| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring only routine calculations: summing probabilities to find k, then applying standard formulas for expectation and variance. The arithmetic is simple (small integers) and all steps are mechanical applications of definitions with no problem-solving or insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4k+6k+6k+4k = 1\), \(20k = 1\), \(k = 0.05\) | M1, A1 NB Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 1\times0.2+2\times0.3+3\times0.3+4\times0.2 = 2.5\) (or by inspection) | M1, A1 CAO | M1 for \(\Sigma rp\) (at least 3 terms correct) |
| \(E(X^2) = 1\times0.2+4\times0.3+9\times0.3+16\times0.2 = 7.3\) | M1 | M1 for \(\Sigma r^2p\) (at least 3 terms correct) |
| \(\text{Var}(X) = 7.3 - 2.5^2 = 1.05\) | M1dep, A1 | M1dep for their \(E(X)^2\); A1 FT their \(E(X)\) provided \(\text{Var}(X)>0\) |
## Question 5:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4k+6k+6k+4k = 1$, $20k = 1$, $k = 0.05$ | M1, A1 NB **Answer given** | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 1\times0.2+2\times0.3+3\times0.3+4\times0.2 = 2.5$ (or by inspection) | M1, A1 CAO | M1 for $\Sigma rp$ (at least 3 terms correct) |
| $E(X^2) = 1\times0.2+4\times0.3+9\times0.3+16\times0.2 = 7.3$ | M1 | M1 for $\Sigma r^2p$ (at least 3 terms correct) |
| $\text{Var}(X) = 7.3 - 2.5^2 = 1.05$ | M1dep, A1 | M1dep for their $E(X)^2$; A1 FT their $E(X)$ provided $\text{Var}(X)>0$ |
---5 The probability distribution of the random variable $X$ is given by the formula
$$\mathrm { P } ( X = r ) = k r ( 5 - r ) \text { for } r = 1,2,3,4$$
(i) Show that $k = 0.05$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 Q5 [7]}}