| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| 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 -0.8 This is a routine application of standard formulas for combined mean and standard deviation. Part (i) uses weighted averages (straightforward calculation), and part (ii) requires the formula σ² = (Σx²)/n - μ², which is given in the formula booklet. The question explicitly guides students to find Σx² first, making it a mechanical exercise with no problem-solving or insight required. Easier than average for A-level. |
| Spec | 2.02g Calculate mean and standard deviation |
| Number of birds | Mean (kg) | Standard deviation (kg) | |
| Turkeys | 9 | 7.1 | 1.45 |
| Geese | 18 | 5.2 | 0.96 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| new mean \(= \frac{9 \times 7.1 + 18 \times 5.2}{27}\) | M1 | Multiply by 9 and 18 and divide by 27 |
| \(= 5.83\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1.45^2 = \frac{\sum x_t^2}{9} - 472.6125\) mm | M1 | Substitute in a correct variance formula; square root or not |
| A1 | Correct \(\sum x_t^2\) (rounding to 470) | |
| \(0.96^2 = \frac{\sum x_g^2}{18} - 5.2^2\), so \(\sum x_g^2 = 503.3088\) | A1 | Correct \(\sum x_g^2\) (rounding to 500) |
| New \(\text{sd}^2 = \frac{472.6\ldots^2 + 503.3\ldots^2}{27} - 5.83\ldots^2 = 2.117\) | M1 | Using \(\sum x_t^2 + \sum x_g^2\), dividing by 27 and subtracting combined mean\(^2\) |
| New sd \(= 1.46\) | A1 [5] | Correct answer |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| new mean $= \frac{9 \times 7.1 + 18 \times 5.2}{27}$ | M1 | Multiply by 9 and 18 and divide by 27 |
| $= 5.83$ | A1 [2] | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1.45^2 = \frac{\sum x_t^2}{9} - 472.6125$ mm | M1 | Substitute in a correct variance formula; square root or not |
| | A1 | Correct $\sum x_t^2$ (rounding to 470) |
| $0.96^2 = \frac{\sum x_g^2}{18} - 5.2^2$, so $\sum x_g^2 = 503.3088$ | A1 | Correct $\sum x_g^2$ (rounding to 500) |
| New $\text{sd}^2 = \frac{472.6\ldots^2 + 503.3\ldots^2}{27} - 5.83\ldots^2 = 2.117$ | M1 | Using $\sum x_t^2 + \sum x_g^2$, dividing by 27 and subtracting combined mean$^2$ |
| New sd $= 1.46$ | A1 [5] | Correct answer |
---5 The table shows the mean and standard deviation of the weights of some turkeys and geese.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Number of birds & Mean (kg) & Standard deviation (kg) \\
\hline
Turkeys & 9 & 7.1 & 1.45 \\
\hline
Geese & 18 & 5.2 & 0.96 \\
\hline
\end{tabular}
\end{center}
(i) Find the mean weight of the 27 birds.\\
(ii) The weights of individual turkeys are denoted by $x _ { t } \mathrm {~kg}$ and the weights of individual geese by $x _ { g } \mathrm {~kg}$. By first finding $\Sigma x _ { t } ^ { 2 }$ and $\Sigma x _ { g } ^ { 2 }$, find the standard deviation of the weights of all 27 birds.
\hfill \mbox{\textit{CAIE S1 2015 Q5 [7]}}