| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2002 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring basic understanding of discrete random variables. Students must identify possible values of A (1, 4, 9, 16), calculate their probabilities from the die faces (P(A=1)=3/6, etc.), then apply standard formulas for E(A) and Var(A). The only mild complication is squaring the scores, but this is explicitly explained in the question. No problem-solving insight needed—purely mechanical application of S1 formulas. |
| Spec | 2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a = 1, 4, 9, 16\); \(P(A = a) = \frac{1}{2}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}\) | M1, A1 | For \(A = 1, 4, 9, 16\) or \(1,1,1,4,9,16\); Any three correct probabilities for 3 different vals of A |
| A1 3 | All correct | |
| (ii) \(E(A) = 1 \times \frac{1}{2} + 4 \times \frac{1}{6} + 9 \times \frac{1}{6} + 16 \times \frac{1}{6} = 5.33\) | M1, A1 | For calculation of \(\sum xp\) where \(\sum p\) must be 1; For correct answer |
| \(\text{Var}(A) = 1^2 \times \frac{1}{2} + 4^2 \times \frac{1}{6} + \ldots - (5.33)^2 = 30.9\) | M1 | For calculation of \(\sum x^2p - (\text{their } E(A))^2\); \(\sum p\) need not be 1 |
| A1 4 | For correct answer |
**(i)** $a = 1, 4, 9, 16$; $P(A = a) = \frac{1}{2}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}$ | M1, A1 | For $A = 1, 4, 9, 16$ or $1,1,1,4,9,16$; Any three correct probabilities for 3 different vals of A
| A1 3 | All correct
**(ii)** $E(A) = 1 \times \frac{1}{2} + 4 \times \frac{1}{6} + 9 \times \frac{1}{6} + 16 \times \frac{1}{6} = 5.33$ | M1, A1 | For calculation of $\sum xp$ where $\sum p$ must be 1; For correct answer
$\text{Var}(A) = 1^2 \times \frac{1}{2} + 4^2 \times \frac{1}{6} + \ldots - (5.33)^2 = 30.9$ | M1 | For calculation of $\sum x^2p - (\text{their } E(A))^2$; $\sum p$ need not be 1
| A1 4 | For correct answerA fair cubical die with faces numbered 1, 1, 1, 2, 3, 4 is thrown and the score noted. The area $A$ of a square of side equal to the score is calculated, so, for example, when the score on the die is 3, the value of $A$ is 9.
\begin{enumerate}[label=(\roman*)]
\item Draw up a table to show the probability distribution of $A$. [3]
\item Find $\text{E}(A)$ and $\text{Var}(A)$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2002 Q3 [7]}}