| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| 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 S1 question requiring routine application of standard formulas: summing probabilities to find k, then calculating E(X) and Var(X) using definitions. The algebra is simple (evaluating r²-1 for four values), and all steps follow directly from basic probability distribution properties with no problem-solving insight needed. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | For \(\sum p\) (at least 3 terms correct). If probs wrong but sum \(= 1\) allow full marks here. If sum \(\neq 1\) allow max M1A0M1 M0A0 (provided all probabilities between 0 and 1). Or it to k. NB \(E(X) = 210k\), \(E(X^2) = 924k\) gets M1A0M1M0A0. \(E(X) = 210k\), Var\((X) = 924k - (210k)^2\) gets M1A0M1M1A0. |
| A1 | cao |
| M1 | For \(\sum p\) (at least 3 terms correct) |
| M1 | dep for – their \(E(X)^2\) |
| A1 | FT their \(E(X)\) provided Var\((X) > 0\) (and of course \(E(X^2)\) is correct) |
| [5] | Use of \(E(X - \mu)^2\) gets M1 for attempt at \((x - \mu)^2\) should see \((-2.2)^2, (-1.2)^2, (-0.2)^2, 0.8^2\) (if \(E(X)\) wrong FT their \(E(X)\)) (all 4 correct for M1), then M1 for \(\sum p(x - \mu)^2\) (at least 3 terms correct with their probabilities). Division by 4 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if \(E(X)\) also divided by 4. Unsupported correct answers get 5 marks |
## Part (ii)
**Answer:** $E(X) = (2 \times 0.06) + (3 \times 0.16) + (4 \times 0.30) + (5 \times 0.48) = 4.2$ or $\frac{21}{5}$
$E(X^2) = (4 \times 0.06) + (9 \times 0.16) + (16 \times 0.30) + (25 \times 0.48) = 18.48$
$\text{Var}(X) = 18.48 - 4.2^2 = 0.84 = \frac{21}{25}$
| **Marks** | **Guidance** |
|-----------|--------------|
| M1 | For $\sum p$ (at least 3 terms correct). If probs wrong but sum $= 1$ allow full marks here. If sum $\neq 1$ allow max M1A0M1 M0A0 (provided all probabilities between 0 and 1). Or it to k. NB $E(X) = 210k$, $E(X^2) = 924k$ gets M1A0M1M0A0. $E(X) = 210k$, Var$(X) = 924k - (210k)^2$ gets M1A0M1M1A0. |
| A1 | cao |
| M1 | For $\sum p$ (at least 3 terms correct) |
| M1 | dep for – their $E(X)^2$ |
| A1 | FT their $E(X)$ provided Var$(X) > 0$ (and of course $E(X^2)$ is correct) |
| **[5]** | Use of $E(X - \mu)^2$ gets M1 for attempt at $(x - \mu)^2$ should see $(-2.2)^2, (-1.2)^2, (-0.2)^2, 0.8^2$ (if $E(X)$ wrong FT their $E(X)$) (all 4 correct for M1), then M1 for $\sum p(x - \mu)^2$ (at least 3 terms correct with their probabilities). Division by 4 or other spurious value at end gives max M1A1M1M1A0, or M1A0M1M1A0 if $E(X)$ also divided by 4. Unsupported correct answers get 5 marks |
---2 The probability distribution of the random variable $X$ is given by the formula
$$\mathrm { P } ( X = r ) = k \left( r ^ { 2 } - 1 \right) \text { for } r = 2,3,4,5 .$$
(i) Show the probability distribution in a table, and find the value of $k$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2013 Q2 [8]}}