| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Vertical line chart construction |
| Difficulty | Easy -1.8 This is a straightforward data handling question requiring basic statistical skills: drawing a simple vertical line chart from given frequencies, identifying obvious negative skewness visually, calculating mean and variance using standard formulas with small integers, and solving a simple linear equation for a target mean. All parts are routine recall and calculation with no problem-solving insight required. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Number correct | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Frequency | 1 | 2 | 3 | 3 | 4 | 7 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Vertical line chart with correct heights at x = 1,2,3,4,5,6,7 with frequencies 1,2,3,3,4,7,5 | B1 for axes labelled, B1 for correct heights | Lines must be vertical, not bars |
| Answer | Marks |
|---|---|
| Negative skew | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = \frac{1(1)+2(2)+3(3)+4(3)+5(4)+6(7)+7(5)}{25} = \frac{126}{25} = 5.04\) | M1 | For attempt at \(\sum fx / 25\) |
| \(\bar{x} = 5.04\) | A1 | |
| \(MSD = \frac{\sum fx^2}{25} - \bar{x}^2 = \frac{1+8+27+48+100+252+245}{25} - 5.04^2\) | M1 | For \(\sum fx^2/n - \bar{x}^2\) |
| \(= \frac{681}{25} - 25.4016 = 27.24 - 25.4016 = 1.84\) | A1 | Accept 1.8384 |
| Answer | Marks |
|---|---|
| Total needed for 31 days: \(31 \times 5 = 155\) | M1 |
| Total for first 25 days = 126, so need \(155 - 126 = 29\) over 6 days | M1 |
| Average = \(\frac{29}{6} = 4.83...\) so need average of \(\frac{29}{6}\) | A1 |
# Question 1:
## Part (i)
| Vertical line chart with correct heights at x = 1,2,3,4,5,6,7 with frequencies 1,2,3,3,4,7,5 | B1 for axes labelled, B1 for correct heights | Lines must be vertical, not bars |
## Part (ii)
| Negative skew | B1 | |
## Part (iii)
| $\bar{x} = \frac{1(1)+2(2)+3(3)+4(3)+5(4)+6(7)+7(5)}{25} = \frac{126}{25} = 5.04$ | M1 | For attempt at $\sum fx / 25$ |
| $\bar{x} = 5.04$ | A1 | |
| $MSD = \frac{\sum fx^2}{25} - \bar{x}^2 = \frac{1+8+27+48+100+252+245}{25} - 5.04^2$ | M1 | For $\sum fx^2/n - \bar{x}^2$ |
| $= \frac{681}{25} - 25.4016 = 27.24 - 25.4016 = 1.84$ | A1 | Accept 1.8384 |
## Part (iv)
| Total needed for 31 days: $31 \times 5 = 155$ | M1 | |
| Total for first 25 days = 126, so need $155 - 126 = 29$ over 6 days | M1 | |
| Average = $\frac{29}{6} = 4.83...$ so need average of $\frac{29}{6}$ | A1 | |
---1 Every day, George attempts the quiz in a national newspaper. The quiz always consists of 7 questions. In the first 25 days of January, the numbers of questions George answers correctly each day are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | }
\hline
Number correct & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
Frequency & 1 & 2 & 3 & 3 & 4 & 7 & 5 \\
\hline
\end{tabular}
\end{center}
(i) Draw a vertical line chart to illustrate the data.\\
(ii) State the type of skewness shown by your diagram.\\
(iii) Calculate the mean and the mean squared deviation of the data.\\
(iv) How many correct answers would George need to average over the next 6 days if he is to achieve an average of 5 correct answers for all 31 days of January?
\hfill \mbox{\textit{OCR MEI S1 2006 Q1 [8]}}