| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| 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 discrete probability distribution question requiring construction of a sample space table and variance calculation using standard formulas. While it involves multiple steps (creating the distribution table, finding E(X), E(X²), then Var(X)), each step uses routine S1 techniques with no conceptual challenges or novel insights required. The unusual die faces add minor complexity but the methodology is entirely standard. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | \(-2\) | \(-1\) |
| \(P(X=x)\) | \(\frac{2}{18}\) | \(\frac{4}{18}\) |
| B1 | \(-2, -1, 0, 1, 2, 3\) seen as top line of table OR attempting to evaluate \(P(-2), P(-1), P(0), P(1), P(2), P(3)\) | |
| B1 | At least 4 probabilities correct (need not be in table) | |
| B1 | All probabilities correct in a table |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{-4-4+0+4+4+3}{18} = \frac{1}{6}\) | M1 | Correct unsimplified expression for mean using their table, \(\Sigma p = 1\), may be implied |
| \(Var(X) = \frac{8+4+0+4+8+9}{18} - \left(\frac{1}{6}\right)^2 = 11/6 - 1/36\) | M1 | Correct unsimplified expression for variance using their table, and their mean² subtracted. Allow \(\Sigma p \neq 1\) |
| \(= 65/36\ (1.81)\) | A1 | Correct answer |
## Question 2(i):
| $x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|---|
| $P(X=x)$ | $\frac{2}{18}$ | $\frac{4}{18}$ | $\frac{5}{18}$ | $\frac{4}{18}$ | $\frac{2}{18}$ | $\frac{1}{18}$ |
| B1 | $-2, -1, 0, 1, 2, 3$ seen as top line of table OR attempting to evaluate $P(-2), P(-1), P(0), P(1), P(2), P(3)$
| B1 | At least 4 probabilities correct (need not be in table)
| B1 | All probabilities correct in a table
**Total: 3 marks**
## Question 2(ii):
$E(X) = \frac{-4-4+0+4+4+3}{18} = \frac{1}{6}$ | M1 | Correct unsimplified expression for mean using their table, $\Sigma p = 1$, may be implied
$Var(X) = \frac{8+4+0+4+8+9}{18} - \left(\frac{1}{6}\right)^2 = 11/6 - 1/36$ | M1 | Correct unsimplified expression for variance using their table, and their mean² subtracted. Allow $\Sigma p \neq 1$
$= 65/36\ (1.81)$ | A1 | Correct answer
---2 A fair 6 -sided die has the numbers $- 1 , - 1,0,0,1,2$ on its faces. A fair 3 -sided spinner has edges numbered $- 1,0,1$. The die is thrown and the spinner is spun. The number on the uppermost face of the die and the number on the edge on which the spinner comes to rest are noted. The sum of these two numbers is denoted by $X$.\\
(i) Draw up a table showing the probability distribution of $X$.\\
(ii) Find $\operatorname { Var } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2018 Q2 [6]}}