| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate mean from coded sums |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for coded data (mean and variance from sums), requiring only algebraic manipulation and basic statistical definitions. Part (i) uses direct formulas, part (ii) is simple rearrangement, parts (iii-iv) involve reading a box plot and applying the outlier definition—all routine S1 techniques with no problem-solving insight required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{-182}{52}\) or \(-3.5\) seen or implied | B1 | \(\Sigma m = 150\times52 - 182\) or \(7618\); B1 |
| Mean \(= 150 - \text{"3.5"}\) | M1 | or \(\frac{-182+150\times52}{52}\) or \(\frac{7800-182}{52}\) B1M1; "7618" \(\div 52\) M1; \(= 146.5\) A1 |
| \(= 146.5\) or \(147\) | A1 | |
| \(\frac{1768}{52} - (\text{"-3.5"})^2\) alone, eg not if \(+150\) | M1 | Allow within \(\sqrt{}\) sign; \((\Sigma(m-150)^2=1768\); \(\Sigma m^2 - 300\Sigma m + 150^2\times52=1768\); \(\Sigma m^2=1768+300\times7618-150^2\times52=1117168)\); \(\frac{1768+300\times\text{'7618'}-150^2\times52}{52}-\text{'146.5'}^2\) |
| \(= 21.75\) or \(21.8\) | A1 [5] | Not ISW, eg \(\sqrt{21.75}\) (or 4.66) M1A0; ans 4.66, no working, M1A0; or \(\frac{\text{'1117168'}}{52}-\text{'146.5'}^2\) fully correct method M1; \(= 21.75\) A1; NB \(\frac{1768}{52}-\text{"146.5"}^2\) or \(1768-(\text{'-3.5'}^2)\) M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\Sigma m^2}{52} - \text{"146.5"}^2 = \text{"21.75"}\); or \(\Sigma m^2 = (\text{'21.75'} + \text{'146.5'}^2)\times52\); ft their mean & +ve var from (i) for M2 | M2 | Allow M1 for \(\frac{\Sigma m^2}{52}-\text{'-3.5'}^2=\text{"21.75"}\); or \(\Sigma m^2=(\text{'21.75'}+\text{'3.5'}^2)\times52\); \(\Sigma(m-150)^2=1768\); \(\Sigma m^2-300\Sigma m+150^2\times52=1768 \geq 2\) terms correct M1; \(\Sigma m^2=1768+300\times\text{'7618'}-150^2\times52\) correct method M1; \(=1117168\) A1 |
| \(\Sigma m^2 = 1117168\) ISW | A1 [3] | Exact; no ft from (i) eg 147 or 21.8; Correct ans, no wking M1M1A1; If incorrect ans given with no wking, possibly M1M1 for (ii) may be obtained by correct method seen in (i), However M1M0 or M0M0 is more likely |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((52+1)\div4 = 13.25\) or \((26+1)\div2 = 13.5\) (\(\Rightarrow\) 13th apple has mass \(< 140\)) | M1 | Allow \(52\div4\) or \(26\div2\ (=13)\) M1; Allow \(52\div4\) or \(26\div2\ (=13)\) M1 |
| \(\Rightarrow\) (no. below 140 \(=\)) 13 | A1 [2] | \(\Rightarrow\) (no. below 140 \(=\)) 13 A1; \(\Rightarrow\) (no. below 140 \(=\)) 12 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(IQR = 15\) seen or implied | B1 | or 22.5 seen or implied |
| \(155+1.5\times15 = 177.5\) (or \(> 176\)); or \(140-1.5\times15 = 117.5\) (or \(< 130\)) | B1 | \(176-155=21\) (or \(< 22.5\)); or \(140-130=10\) (or \(< 22.5\)); \(\frac{176-155}{15}=1.4\) (or \(<1.5\)); or \(\frac{140-130}{15}=\frac{2}{3}\) (or \(<1.5\)); Equivalent correct methods may be seen |
| No outliers | B1 [3] | Ignore method; For 2nd B1 allow \(14\leq IQR\leq16\) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{-182}{52}$ or $-3.5$ seen or implied | B1 | $\Sigma m = 150\times52 - 182$ or $7618$; B1 |
