at least 3 heights correct then max M1A0G1G1G0. Allow restart with correct heights if given fd wrong (for last three marks only) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mean} = \frac{(40\times11)+(55\times10)+(65\times18)+(75\times14)+(85\times7)}{60} = \frac{3805}{60}\) | M1 | For midpoints (at least 3 correct): products are 440, 550, 1170, 1050, 595. No marks for mean or sd unless using midpoints |
| \(= 63.4\) (or 63.42) | A1 | CAO (exact answer 63.41666…). Answer must NOT be left as improper fraction as this is an estimate |
| \(\sum x^2 f = (40^2\times11)+(55^2\times10)+(65^2\times18)+(75^2\times14)+(85^2\times7) = 253225\) | ||
| \(S_{xx} = 253225 - \frac{3805^2}{60} = 11924.6\) | M1 | For attempt at \(S_{xx}\). Should include sum of at least 3 correct multiples \(fx^2 - \frac{\sum x^2}{n}\). Allow M1 for anything which rounds to 11900. Use of mean 63.4 leading to answer of 14.29199… with \(S_{xx} = 12051.4\) gets full credit |
| \(s = \sqrt{\frac{11924.6}{59}} = \sqrt{202.11} = 14.2\) | A1 | At least 1dp required. 63.42 leads to 14.2014… Do not FT their incorrect mean (exact answer 14.2166…). Allow SC1 for RMSD 14.1 (14.0976…) from calculator. If using \((x-\bar{x})^2\) method, B2 if 14.2 or better (14.3 if use of 63.4), otherwise B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} - 2s = 63.4 - (2 \times 14.2) = 35\) | M1 | For either; No marks in (iii) unless using \(\bar{x} + 2s\) or \(x - 2s\); Only follow through numerical values, not variables such as \(s\); if candidate does not find \(s\) but writes 'limit is \(63.4 + 2 \times\) standard deviation', do NOT award M1; Do not penalise for over-specification |
| \(\bar{x} + 2s = 63.4 + (2 \times 14.2) = 91.8\) | A1 | For both (FT); Must have correct limits to get this mark |
| So there are probably some outliers at the lower end, but none at the upper end | E1 | Must include an element of doubt and must mention both ends |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \dfrac{3624.5}{50} = 72.5\text{g}\) (or exact answer 72.49g) | B1 | CAO Ignore units |
| \(S_{xx} = 265416 - \dfrac{3624.5^2}{50} = 2676\) | M1 | For \(S_{xx}\); M1 for \(265416 - 50 \times \text{their mean}^2\); BUT NOTE M0 if their \(S_{xx} < 0\); For \(s^2\) of 54.6 (or better) allow M1A0 with or without working |
| \(s = \sqrt{\dfrac{2676}{49}} = \sqrt{54.61} = 7.39\text{g}\) | A1 | CAO ignore units; Allow 7.4 but NOT 7.3 (unless RMSD with working); For RMSD of 7.3 (or better) allow M1A0 provided working seen; For RMSD\(^2\) of 53.5 (or better) allow M1A0 provided working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Variety A have lower average than Variety B | E1 | FT their means; Do not condone 'lower central tendency' or 'lower mean'; Allow 'on the whole' or similar in place of 'average' |
| Variety A have higher variation than Variety B | E1 | FT their sd; Allow 'more spread' or similar but not 'higher range' or 'higher variance'; Condone 'less consistent' |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Frequency densities: $30 \le w < 50$: fd $= 0.55$; $50 \le w < 60$: fd $= 1$; $60 \le w < 70$: fd $= 1.8$; $70 \le w < 80$: fd $= 1.4$; $80 \le w < 90$: fd $= 0.7$ | M1 | For fd's — at least 3 correct. Accept any suitable unit for fd such as freq per 10g. M1 can also be gained from freq per: 5.5, 10, 18, 14, 7 (at least 3 correct) or similar. If fd not explicitly given, M1 and A1 can be gained from all heights correct (within half a square) on histogram |
| All fd's correct | A1 | |
| Linear scales on both axes and labels | G1 | Linear scale and label on vertical axis in relation to first M1 mark, i.e. fd or frequency density, or if relevant freq/10, etc (NOT e.g. fd/10). Allow scale given as fd$\times$10 or similar. Accept f/w or f/cw. Ignore horizontal label |
| Width of bars correct (no gaps, drawn at 30, 50, etc.) | G1 | Must be drawn at 30, 50, 60 etc. NOT 29.5 or 30.5 etc. No gaps allowed. No inequality labels on their own such as $30 \le W < 50$ etc but allow if 30, 50, 60 etc occur at correct boundary position |
