| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Standard combined mean and SD |
| Difficulty | Moderate -0.3 This is a standard textbook exercise on combining means and standard deviations from two groups. Part (i) is straightforward weighted mean calculation. Part (ii) requires the formula σ² = Σx²/n - μ², which is routine for S1 students who have practiced this technique. The question is slightly easier than average because it's purely procedural with clear steps and no conceptual challenges. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
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| |||||||
| Farfield Travel | 30 | 1500 | 230 | ||||||
| Lacket Travel | 21 | 2400 | 160 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Mean} = (30 \times 1500 + 21 \times 2400)/51\) | M1 | Multiply by 30 and 21, summing and dividing total by 51; \(\left(\frac{45000+50400}{51}\right)\) |
| \(= 1870\) (1870.59) | A1 | Correct answer (to 3sf) |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(230^2 = \frac{\Sigma x_F^2}{30} - 1500^2\) so \(\Sigma x_F^2 = 69\,087\,000\) | M1 | One correct substitution into a correct variance formula |
| A1 | Correct \(\Sigma x_F^2\) (rounding to 69 000 000 2sf) | |
| \(160^2 = \frac{\Sigma x_L^2}{21} - 2400^2\) so \(\Sigma x_L^2 = 121\,497\,600\) | A1 | Correct \(\Sigma x_L^2\) (rounding to 121 000 000 3sf) |
| \(\text{New var} = \frac{69\,087\,000 + 121\,497\,600}{51} - 1870.588^2 = 237\,853\) | M1 | Using \(`\Sigma x_F^{2'} + ``\Sigma x_L^2\)' dividing by 51 and subtracting \(`i'\) squared. (Correct \(`\Sigma x_F^{2'} + ``\Sigma x_L^2\)' \(= 190\,584\,600\)) |
| New \(\text{sd} = 488\) | A1 | Correct answer; accept anything between 486 and 490 |
| Total: 5 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = (30 \times 1500 + 21 \times 2400)/51$ | M1 | Multiply by 30 and 21, summing and dividing total by 51; $\left(\frac{45000+50400}{51}\right)$ |
| $= 1870$ (1870.59) | A1 | Correct answer (to 3sf) |
| **Total: 2** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $230^2 = \frac{\Sigma x_F^2}{30} - 1500^2$ so $\Sigma x_F^2 = 69\,087\,000$ | M1 | One correct substitution into a correct variance formula |
| | A1 | Correct $\Sigma x_F^2$ (rounding to 69 000 000 2sf) |
| $160^2 = \frac{\Sigma x_L^2}{21} - 2400^2$ so $\Sigma x_L^2 = 121\,497\,600$ | A1 | Correct $\Sigma x_L^2$ (rounding to 121 000 000 3sf) |
| $\text{New var} = \frac{69\,087\,000 + 121\,497\,600}{51} - 1870.588^2 = 237\,853$ | M1 | Using $`\Sigma x_F^{2'} + ``\Sigma x_L^2$' dividing by 51 and subtracting $`i'$ squared. (Correct $`\Sigma x_F^{2'} + ``\Sigma x_L^2$' $= 190\,584\,600$) |
| New $\text{sd} = 488$ | A1 | Correct answer; accept anything between 486 and 490 |
| **Total: 5** | | |4 Farfield Travel and Lacket Travel are two travel companies which arrange tours abroad. The numbers of holidays arranged in a certain week are recorded in the table below, together with the means and standard deviations of the prices.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& \begin{tabular}{ c }
Number of \\
holidays \\
\end{tabular} & \begin{tabular}{ c }
Mean price \\
$( \$ )$ \\
\end{tabular} & \begin{tabular}{ c }
Standard \\
deviation $( \$ )$ \\
\end{tabular} \\
\hline
Farfield Travel & 30 & 1500 & 230 \\
\hline
Lacket Travel & 21 & 2400 & 160 \\
\hline
\end{tabular}
\end{center}
(i) Calculate the mean price of all 51 holidays.\\
(ii) The prices of individual holidays with Farfield Travel are denoted by $\$ x _ { F }$ and the prices of individual holidays with Lacket Travel are denoted by $\$ x _ { L }$. By first finding $\Sigma x _ { F } ^ { 2 }$ and $\Sigma x _ { L } ^ { 2 }$, find the standard deviation of the prices of all 51 holidays.\\
\hfill \mbox{\textit{CAIE S1 2018 Q4 [7]}}