| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sum or difference of two spinners/dice |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes (4×3=12 equally likely cases), constructing a probability table, then applying standard variance formula. The arithmetic is simple and the method is routine for S1, making it slightly easier than average but not trivial due to the multi-step calculation required. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): \(-1, 0, 1, 2, 3, 4\); \(p\): \(\frac{1}{12}, \frac{1}{12}, \frac{3}{12}, \frac{2}{12}, \frac{3}{12}, \frac{2}{12}\) | B1 | Table with correct values of \(x\), at least 1 probability, all probabilities \(\leqslant 1\) |
| B1 | 2 probabilities correct, may not be in table | |
| B1 | 2 more probabilities correct, may not be in table | |
| B1 | All correct, values in table. SC1 No more than 1 correct probability and at least 5 probabilities summing to 1 in table |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([E(X)] = \left(\frac{-1+0+3+4+9+8}{12}\right) = \frac{23}{12}\) | M1 | May be implied by use in variance. Allow unsimplified expression |
| \([Var(X)] = \frac{1+0+3+8+27+32\ (=71)}{12} - \left(\frac{23}{12}\right)^2\) | M1 | Appropriate variance formula using *their* \(E(X)^2\) |
| \(2.24\) or \(\frac{323}{144}\) or \(2\frac{35}{144}\) | A1 | CAO |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: $-1, 0, 1, 2, 3, 4$; $p$: $\frac{1}{12}, \frac{1}{12}, \frac{3}{12}, \frac{2}{12}, \frac{3}{12}, \frac{2}{12}$ | B1 | Table with correct values of $x$, at least 1 probability, all probabilities $\leqslant 1$ |
| | B1 | 2 probabilities correct, may not be in table |
| | B1 | 2 more probabilities correct, may not be in table |
| | B1 | All correct, values in table. SC1 No more than 1 correct probability and at least 5 probabilities summing to 1 in table |
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## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[E(X)] = \left(\frac{-1+0+3+4+9+8}{12}\right) = \frac{23}{12}$ | M1 | May be implied by use in variance. Allow unsimplified expression |
| $[Var(X)] = \frac{1+0+3+8+27+32\ (=71)}{12} - \left(\frac{23}{12}\right)^2$ | M1 | Appropriate variance formula using *their* $E(X)^2$ |
| $2.24$ or $\frac{323}{144}$ or $2\frac{35}{144}$ | A1 | CAO |
---5 A fair red spinner has four sides, numbered 1, 2, 3, 3. A fair blue spinner has three sides, numbered $- 1,0,2$. When a spinner is spun, the score is the number on the side on which it lands. The spinners are spun at the same time. The random variable $X$ denotes the score on the red spinner minus the score on the blue spinner.\\
(i) Draw up the probability distribution table for $X$.\\
(ii) Find $\operatorname { Var } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q5 [7]}}