| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from grouped frequency table |
| Difficulty | Moderate -0.8 This is a routine grouped frequency statistics question requiring standard textbook procedures: calculating mean/SD using midpoints and class widths, then drawing a histogram with unequal class widths. All techniques are algorithmic with no problem-solving or conceptual insight needed, making it easier than average but not trivial due to the computational work involved. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Percentage of meat | \(1 - 5\) | \(6 - 10\) | \(11 - 20\) | \(21 - 30\) | \(31 - 50\) |
| Frequency | 59 | 67 | 38 | 18 | 11 |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1\sin(APN) = T_2\sin(BPN)]\) | M1 | For resolving forces horizontally |
| \((12 \div 13)T_1 = (15 \div 25)T_2\) or \(T_1\sin67.4° = T_2\sin36.9°\) | A1 | AEF |
| \([T_1\cos(APN) + T_2\cos(BPN) = 21]\) | M1 | For resolving forces vertically |
| \((5 \div 13)T_1 + (20 \div 25)T_2 = 21\) or \(T_1\cos67.4° + T_2\cos36.9° = 21\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1/\sin(180 - BPN) = 21/\sin(APN + BPN)]\) | M1 | For using Lami's Theorem to form an equation in \(T_1\) |
| \(T_1/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_1/\sin(180 - 36.9) = 21/\sin(36.9 + 67.4)\) | A1 | AEF |
| \([T_2/\sin(180 - APN) = 21/\sin(APN + BPN)]\) | M1 | For using Lami's Theorem to form an equation in \(T_2\) |
| \(T_2/\sin(180 - \cos^{-1}(20/52)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_2/\sin(180-67.4) = 21/\sin(36.9 + 67.4)\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1/\sin(BPN) = 21/\sin(180 - (APN + BPN))]\) | M1 | For using the Sine Rule on a triangle of forces to form an equation in \(T_1\) |
| \(T_1/(15/25) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_1/\sin36.9° = 21/\sin(180 - (36.9 + 67.4))\) | A1 | AEF |
| \([T_2/\sin(APN) = 21/\sin(180 - (APN + BPN))]\) | M1 | For using the Sine Rule to form an equation in \(T_2\) |
| \(T_2/(12/13) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_2/\sin67.4° = 21/\sin(180 - (36.9 + 67.4))\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
$[T_1\sin(APN) = T_2\sin(BPN)]$ | M1 | For resolving forces horizontally
$(12 \div 13)T_1 = (15 \div 25)T_2$ or $T_1\sin67.4° = T_2\sin36.9°$ | A1 | AEF
$[T_1\cos(APN) + T_2\cos(BPN) = 21]$ | M1 | For resolving forces vertically
$(5 \div 13)T_1 + (20 \div 25)T_2 = 21$ or $T_1\cos67.4° + T_2\cos36.9° = 21$ | A1 | AEF
| M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
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## Alternative solution using Lami's Theorem
$[T_1/\sin(180 - BPN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_1$
$T_1/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin(180 - 36.9) = 21/\sin(36.9 + 67.4)$ | A1 | AEF
$[T_2/\sin(180 - APN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_2$
$T_2/\sin(180 - \cos^{-1}(20/52)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin(180-67.4) = 21/\sin(36.9 + 67.4)$ | A1 | AEF
| M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
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## Alternative solution using Sine Rule
$[T_1/\sin(BPN) = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule on a triangle of forces to form an equation in $T_1$
$T_1/(15/25) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin36.9° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF
$[T_2/\sin(APN) = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule to form an equation in $T_2$
$T_2/(12/13) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin67.4° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF
| M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
---4 In a survey, the percentage of meat in a certain type of take-away meal was found. The results, to the nearest integer, for 193 take-away meals are summarised in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Percentage of meat & $1 - 5$ & $6 - 10$ & $11 - 20$ & $21 - 30$ & $31 - 50$ \\
\hline
Frequency & 59 & 67 & 38 & 18 & 11 \\
\hline
\end{tabular}
\end{center}
(i) Calculate estimates of the mean and standard deviation of the percentage of meat in these take-away meals.\\
(ii) Draw, on graph paper, a histogram to illustrate the information in the table.
\hfill \mbox{\textit{CAIE S1 2012 Q4 [9]}}