| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Easy -1.2 This is a straightforward application of standard formulas for mean, standard deviation, median, and quartiles from a discrete frequency table. The only minor complication is treating '5 or more' as '5 or 6' (which is explicitly told), but otherwise requires only routine calculation with no problem-solving or conceptual insight. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| 1 | 2 | 3 | 4 | 5 or more | ||
| 86900 | 92500 | 45000 | 37100 | 19400 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\Sigma fx}{\Sigma f}\) attempted \(\left(= \frac{662000}{280900}\right)\) | M1 | 3 terms of \(\Sigma fx\) correct and \(\div \Sigma f\); Allow incorrect \(\Sigma f\) NOT \(\Sigma x\); \(\div 5\) or \(\div 6\) M0A0 |
| \(= 2.36\) (3 sf) | A1 | |
| \(\frac{\Sigma fx^2}{\Sigma f}\) attempted \(\left(= \frac{2042350}{280900} = 7.270737\right)\) | M1 | 3 terms of \(\Sigma fx^2\) correct and \(\div \Sigma f\); Allow incorrect \(\Sigma f\) NOT \(\Sigma x\) |
| \(- \text{"2.36"}^2\) \((= 1.70\) to \(1.72\), 3 sf) | M1 | dep +ve result; \(\div 5\) or \(\div 6\) M0M0A0 |
| s.d. \(= 1.31\) or \(1.30\) (3 sf) | A1 | allow 1.3; Correct answer without working scores full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 2 | B1 | Ignore working for both, even if incorrect |
| 3 | B1 | allow \(IQR = 3 - 1 = 2\), ie \(UQ = 3\) implied; NB 3, 2 B0B0 unless labelled correctly |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\Sigma fx}{\Sigma f}$ attempted $\left(= \frac{662000}{280900}\right)$ | M1 | 3 terms of $\Sigma fx$ correct and $\div \Sigma f$; Allow incorrect $\Sigma f$ NOT $\Sigma x$; $\div 5$ or $\div 6$ M0A0 |
| $= 2.36$ (3 sf) | A1 | |
| $\frac{\Sigma fx^2}{\Sigma f}$ attempted $\left(= \frac{2042350}{280900} = 7.270737\right)$ | M1 | 3 terms of $\Sigma fx^2$ correct and $\div \Sigma f$; Allow incorrect $\Sigma f$ NOT $\Sigma x$ |
| $- \text{"2.36"}^2$ $(= 1.70$ to $1.72$, 3 sf) | M1 | dep +ve result; $\div 5$ or $\div 6$ M0M0A0 |
| s.d. $= 1.31$ or $1.30$ (3 sf) | A1 | allow 1.3; Correct answer without working scores full marks |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 2 | B1 | Ignore working for both, even if incorrect |
| 3 | B1 | allow $IQR = 3 - 1 = 2$, ie $UQ = 3$ implied; NB 3, 2 B0B0 unless labelled correctly |3 The table shows information about the numbers of people per household in 280900 households in the northwest of England in 2001.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of \\
people \\
\end{tabular} & 1 & 2 & 3 & 4 & 5 or more \\
\hline
\begin{tabular}{ l }
Number of \\
households \\
\end{tabular} & 86900 & 92500 & 45000 & 37100 & 19400 \\
\hline
\end{tabular}
\end{center}
(i) Taking ' 5 or more' to mean ' 5 or 6 ', calculate estimates of the mean and standard deviation of the number of people per household.\\
(ii) State the values of the median and upper quartile of the number of people per household.
\hfill \mbox{\textit{OCR S1 2014 Q3 [7]}}