| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| 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 -0.3 This is a straightforward application of standard formulas for combined means and the relationship between variance, mean, and sum of squares. Part (i) requires weighted average calculation, while part (ii) uses the formula σ² = (Σx²/n) - μ² to find Σx² for each group, then combines them. The question is slightly easier than average because it's entirely procedural with clear steps and no conceptual traps, though it does require careful arithmetic across multiple calculations. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((220 \times 20 + 118 \times 25)/45 = 163\) | M1 | Mult by 20 and 25 and dividing their sum by 45 |
| A1 2 | Correct answer, 163.3 or 490/3 oe acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma x_o^2 = 988480\) | M1 | Subst in correct variance formula |
| A1 | Correct \(\Sigma x_o^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma x_1^2 = 351700\) | A1 | correct \(\Sigma x_1^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| New var \(= 1340180/45 - (7350/45)^2 = 3100 - 3120\) | M1 | Subst their combined results in correct var formula |
| A1 5 | Correct answer |
(i) $(220 \times 20 + 118 \times 25)/45 = 163$ | M1 | Mult by 20 and 25 and dividing their sum by 45
| A1 2 | Correct answer, 163.3 or 490/3 oe acceptable
(ii) $\Sigma x_o^2/20 - 220^2 = 32^2$
$\Sigma x_o^2 = 988480$ | M1 | Subst in correct variance formula
| A1 | Correct $\Sigma x_o^2$
$\Sigma x_1^2/25 - 118^2 = 12^2$
$\Sigma x_1^2 = 351700$ | A1 | correct $\Sigma x_1^2$
$\Sigma x_o^2 + \Sigma x_1^2 = 1340180$
New var $= 1340180/45 - (7350/45)^2 = 3100 - 3120$ | M1 | Subst their combined results in correct var formula
| A1 5 | Correct answer4 Barry weighs 20 oranges and 25 lemons. For the oranges, the mean weight is 220 g and the standard deviation is 32 g . For the lemons, the mean weight is 118 g and the standard deviation is 12 g .\\
(i) Find the mean weight of the 45 fruits.\\
(ii) The individual weights of the oranges in grams are denoted by $x _ { o }$, and the individual weights of the lemons in grams are denoted by $x _ { l }$. By first finding $\Sigma x _ { o } ^ { 2 }$ and $\Sigma x _ { l } ^ { 2 }$, find the variance of the weights of the 45 fruits.
\hfill \mbox{\textit{CAIE S1 2013 Q4 [7]}}