| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| 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 routine application of standard formulas connecting mean/SD to sums. Part (i) requires direct substitution into memorized formulas (Σx = n×mean, Σx² from SD formula). Part (ii) combines two datasets using the same formulas in reverse. No problem-solving or conceptual insight needed—purely mechanical calculation with well-practiced techniques. |
| Spec | 2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum x = 60 \times 20 = 1200\) | B1 | |
| \(\dfrac{\sum x^2}{20} - 60^2 = 4^2\) | M1 | Correct variance formula used, condone \(= 4\) |
| \(\sum x^2 = 3616 \times 20 = 72320\) | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum x = 1200 + 550 = 1750\), \(\sum x^2 = 72320 + 40500 = 112800\) | M1 | Summing both values of \(\sum x\) and \(\sum x^2\) |
| \(\text{Mean} = \frac{1750}{30} = 58.3\) | B1FT | FT *their* 1750 (not 550 or 1200)/*their*(20+10), accept unsimplified |
| \(\text{Variance} = \frac{112820}{30} - \left(\frac{1750}{30}\right)^2 = 357.89\) | M1 | Substitute *their* \(\sum x\) and \(\sum x^2\) into correct variance formula |
| \(\text{s.d.} = 18.9\) | A1 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum x = 60 \times 20 = 1200$ | B1 | |
| $\dfrac{\sum x^2}{20} - 60^2 = 4^2$ | M1 | Correct variance formula used, condone $= 4$ |
| $\sum x^2 = 3616 \times 20 = 72320$ | A1 | Exact value |
**Total: 3 marks**
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum x = 1200 + 550 = 1750$, $\sum x^2 = 72320 + 40500 = 112800$ | M1 | Summing both values of $\sum x$ and $\sum x^2$ |
| $\text{Mean} = \frac{1750}{30} = 58.3$ | B1FT | FT *their* 1750 (not 550 or 1200)/*their*(20+10), accept unsimplified |
| $\text{Variance} = \frac{112820}{30} - \left(\frac{1750}{30}\right)^2 = 357.89$ | M1 | Substitute *their* $\sum x$ and $\sum x^2$ into correct variance formula |
| $\text{s.d.} = 18.9$ | A1 | |
---3 The mean and standard deviation of 20 values of $x$ are 60 and 4 respectively.\\
(i) Find the values of $\Sigma x$ and $\Sigma x ^ { 2 }$.\\
Another 10 values of $x$ are such that their sum is 550 and the sum of their squares is 40500 .\\
(ii) Find the mean and standard deviation of all these 30 values of $x$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q3 [7]}}