| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate variance/SD from coded sums |
| Difficulty | Moderate -0.8 This is a straightforward two-part question testing standard formulas for mean/standard deviation from coded data and basic normal distribution calculations. Part (i) requires direct application of well-known formulas with coding, and part (ii) is a routine standardization and table lookup. Both are textbook exercises with no problem-solving or insight required, making this easier than average. |
| Spec | 2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(-47.2/30 = -1.573\) OR \(\sum x = \sum 110 = -47.2\) and \(\sum 110 = 3300\) | B1 | |
| \(\bar{x} = 110 - 1.573 = 108\) (108.4) | B1 | For correct answer |
| standard deviation \(= \sqrt{\frac{5460}{30} - (-1.573)^2} = 13.4\) | M1, A1 4 | For \(\frac{5460}{30} - (\text{their coded mean})^2\); For correct answer |
| (ii) \(z = \frac{110 - 107.6}{13.8} = 0.174\) | M1 | For standardising, can have \(\sqrt{13.8}\) on denom not 13.8² |
| \(P(X > 110) = 1 - \Phi(0.174) = 1 - 0.5691 = 0.431\) | M1, A1 3 | For using tables correctly and finding a correct area from their z; For correct answer |
**(i)** $-47.2/30 = -1.573$ OR $\sum x = \sum 110 = -47.2$ and $\sum 110 = 3300$ | B1 |
$\bar{x} = 110 - 1.573 = 108$ (108.4) | B1 | For correct answer
standard deviation $= \sqrt{\frac{5460}{30} - (-1.573)^2} = 13.4$ | M1, A1 4 | For $\frac{5460}{30} - (\text{their coded mean})^2$; For correct answer
**(ii)** $z = \frac{110 - 107.6}{13.8} = 0.174$ | M1 | For standardising, can have $\sqrt{13.8}$ on denom not 13.8²
$P(X > 110) = 1 - \Phi(0.174) = 1 - 0.5691 = 0.431$ | M1, A1 3 | For using tables correctly and finding a correct area from their z; For correct answer\begin{enumerate}[label=(\roman*)]
\item In a spot check of the speeds $x \text{ km h}^{-1}$ of 30 cars on a motorway, the data were summarised by $\Sigma(x - 110) = -47.2$ and $\Sigma(x - 110)^2 = 5460$. Calculate the mean and standard deviation of these speeds. [4]
\item On another day the mean speed of cars on the motorway was found to be $107.6 \text{ km h}^{-1}$ and the standard deviation was $13.8 \text{ km h}^{-1}$. Assuming these speeds follow a normal distribution and that the speed limit is $110 \text{ km h}^{-1}$, find what proportion of cars exceed the speed limit. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2002 Q4 [7]}}