CAIE S1 2017 November — Question 4 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2017
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeStandard combined mean and SD
DifficultyModerate -0.3 This is a standard textbook exercise on combining means and standard deviations from two groups. Part (i) requires straightforward weighted mean calculation. Part (ii) involves routine application of the formula relating variance to sum of squares, requiring students to work backwards from SD to find Σx² and Σy², then combine and convert back. While multi-step, it follows a well-practiced algorithm with no novel insight required, making it slightly easier than average.
Spec2.02g Calculate mean and standard deviation

4 The ages of a group of 12 people at an Art class have mean 48.7 years and standard deviation 7.65 years. The ages of a group of 7 people at another Art class have mean 38.1 years and standard deviation 4.2 years.
  1. Find the mean age of all 19 people.
  2. The individual ages in years of people in the first Art class are denoted by \(x\) and those in the second Art class by \(y\). By first finding \(\Sigma x ^ { 2 }\) and \(\Sigma y ^ { 2 }\), find the standard deviation of the ages of all 19 people.

Question 4:
Part 4(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{48.7 \times 12 + 38.1 \times 7}{19}\)M1 Accept unsimplified (may be separate calculations)
\(= 44.8\)A1
2
Part 4(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(7.65^2 = \frac{\Sigma x^2}{12} - 48.7^2 \Rightarrow \Sigma x^2 = 29162.55\)M1 Substitution in one correct variance formula
\(4.2^2 = \frac{\Sigma y^2}{7} - 38.1^2 \Rightarrow \Sigma y^2 = 10284.75\)A1 One \(\Sigma x^2\) or \(\Sigma y^2\) correct (can be rounded to 4sf)
Combined var \(= \frac{(29162.55 + 10284.75)}{19} - 44.79^2\)M1 Using their \(\Sigma x^2\) and \(\Sigma y^2\) and their 4(i) in the variance formula
\(= \frac{39447.3}{19} - 44.79^2\)
Combined \(\sigma = 8.37\) or \(8.36\)A1
4
## Question 4:

### Part 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{48.7 \times 12 + 38.1 \times 7}{19}$ | M1 | Accept unsimplified (may be separate calculations) |
| $= 44.8$ | A1 | |
| | **2** | |

### Part 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $7.65^2 = \frac{\Sigma x^2}{12} - 48.7^2 \Rightarrow \Sigma x^2 = 29162.55$ | M1 | Substitution in one correct variance formula |
| $4.2^2 = \frac{\Sigma y^2}{7} - 38.1^2 \Rightarrow \Sigma y^2 = 10284.75$ | A1 | One $\Sigma x^2$ or $\Sigma y^2$ correct (can be rounded to 4sf) |
| Combined var $= \frac{(29162.55 + 10284.75)}{19} - 44.79^2$ | M1 | Using their $\Sigma x^2$ and $\Sigma y^2$ and their **4(i)** in the variance formula |
| $= \frac{39447.3}{19} - 44.79^2$ | | |
| Combined $\sigma = 8.37$ or $8.36$ | A1 | |
| | **4** | |

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4 The ages of a group of 12 people at an Art class have mean 48.7 years and standard deviation 7.65 years. The ages of a group of 7 people at another Art class have mean 38.1 years and standard deviation 4.2 years.\\
(i) Find the mean age of all 19 people.\\

(ii) The individual ages in years of people in the first Art class are denoted by $x$ and those in the second Art class by $y$. By first finding $\Sigma x ^ { 2 }$ and $\Sigma y ^ { 2 }$, find the standard deviation of the ages of all 19 people.\\

\hfill \mbox{\textit{CAIE S1 2017 Q4 [6]}}