CAIE S1 2017 November — Question 2 3 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2017
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeFind coded sums from raw data
DifficultyEasy -1.2 This is a straightforward application of coding formulas for sums and sum of squares. Students need to recall that Σ(x-a) = Σx - na and Σ(x-a)² = Σx² - 2aΣx + na², combined with the variance formula to find Σx². The calculations are routine with no problem-solving or conceptual insight required, making it easier than average.
Spec2.02g Calculate mean and standard deviation
2 Tien measured the arm lengths, \(x \mathrm {~cm}\), of 20 people in his class. He found that \(\Sigma x = 1218\) and the standard deviation of \(x\) was 4.2. Calculate \(\Sigma ( x - 45 )\) and \(\Sigma ( x - 45 ) ^ { 2 }\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\Sigma(x-45) = 1218 - 20\times45 = 318\)B1
\(\frac{\Sigma(x-45)^2}{20} - \left(\frac{\Sigma(x-45)}{20}\right)^2 = 4.2^2\)M1 Fully correct substitution in coded variance formula with their \(\Sigma(x-45)\); OR valid method for \(\Sigma x^2 = 74529\) \(\left(4.2^2 = \frac{\Sigma x^2}{20} - \left(\frac{1218}{20}\right)^2\right)\) and expanding \(\Sigma(x-45)^2\) correctly \(= \Sigma x^2 - 90\Sigma x + 20\times45^2 = 74529 - 90\times1218 + 40500 = 5409\)
\(\Sigma(x-45)^2 = 5409\)A1
Total: 3 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Sigma(x-45) = 1218 - 20\times45 = 318$ | B1 | |
| $\frac{\Sigma(x-45)^2}{20} - \left(\frac{\Sigma(x-45)}{20}\right)^2 = 4.2^2$ | M1 | Fully correct substitution in coded variance formula with their $\Sigma(x-45)$; **OR** valid method for $\Sigma x^2 = 74529$ $\left(4.2^2 = \frac{\Sigma x^2}{20} - \left(\frac{1218}{20}\right)^2\right)$ and expanding $\Sigma(x-45)^2$ correctly $= \Sigma x^2 - 90\Sigma x + 20\times45^2 = 74529 - 90\times1218 + 40500 = 5409$ |
| $\Sigma(x-45)^2 = 5409$ | A1 | |
| **Total: 3** | | |

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2 Tien measured the arm lengths, $x \mathrm {~cm}$, of 20 people in his class. He found that $\Sigma x = 1218$ and the standard deviation of $x$ was 4.2. Calculate $\Sigma ( x - 45 )$ and $\Sigma ( x - 45 ) ^ { 2 }$.\\

\hfill \mbox{\textit{CAIE S1 2017 Q2 [3]}}
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