| Height of bars correct | G1 | Height of bars — must be linear vertical scale. FT of heights dep on at least 3 heights correct and all must agree with their fds. If fds not given and at least 3 heights correct then max M1A0G1G1G0. Allow restart with correct heights if given fd wrong (for last three marks only) |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = \frac{(40\times11)+(55\times10)+(65\times18)+(75\times14)+(85\times7)}{60} = \frac{3805}{60}$ | M1 | For midpoints (at least 3 correct): products are 440, 550, 1170, 1050, 595. No marks for mean or sd unless using midpoints |
| $= 63.4$ (or 63.42) | A1 | CAO (exact answer 63.41666…). Answer must NOT be left as improper fraction as this is an estimate |
| $\sum x^2 f = (40^2\times11)+(55^2\times10)+(65^2\times18)+(75^2\times14)+(85^2\times7) = 253225$ | | |
| $S_{xx} = 253225 - \frac{3805^2}{60} = 11924.6$ | M1 | For attempt at $S_{xx}$. Should include sum of at least 3 correct multiples $fx^2 - \frac{\sum x^2}{n}$. Allow M1 for anything which rounds to 11900. Use of mean 63.4 leading to answer of 14.29199… with $S_{xx} = 12051.4$ gets full credit |
| $s = \sqrt{\frac{11924.6}{59}} = \sqrt{202.11} = 14.2$ | A1 | At least 1dp required. 63.42 leads to 14.2014… Do not FT their incorrect mean (exact answer 14.2166…). Allow SC1 for RMSD 14.1 (14.0976…) from calculator. If using $(x-\bar{x})^2$ method, B2 if 14.2 or better (14.3 if use of 63.4), otherwise B0 |
## Question 4(iii):
$\bar{x} - 2s = 63.4 - (2 \times 14.2) = 35$ | M1 | For either; No marks in (iii) unless using $\bar{x} + 2s$ or $x - 2s$; Only follow through numerical values, not variables such as $s$; if candidate does not find $s$ but writes 'limit is $63.4 + 2 \times$ standard deviation', do NOT award M1; Do not penalise for over-specification
$\bar{x} + 2s = 63.4 + (2 \times 14.2) = 91.8$ | A1 | For both (FT); Must have correct limits to get this mark
So there are probably some outliers at the lower end, but none at the upper end | E1 | Must include an element of doubt and must mention both ends
**Total: [3]**
---
## Question 4(iv):
$\text{Mean} = \dfrac{3624.5}{50} = 72.5\text{g}$ (or exact answer 72.49g) | B1 | CAO Ignore units
$S_{xx} = 265416 - \dfrac{3624.5^2}{50} = 2676$ | M1 | For $S_{xx}$; M1 for $265416 - 50 \times \text{their mean}^2$; BUT NOTE M0 if their $S_{xx} < 0$; For $s^2$ of 54.6 (or better) allow M1A0 with or without working
$s = \sqrt{\dfrac{2676}{49}} = \sqrt{54.61} = 7.39\text{g}$ | A1 | CAO ignore units; Allow 7.4 but NOT 7.3 (unless RMSD with working); For RMSD of 7.3 (or better) allow M1A0 provided working seen; For RMSD$^2$ of 53.5 (or better) allow M1A0 provided working seen
**Total: [3]**
---
## Question 4(v):
Variety A have lower average than Variety B | E1 | FT their means; Do not condone 'lower central tendency' or 'lower mean'; Allow 'on the whole' or similar in place of 'average'
Variety A have higher variation than Variety B | E1 | FT their sd; Allow 'more spread' or similar but not 'higher range' or 'higher variance'; Condone 'less consistent'
**Total: [2]**
4 The weights, $w$ grams, of a random sample of 60 carrots of variety A are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Weight & $30 \leqslant w < 50$ & $50 \leqslant w < 60$ & $60 \leqslant w < 70$ & $70 \leqslant w < 80$ & $80 \leqslant w < 90$ \\
\hline
Frequency & 11 & 10 & 18 & 14 & 7 \\
\hline
\end{tabular}
\end{center}
(i) Draw a histogram to illustrate these data.\\
(ii) Calculate estimates of the mean and standard deviation of $w$.\\
(iii) Use your answers to part (ii) to investigate whether there are any outliers.
The weights, $x$ grams, of a random sample of 50 carrots of variety B are summarised as follows.
$$n = 50 \quad \sum x = 3624.5 \quad \sum x ^ { 2 } = 265416$$
(iv) Calculate the mean and standard deviation of $x$.\\
(v) Compare the central tendency and variation of the weights of varieties A and B .
\hfill \mbox{\textit{OCR MEI S1 Q4 [17]}}