| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from raw data |
| Difficulty | Moderate -0.8 This is a straightforward statistics question requiring basic calculations: mean from raw data, histogram drawing, mean/SD from grouped data, and simple linear transformations. All techniques are standard S1 content with no problem-solving insight needed—purely procedural application of formulas. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Grade | A* | A | B | C | D | E | F | G | U |
| Points | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
| Mean GCSE score | Number of students |
| \(4.5 \leqslant X < 5.5\) | 8 |
| \(5.5 \leqslant X < 6.0\) | 14 |
| \(6.0 \leqslant X < 6.5\) | 19 |
| \(6.5 \leqslant X < 7.0\) | 13 |
| \(7.0 \leqslant X \leqslant 8.0\) | 6 |
| AS Grade | A | B | C | D | E | U |
| Score | 60 | 50 | 40 | 30 | 20 | 0 |
| Answer | Marks |
|---|---|
| \(X = \frac{8+8+7+7+7+6+6+6+6+5+4}{11} = \frac{70}{11} \approx 6.36\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Histogram with frequency density on \(y\)-axis | B1 | Correct axis label |
| All 5 bars correct heights (fd: 16, 28, 38, 26, 12 — divide by class width 0.5 gives 16, 28, 38, 26, 12) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{8(5)+14(5.75)+19(6.25)+13(6.75)+6(7.5)}{60}\) | M1 | Use of midpoints |
| \(= \frac{40+80.5+118.75+87.75+45}{60} = \frac{372}{60} = 6.2\) | A1 | |
| \(s = \sqrt{\frac{\sum fx^2}{60} - \bar{x}^2}\), \(\sum fx^2 = 8(25)+14(33.0625)+19(39.0625)+13(45.5625)+6(56.25)\) | M1 |
| Answer | Marks |
|---|---|
| \(s = \sqrt{\frac{2334.875}{60} - 6.2^2} = \sqrt{38.914... - 38.44} = \sqrt{0.474...} \approx 0.689\) | A1 A1 |
| Answer | Marks |
|---|---|
| \(Y = 13(7.4) - 46 = 96.2 - 46 = 50.2 \Rightarrow\) Grade B | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Predict grade E (score 20) or U | B1 M1 A1 | With valid reason (model unreliable at extremes) |
| Answer | Marks | Guidance |
|---|---|---|
| Mean of \(Y\): \(\bar{Y} = 13\bar{X} - 46 = 13(6.2) - 46 = 80.6 - 46 = 34.6\) | B1 | ft their mean |
| \(s_Y = 13 s_X = 13(0.689) \approx 8.96\) | M1 A1 | ft their sd |
# Question 7:
**(i)**
$X = \frac{8+8+7+7+7+6+6+6+6+5+4}{11} = \frac{70}{11} \approx 6.36$ | M1 A1 |
**(ii)**
Histogram with frequency density on $y$-axis | B1 | Correct axis label
All 5 bars correct heights (fd: 16, 28, 38, 26, 12 — divide by class width 0.5 gives 16, 28, 38, 26, 12) | B1 B1 |
**(iii)**
$\bar{x} = \frac{8(5)+14(5.75)+19(6.25)+13(6.75)+6(7.5)}{60}$ | M1 | Use of midpoints
$= \frac{40+80.5+118.75+87.75+45}{60} = \frac{372}{60} = 6.2$ | A1 |
$s = \sqrt{\frac{\sum fx^2}{60} - \bar{x}^2}$, $\sum fx^2 = 8(25)+14(33.0625)+19(39.0625)+13(45.5625)+6(56.25)$ | M1 |
$= 200 + 462.875 + 742.1875 + 592.3125 + 337.5 = 2334.875$
$s = \sqrt{\frac{2334.875}{60} - 6.2^2} = \sqrt{38.914... - 38.44} = \sqrt{0.474...} \approx 0.689$ | A1 A1 |
**(iv)**
$Y = 13(7.4) - 46 = 96.2 - 46 = 50.2 \Rightarrow$ Grade B | M1 A1 |
**(v)**
$Y = 13(5.5) - 46 = 71.5 - 46 = 25.5$, but this is below the minimum score of 20 for grade E
Predict grade E (score 20) or U | B1 M1 A1 | With valid reason (model unreliable at extremes)
**(vi)**
Mean of $Y$: $\bar{Y} = 13\bar{X} - 46 = 13(6.2) - 46 = 80.6 - 46 = 34.6$ | B1 | ft their mean
$s_Y = 13 s_X = 13(0.689) \approx 8.96$ | M1 A1 | ft their sd
---7 At East Cornwall College, the mean GCSE score of each student is calculated. This is done by allocating a number of points to each GCSE grade in the following way.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | }
\hline
Grade & A* & A & B & C & D & E & F & G & U \\
\hline
Points & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
(i) Calculate the mean GCSE score, $X$, of a student who has the following GCSE grades:
$$\mathrm { A } ^ { * } , \mathrm {~A} ^ { * } , \mathrm {~A} , \mathrm {~A} , \mathrm {~A} , \mathrm {~B} , \mathrm {~B} , \mathrm {~B} , \mathrm {~B} , \mathrm { C } , \mathrm { D } .$$
60 students study AS Mathematics at the college. The mean GCSE scores of these students are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Mean GCSE score & Number of students \\
\hline
$4.5 \leqslant X < 5.5$ & 8 \\
\hline
$5.5 \leqslant X < 6.0$ & 14 \\
\hline
$6.0 \leqslant X < 6.5$ & 19 \\
\hline
$6.5 \leqslant X < 7.0$ & 13 \\
\hline
$7.0 \leqslant X \leqslant 8.0$ & 6 \\
\hline
\end{tabular}
\end{center}
(ii) Draw a histogram to illustrate this information.\\
(iii) Calculate estimates of the sample mean and the sample standard deviation.
The scoring system for AS grades is shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
AS Grade & A & B & C & D & E & U \\
\hline
Score & 60 & 50 & 40 & 30 & 20 & 0 \\
\hline
\end{tabular}
\end{center}
The Mathematics department at the college predicts each student's AS score, $Y$, using the formula $Y = 13 X - 46$, where $X$ is the student's average GCSE score.\\
(iv) What AS grade would the department predict for a student with an average GCSE score of 7.4 ?\\
(v) What do you think the prediction should be for a student with an average GCSE score of 5.5? Give a reason for your answer.\\
(vi) Using your answers to part (iii), estimate the sample mean and sample standard deviation of the predicted AS scores of the 60 students in the department.
\hfill \mbox{\textit{OCR MEI S1 2006 Q7 [18]}}