| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from discrete rounded data |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question requiring standard procedures: plotting a cumulative frequency graph, reading a percentile from the graph, and calculating mean and standard deviation from grouped data using mid-interval values. All techniques are straightforward textbook exercises with no problem-solving or conceptual challenges beyond basic statistical literacy. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| \(10 - 19\) | \(20 - 29\) | \(30 - 39\) | \(40 - 44\) | \(45 - 49\) | \(50 - 54\) | \(55 - 59\) | ||
| Frequency | 10 | 18 | 32 | 42 | 28 | 14 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| UB: 19.5, 29.5, 39.5, 44.5, 49.5, 54.5, 59.5 and cf: [10], 28, 60, 102, 130, 144, 150 | B1 | Cf values 28, 60, 102, 130, 144, 150 seen. Condone omission of 10. May be implied by accurate plotting (scale no less than 1cm = 10). May be by data table |
| Linearly scaled axes correctly labelled | B1 | Cumulative frequency (cf) from 0 to 150 and height (h) in centimetres from 9.5 to 59.5 with at least 3 values identified on each. Axes can be the other way round |
| At least 4 points plotted at upper boundary \(\pm 0.5\) | M1 | e.g. allow (19, 19.5 or 20, 10) etc. on correctly scaled axes. \((9.5, 0), (19.5, 10), (29.5, 28), (39.5, 60), (44.5, 102), (49.5, 130), (54.5, 144), (59.5, 150)\) |
| All points plotted correctly, curve drawn | A1 | Within tolerance, joined to \((9.5, 0)\) and not going beyond 150 vertically. A0 if straight line segments used |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([150 \times 0.3 = 45]\) | M1 | Use of graph must be seen |
| Line drawn from 45 on cf axis to meet graph at \(h = 36\) | A1 FT | Must be an increasing cf graph. Expect answer in range \(35 \leqslant h \leqslant 37\) for a correct graph |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Midpoints: 14.5, 24.5, 34.5, 42, 47, 52, 57 | B1 | At least 6 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table |
| \(\text{Mean} = \frac{10\times14.5 + 18\times24.5 + 32\times34.5 + 42\times42 + 28\times47 + 14\times52 + 6\times57}{150}\) \(\left[= \frac{145 + 441 + 1104 + 1764 + 1316 + 728 + 342}{150}\right]\) | M1 | Correct unsimplified mean formula with *their* midpoints (not ub, lb, upper limits, lower limits, cw, fd, f or cf and must be within class). If midpoints correct, accept partially evaluated |
| \(= \frac{5840}{150},\ \frac{584}{15},\ 38\frac{14}{15},\ 38.9\) | A1 | Accept answers wrt 38.9 WWW. If M1 withheld, SC B1 for \(\frac{5840}{150}, \frac{584}{15}, 38\frac{14}{15}, 38.9\) |
| \(\text{sd}^2 = \frac{10\times14.5^2 + 18\times24.5^2 + 32\times34.5^2 + 42\times42^2 + 28\times47^2 + 14\times52^2 + 6\times57^2}{150} - \left(\textit{their}\ \frac{5840}{150}\right)^2\) \(\left[= \frac{244285}{150} - \left(\textit{their}\ \frac{5840}{150}\right)^2\right]\) | M1 | Correct unsimplified variance formula with *their* midpoints (not ub, lb, upper limits, lower limits, cw, fd, f or cf and must be within class). If midpoints correct, accept partially evaluated |
| \([= 112.76]\), standard deviation \([\sqrt{112.76}] = 10.6\) | A1 | AWRT 10.6 WWW. If second M1 withheld, SC B1 for 10.6 WWW |
| 5 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| UB: 19.5, 29.5, 39.5, 44.5, 49.5, 54.5, 59.5 and cf: [10], 28, 60, 102, 130, 144, 150 | **B1** | Cf values 28, 60, 102, 130, 144, 150 seen. Condone omission of 10. May be implied by accurate plotting (scale no less than 1cm = 10). May be by data table |
| Linearly scaled axes correctly labelled | **B1** | Cumulative frequency (cf) from 0 to 150 and height (h) in centimetres from 9.5 to 59.5 with at least 3 values identified on each. Axes can be the other way round |
| At least 4 points plotted at upper boundary $\pm 0.5$ | **M1** | e.g. allow (19, 19.5 or 20, 10) etc. on correctly scaled axes. $(9.5, 0), (19.5, 10), (29.5, 28), (39.5, 60), (44.5, 102), (49.5, 130), (54.5, 144), (59.5, 150)$ |
| All points plotted correctly, curve drawn | **A1** | Within tolerance, joined to $(9.5, 0)$ and not going beyond 150 vertically. A0 if straight line segments used |
| | **4** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[150 \times 0.3 = 45]$ | **M1** | Use of graph must be seen |
| Line drawn from 45 on cf axis to meet graph at $h = 36$ | **A1 FT** | Must be an increasing cf graph. Expect answer in range $35 \leqslant h \leqslant 37$ for a correct graph |
| | **2** | |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoints: 14.5, 24.5, 34.5, 42, 47, 52, 57 | **B1** | At least 6 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table |
| $\text{Mean} = \frac{10\times14.5 + 18\times24.5 + 32\times34.5 + 42\times42 + 28\times47 + 14\times52 + 6\times57}{150}$ $\left[= \frac{145 + 441 + 1104 + 1764 + 1316 + 728 + 342}{150}\right]$ | **M1** | Correct unsimplified mean formula with *their* midpoints (not ub, lb, upper limits, lower limits, cw, fd, f or cf and must be within class). If midpoints correct, accept partially evaluated |
| $= \frac{5840}{150},\ \frac{584}{15},\ 38\frac{14}{15},\ 38.9$ | **A1** | Accept answers wrt 38.9 WWW. If M1 withheld, **SC B1** for $\frac{5840}{150}, \frac{584}{15}, 38\frac{14}{15}, 38.9$ |
| $\text{sd}^2 = \frac{10\times14.5^2 + 18\times24.5^2 + 32\times34.5^2 + 42\times42^2 + 28\times47^2 + 14\times52^2 + 6\times57^2}{150} - \left(\textit{their}\ \frac{5840}{150}\right)^2$ $\left[= \frac{244285}{150} - \left(\textit{their}\ \frac{5840}{150}\right)^2\right]$ | **M1** | Correct unsimplified variance formula with *their* midpoints (not ub, lb, upper limits, lower limits, cw, fd, f or cf and must be within class). If midpoints correct, accept partially evaluated |
| $[= 112.76]$, standard deviation $[\sqrt{112.76}] = 10.6$ | **A1** | AWRT 10.6 WWW. If second M1 withheld, **SC B1** for 10.6 WWW |
| | **5** | |4 On a certain day, the heights of 150 sunflower plants grown by children at a local school are measured, correct to the nearest cm . These heights are summarised in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Height \\
$( \mathrm { cm } )$ \\
\end{tabular} & $10 - 19$ & $20 - 29$ & $30 - 39$ & $40 - 44$ & $45 - 49$ & $50 - 54$ & $55 - 59$ \\
\hline
Frequency & 10 & 18 & 32 & 42 & 28 & 14 & 6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a cumulative frequency graph to illustrate the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{915661eb-2544-4293-af72-608fedb43d70-06_1600_1301_760_383}
\item Use your graph to estimate the 30th percentile of the heights of the sunflower plants.\\
\includegraphics[max width=\textwidth, alt={}, center]{915661eb-2544-4293-af72-608fedb43d70-07_2723_35_101_20}
\item Calculate estimates for the mean and the standard deviation of the heights of the 150 sunflower plants.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q4 [11]}}