| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Standard combined mean and SD |
| Difficulty | Moderate -0.8 This is a straightforward application of combining means and standard deviations from two groups using standard formulas. It requires only direct substitution into well-practiced formulas (combined mean = total sum/total n, combined variance uses pooled sums) with no conceptual difficulty or problem-solving insight—purely mechanical calculation that is easier than a typical A-level question. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean height \(= \frac{\Sigma x + \Sigma y}{6+11} = \frac{1050+1991}{6+11} = \frac{3041}{17}\) | M1 | Use of appropriate formula with values substituted, accept unsimplified |
| 178.9 | A1 | Allow 178.88, \(178\frac{15}{17}\), 179 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\Sigma x^2 + \Sigma y^2}{6+11} = \frac{193700+366400}{6+11}\) | M1 | Use of appropriate formula with values substituted, accept unsimplified |
| \(\text{Sd}^2 = \frac{560100}{17} - their\ 178.88^2\ [= 948.289]\) | M1 | Appropriate variance formula using *their* mean², accept unsimplified expression |
| Standard deviation \(= 30.8\) | A1 | Accept 30.7 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean height $= \frac{\Sigma x + \Sigma y}{6+11} = \frac{1050+1991}{6+11} = \frac{3041}{17}$ | M1 | Use of appropriate formula with values substituted, accept unsimplified |
| 178.9 | A1 | Allow 178.88, $178\frac{15}{17}$, 179 |
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## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\Sigma x^2 + \Sigma y^2}{6+11} = \frac{193700+366400}{6+11}$ | M1 | Use of appropriate formula with values substituted, accept unsimplified |
| $\text{Sd}^2 = \frac{560100}{17} - their\ 178.88^2\ [= 948.289]$ | M1 | Appropriate variance formula using *their* mean², accept unsimplified expression |
| Standard deviation $= 30.8$ | A1 | Accept 30.7 |
---3 A sports club has a volleyball team and a hockey team. The heights of the 6 members of the volleyball team are summarised by $\Sigma x = 1050$ and $\Sigma x ^ { 2 } = 193700$, where $x$ is the height of a member in cm . The heights of the 11 members of the hockey team are summarised by $\Sigma y = 1991$ and $\Sigma y ^ { 2 } = 366400$, where $y$ is the height of a member in cm .
\begin{enumerate}[label=(\alph*)]
\item Find the mean height of all 17 members of the club.
\item Find the standard deviation of the heights of all 17 members of the club.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q3 [5]}}