| Mean $= 150 - \text{"3.5"}$ | M1 | or $\frac{-182+150\times52}{52}$ or $\frac{7800-182}{52}$ B1M1; "7618" $\div 52$ M1; $= 146.5$ A1 |
| $= 146.5$ or $147$ | A1 | |
| $\frac{1768}{52} - (\text{"-3.5"})^2$ alone, eg not if $+150$ | M1 | Allow within $\sqrt{}$ sign; $(\Sigma(m-150)^2=1768$; $\Sigma m^2 - 300\Sigma m + 150^2\times52=1768$; $\Sigma m^2=1768+300\times7618-150^2\times52=1117168)$; $\frac{1768+300\times\text{'7618'}-150^2\times52}{52}-\text{'146.5'}^2$ |
| $= 21.75$ or $21.8$ | A1 **[5]** | Not ISW, eg $\sqrt{21.75}$ (or 4.66) M1A0; ans 4.66, no working, M1A0; or $\frac{\text{'1117168'}}{52}-\text{'146.5'}^2$ fully correct method M1; $= 21.75$ A1; NB $\frac{1768}{52}-\text{"146.5"}^2$ or $1768-(\text{'-3.5'}^2)$ M0A0 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\Sigma m^2}{52} - \text{"146.5"}^2 = \text{"21.75"}$; or $\Sigma m^2 = (\text{'21.75'} + \text{'146.5'}^2)\times52$; ft their mean & +ve var from (i) for M2 | M2 | Allow M1 for $\frac{\Sigma m^2}{52}-\text{'-3.5'}^2=\text{"21.75"}$; or $\Sigma m^2=(\text{'21.75'}+\text{'3.5'}^2)\times52$; $\Sigma(m-150)^2=1768$; $\Sigma m^2-300\Sigma m+150^2\times52=1768 \geq 2$ terms correct M1; $\Sigma m^2=1768+300\times\text{'7618'}-150^2\times52$ correct method M1; $=1117168$ A1 |
| $\Sigma m^2 = 1117168$ ISW | A1 **[3]** | Exact; no ft from (i) eg 147 or 21.8; Correct ans, no wking M1M1A1; If incorrect ans given with no wking, possibly M1M1 for (ii) may be obtained by correct method seen in (i), However M1M0 or M0M0 is more likely |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(52+1)\div4 = 13.25$ or $(26+1)\div2 = 13.5$ ($\Rightarrow$ 13th apple has mass $< 140$) | M1 | Allow $52\div4$ or $26\div2\ (=13)$ M1; Allow $52\div4$ or $26\div2\ (=13)$ M1 |
| $\Rightarrow$ (no. below 140 $=$) 13 | A1 **[2]** | $\Rightarrow$ (no. below 140 $=$) 13 A1; $\Rightarrow$ (no. below 140 $=$) 12 A0 |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $IQR = 15$ seen or implied | B1 | or 22.5 seen or implied |
| $155+1.5\times15 = 177.5$ (or $> 176$); or $140-1.5\times15 = 117.5$ (or $< 130$) | B1 | $176-155=21$ (or $< 22.5$); or $140-130=10$ (or $< 22.5$); $\frac{176-155}{15}=1.4$ (or $<1.5$); or $\frac{140-130}{15}=\frac{2}{3}$ (or $<1.5$); Equivalent correct methods may be seen |
| No outliers | B1 **[3]** | Ignore method; For 2nd B1 allow $14\leq IQR\leq16$ |
---3 The masses, $m$ grams, of 52 apples of a certain variety were found and summarised as follows.
$$n = 52 \quad \Sigma ( m - 150 ) = - 182 \quad \Sigma ( m - 150 ) ^ { 2 } = 1768$$
(i) Find the mean and variance of the masses of these 52 apples.\\
(ii) Use your answers from part (i) to find the exact value of $\Sigma m ^ { 2 }$.
The masses of the apples are illustrated in the box-and-whisker plot below.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5ce3230-7528-439c-9e85-ef159a49cba3-3_250_1310_662_383}\\
(iii) How many apples have masses in the interval $130 \leqslant m < 140$ ?\\
(iv) An 'outlier' is a data item that lies more than 1.5 times the interquartile range above the upper quartile, or more than 1.5 times the interquartile range below the lower quartile. Explain whether any of the masses of these apples are outliers.
\hfill \mbox{\textit{OCR S1 2016 Q3 [13]}}