| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then estimate mean/standard deviation |
| Difficulty | Moderate -0.3 This is a standard S1 statistics question requiring routine histogram drawing with unequal class widths, followed by straightforward mean/standard deviation calculations from grouped data using mid-interval values. Part (iv) involves basic outlier detection using the mean ± 2SD rule, and part (v) applies standard linear transformation formulas. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02j Clean data: missing data, errors |
| \(0 \leqslant x \leqslant 20\) | \(20 < x \leqslant 40\) | \(40 < x \leqslant 60\) | \(60 < x \leqslant 100\) | \(100 < x \leqslant 200\) | ||
| Frequency | 238 | 365 | 142 | 128 | 45 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Frequency densities: \(0\leq x\leq 20\): 11.9; \(20< x\leq 40\): 18.25; \(40< x\leq 60\): 7.1; \(60< x\leq 100\): 3.2; \(100< x\leq 200\): 0.45 | M1 for fds (at least 4 correct); A1 CAO | Accept any suitable unit for fd such as freq per £1000. NOT fd per £1000 used as label if fd per £10K intended. Allow fds correct to at least 1 dp |
| Linear scale and label on vertical axis | L1 | Label required: fd or frequency density or freq/£10K etc (NOT fd/£10K) |
| Linear scale on horizontal axis and correct width of bars | W1 | Must be drawn at 0, 20, 40 etc. NO GAPS ALLOWED. Must have linear scale |
| Correct heights of bars | H1 | Visual check only (within one square) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Mean} = \frac{10\times238 + 30\times365 + 50\times142 + 80\times128 + 150\times45}{918} = \frac{37420}{918} = 40.8\) | M1 for midpoints; M1 for midpoints \(\times\) frequencies with divisor 918; A1 CAO | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum fx^2 = 238\times10^2 + 365\times30^2 + 142\times50^2 + 128\times80^2 + 45\times150^2 = 2539000\) | M1 for at least 3 multiples \(fx^2\); A1 for \(\sum fx^2\) | For A1, all midpoints and frequencies correct |
| \(S_{xx} = 2539000 - \frac{37420^2}{918} = 1013666\) | M1 for attempt at \(S_{xx}\); Dep on first M1; NOTE M1M0 if \(S_{xx} < 0\) | Or \(S_{xx} = 2539000 - 918\times40.76^2 = 1013855\); \(s=33.25\). Using mean 40.8 leads to \(S_{xx}=1010861\), \(s=33.20\). Using mean \(= 41\) leads to \(S_{xx} = 995844\), \(s=32.95\) |
| \(s = \sqrt{\frac{1013666}{917}} = 33.2\) | A1 CAO | Allow answers 33.25 or 33.3 or 33.2; rmsd \(= \sqrt{(1013666/918)}\) (\(=33.23\)) gets M1A1M1A0; WATCH FOR DIVISOR OF 918; Allow max 4 sf in final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} - 2s = 40.76 - 2\times33.25 = -25.74\) | M1 for \(\bar{x}+2s\) or \(\bar{x}-2s\) | FT any positive mean and positive sd for M1 |
| \(\bar{x} + 2s = 40.76 + 2\times33.25 = 107.26\) | A1 for 107.26 (FT) | Award E0E0 if upper limit \(> 200\) |
| Comment that there are almost certainly some outliers | E1 | Allow 'Must be some outliers'; allow any comment implying there are outliers |
| 'No, since there is nothing to indicate that these high earners represent a separate population.' | E1 Dep on upper limit in range 106–108 | No marks in (iv) unless using \(\bar{x}+2s\) or \(\bar{x}-2s\) |
| Answer | Marks | Guidance |
|---|---|---|
| New mean \(= 1.15\times40.76 = 46.87\) | B1 FT | FT their mean (if given to \(\geq 2\) s.f.) |
| New variance \(= 1.15^2\times33.25^2 = 1462\) | M1A1 FT | FT their \(s\) (if given to \(\geq 2\) s.f.) provided \(s>0\); for new SD \(= 38.24\) found instead of variance give M1A0 even if called variance; M0A0 for \(1.15\times33.25^2 = 1271\); allow max 4 sf in final answers |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Frequency densities: $0\leq x\leq 20$: 11.9; $20< x\leq 40$: 18.25; $40< x\leq 60$: 7.1; $60< x\leq 100$: 3.2; $100< x\leq 200$: 0.45 | M1 for fds (at least 4 correct); A1 CAO | Accept any suitable unit for fd such as freq per £1000. NOT fd per £1000 used as label if fd per £10K intended. Allow fds correct to at least 1 dp |
