| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward probability distribution question requiring only basic recall: probabilities sum to 1 gives a simple linear equation (6p = 1), then standard application of expectation and variance formulas. No problem-solving insight needed, just routine calculation with given formulas. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2p + p + 3p = 1\) | M1 | Equation involving \(p\)s and summing to 1 |
| \(p = 1/6\ (= 0.167)\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = -2 \times 2/6 + 0 + 4 \times 3/6\) | M1 | Using correct formula for \(E(X)\), in terms of \(p\) or their \(p < 1\) |
| \(= 4/3\ (= 1.33)\) | A1ft | Correct expectation ft on their \(p\) if \(p \leq 1/3\) |
| \(\text{Var}(X) = 4 \times 2/6 + 0 + 16 \times 3/6 - (4/3)^2\) | M1 | Substitution in their \(\Sigma px^2\) – their \(E^2(X)\), need 2 terms |
| \(= 7.56\ (68/9)\) | A1 [4] | Correct answer |
## Question 2:
**Part (i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2p + p + 3p = 1$ | M1 | Equation involving $p$s and summing to 1 |
| $p = 1/6\ (= 0.167)$ | A1 **[2]** | Correct answer |
**Part (ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = -2 \times 2/6 + 0 + 4 \times 3/6$ | M1 | Using correct formula for $E(X)$, in terms of $p$ or their $p < 1$ |
| $= 4/3\ (= 1.33)$ | A1ft | Correct expectation ft on their $p$ if $p \leq 1/3$ |
| $\text{Var}(X) = 4 \times 2/6 + 0 + 16 \times 3/6 - (4/3)^2$ | M1 | Substitution in their $\Sigma px^2$ – their $E^2(X)$, need 2 terms |
| $= 7.56\ (68/9)$ | A1 **[4]** | Correct answer |
---2 The random variable $X$ takes the values $- 2,0$ and 4 only. It is given that $\mathrm { P } ( X = - 2 ) = 2 p , \mathrm { P } ( X = 0 ) = p$ and $\mathrm { P } ( X = 4 ) = 3 p$.\\
(i) Find $p$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{CAIE S1 2007 Q2 [6]}}