| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate using histogram bar dimensions |
| Difficulty | Moderate -0.8 This is a straightforward histogram interpretation question requiring basic skills: reading frequency density scale, calculating frequencies from areas, and finding the median from grouped data. These are standard S1 techniques with no conceptual challenges, making it easier than average but not trivial due to the multi-step calculation required. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt find total area, (even if includes \(a^2\)) eg 20×1.4a+10×3.4a+6×a.6a+4×2.6a+10×3a+30 or 28a+34a+27.6a+10.4a+30a+30 or 28+34+27.6+10.4a+30a+30 or 7×20+17×10+23×6 + .... or 160a or 160 or 16 or 16a oa (if area, not ht) | M1 | eg tot area = 16cm² or 16a; 800/16 (= 50); a × 10 = 50 a = 5; M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}\) total area or \(\frac{1}{2}\) total no. apples ft their 6(i); Median is in 50 − 56 class stated or implied; Calculate (approx) \(\frac{2}{3}\) of way along class or \(\frac{1}{3}\) of way from top of class; Median = 53.9 to 54; Not eg 54.2 | B1f, M1, M1, A1, [4] | Examples of correct methods: 400 − (7×20+17×10) (= 90); 50 + \(\frac{\text{"90"}}{23×6}\) × 6 = 54; 200 − (70+85) (= 45); 50 + \(\frac{\text{"45"}}{69}\) × 6 = 54; 400.5 − (7×20+17×10) (= 90.5); 50 + \(\frac{\text{"90.5"}}{23×6}\) × 6 = 54; Use of LB = 49.5: eg median = 49.5 +appr \(\frac{2}{3}\) × 6 = 53.4; B1M1A1A0 |
## (i)
Attempt find total area, (even if includes $a^2$) eg 20×1.4a+10×3.4a+6×a.6a+4×2.6a+10×3a+30 or 28a+34a+27.6a+10.4a+30a+30 or 28+34+27.6+10.4a+30a+30 or 7×20+17×10+23×6 + .... or 160a or 160 or 16 or 16a oa (if area, not ht) | M1 | eg tot area = 16cm² or 16a; 800/16 (= 50); a × 10 = 50 a = 5; M1, A1 | $a = 5$ gives $7×20+17×10+23×6 + .... = 800$; M1; But no of apples = 800; M1; Hence $a = 5$; A1 | 800 = their total (must involve area, not ht); M1dep; eg 160a = 800, 800÷; a = 5; A1 | eg tot area = 400 (sqs); 800/400 (= 2); 1.4a ×20 = 70 × 2; a = 5; A1; [3] | Correct ans with nothing incorrect seen: M1M1A1; But where the correct answer clearly results from incorrect working, eg $a = 800/167 = 4.8$ rounded to $a = 5$, then max M1M1A0 | NOT "1cm = 5" (because may just come from counting squares); NB total ht = 16cm so if 16 seen, must clearly be area eg 800/16 may score 0 or 2
## (ii)
$\frac{1}{2}$ total area or $\frac{1}{2}$ total no. apples ft their 6(i); Median is in 50 − 56 class stated or implied; Calculate (approx) $\frac{2}{3}$ of way along class or $\frac{1}{3}$ of way from top of class; Median = 53.9 to 54; Not eg 54.2 | B1f, M1, M1, A1, [4] | Examples of correct methods: 400 − (7×20+17×10) (= 90); 50 + $\frac{\text{"90"}}{23×6}$ × 6 = 54; 200 − (70+85) (= 45); 50 + $\frac{\text{"45"}}{69}$ × 6 = 54; 400.5 − (7×20+17×10) (= 90.5); 50 + $\frac{\text{"90.5"}}{23×6}$ × 6 = 54; Use of LB = 49.5: eg median = 49.5 +appr $\frac{2}{3}$ × 6 = 53.4; B1M1A1A0 | Correct ans with nothing incorrect seen: M1M1A1; But where the correct answer clearly results from incorrect working, eg $a = 800/167 = 4.8$ rounded to $a = 5$, then max M1M1A0
---The masses, $x$ grams, of 800 apples are summarised in the histogram.
\includegraphics{figure_6}
\begin{enumerate}[label=(\roman*)]
\item On the frequency density axis, 1 cm represents $a$ units. Find the value of $a$. [3]
\item Find an estimate of the median mass of the apples. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q6 [7]}}