| Linear scale and label on vertical axis | L1 | Label required: fd or frequency density or freq/£10K etc (NOT fd/£10K) |
| Linear scale on horizontal axis and correct width of bars | W1 | Must be drawn at 0, 20, 40 etc. NO GAPS ALLOWED. Must have linear scale |
| Correct heights of bars | H1 | Visual check only (within one square) |
## Question (Mean calculation):
| $\text{Mean} = \frac{10\times238 + 30\times365 + 50\times142 + 80\times128 + 150\times45}{918} = \frac{37420}{918} = 40.8$ | M1 for midpoints; M1 for midpoints $\times$ frequencies with divisor 918; A1 CAO | **3 marks** |
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## Question (iii):
| $\sum fx^2 = 238\times10^2 + 365\times30^2 + 142\times50^2 + 128\times80^2 + 45\times150^2 = 2539000$ | M1 for at least 3 multiples $fx^2$; A1 for $\sum fx^2$ | For A1, all midpoints and frequencies correct |
| $S_{xx} = 2539000 - \frac{37420^2}{918} = 1013666$ | M1 for attempt at $S_{xx}$; Dep on first M1; NOTE M1M0 if $S_{xx} < 0$ | Or $S_{xx} = 2539000 - 918\times40.76^2 = 1013855$; $s=33.25$. Using mean 40.8 leads to $S_{xx}=1010861$, $s=33.20$. Using mean $= 41$ leads to $S_{xx} = 995844$, $s=32.95$ |
| $s = \sqrt{\frac{1013666}{917}} = 33.2$ | A1 CAO | Allow answers 33.25 or 33.3 or 33.2; rmsd $= \sqrt{(1013666/918)}$ ($=33.23$) gets M1A1M1A0; **WATCH FOR DIVISOR OF 918**; Allow max 4 sf in final answer | **4 marks** |
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## Question (iv):
| $\bar{x} - 2s = 40.76 - 2\times33.25 = -25.74$ | M1 for $\bar{x}+2s$ or $\bar{x}-2s$ | FT any positive mean and positive sd for M1 |
| $\bar{x} + 2s = 40.76 + 2\times33.25 = 107.26$ | A1 for 107.26 (FT) | Award E0E0 if upper limit $> 200$ |
| Comment that there are almost certainly some outliers | E1 | Allow 'Must be some outliers'; allow any comment implying there are outliers |
| 'No, since there is nothing to indicate that these high earners represent a separate population.' | E1 Dep on upper limit in range 106–108 | No marks in (iv) unless using $\bar{x}+2s$ or $\bar{x}-2s$ | **4 marks** |
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## Question (v):
| New mean $= 1.15\times40.76 = 46.87$ | B1 FT | FT their mean (if given to $\geq 2$ s.f.) |
| New variance $= 1.15^2\times33.25^2 = 1462$ | M1A1 FT | FT their $s$ (if given to $\geq 2$ s.f.) provided $s>0$; for new SD $= 38.24$ found instead of variance give M1A0 even if called variance; M0A0 for $1.15\times33.25^2 = 1271$; allow max 4 sf in final answers | **3 marks** |4 The incomes of a sample of 918 households on an island are given in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Income \\
$( x$ thousand pounds $)$ \\
\end{tabular} & $0 \leqslant x \leqslant 20$ & $20 < x \leqslant 40$ & $40 < x \leqslant 60$ & $60 < x \leqslant 100$ & $100 < x \leqslant 200$ \\
\hline
Frequency & 238 & 365 & 142 & 128 & 45 \\
\hline
\end{tabular}
\end{center}
(i) Draw a histogram to illustrate the data.\\
(ii) Calculate an estimate of the mean income.\\
(iii) Calculate an estimate of the standard deviation of the incomes.\\
(iv) Use your answers to parts (ii) and (iii) to show there are almost certainly some outliers in the sample. Explain whether or not it would be appropriate to exclude the outliers from the calculation of the mean and the standard deviation.\\
(v) The incomes were converted into another currency using the formula $y = 1.15 x$. Calculate estimates of the mean and variance of the incomes in the new currency.
\hfill \mbox{\textit{OCR MEI S1 Q4 [19]}}