OCR MEI S1 — Question 4 17 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeHistogram from continuous grouped data
DifficultyModerate -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.
Spec2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers

4 The weights, \(w\) grams, of a random sample of 60 carrots of variety A are summarised in the table below.
Weight\(30 \leqslant w < 50\)\(50 \leqslant w < 60\)\(60 \leqslant w < 70\)\(70 \leqslant w < 80\)\(80 \leqslant w < 90\)
Frequency111018147
  1. Draw a histogram to illustrate these data.
  2. Calculate estimates of the mean and standard deviation of \(w\).
  3. 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. $$n = 50 \quad \sum x = 3624.5 \quad \sum x ^ { 2 } = 265416$$
  4. Calculate the mean and standard deviation of \(x\).
  5. Compare the central tendency and variation of the weights of varieties A and B .

Question 4:
Part (i)
AnswerMarks Guidance
AnswerMarks 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 correctA1
Linear scales on both axes and labelsG1 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 correctG1 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)
AnswerMarks Guidance
AnswerMarks 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):
AnswerMarks 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 endE1 Must include an element of doubt and must mention both ends
Total: [3]
Question 4(iv):
AnswerMarks 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
Total: [3]
Question 4(v):
AnswerMarks Guidance
Variety A have lower average than Variety BE1 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 BE1 FT their sd; Allow 'more spread' or similar but not 'higher range' or 'higher variance'; Condone 'less consistent'
Total: [2]
## 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]}}