| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.3 This is a straightforward S1 probability distribution question requiring routine algebraic manipulation to find k by summing probabilities to 1, then standard application of expectation and variance formulas. The partial fractions structure makes calculations simpler than they initially appear, but no novel insight is required—just careful arithmetic and formula recall. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{k}{2} + \frac{k}{6} + \frac{k}{12} + \frac{k}{20} + \frac{k}{30} = 1\) | M1 | For correct equation including \(=1\); Allow substitution of \(k=1.2\) to show probabilities add to 1 with convincing working |
| \((30+10+5+3+2)\frac{k}{60} = 1\) | A1 | Need one further intermediate step after equation |
| \(50k = 60\) | NB Answer Given | |
| \(k = 1.2\) | ||
| Table: \(r\): 2, 3, 4, 5, 6; \(P(X=r)\): \(0.6, 0.2, 0.1, 0.06, 0.04\) (i.e. \(\frac{3}{5}, \frac{1}{5}, \frac{1}{10}, \frac{3}{50}, \frac{1}{25}\)) | B1 [3] | Complete correct table in fraction or decimal form NOT in terms of \(k\); Must tabulate probabilities; If fractions any denominator is ok provided numerators are integers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (2\times0.6)+(3\times0.2)+(4\times0.1)+(5\times0.06)+(6\times0.04)\) | M1 | For \(\Sigma rp\) (at least 3 terms correct) provided 5 reasonable probabilities seen; CAO; Use of \(E(X-\mu)^2\) gets M1 for attempt at \((x-\mu)^2\) |
| \(E(X) = 2.74\) or \(\frac{137}{50}\) | A1 | If probs wrong but sum \(=1\) allow max M1A0M1A1; If sum \(\neq 1\) allow max M1A0M1A0 (provided all probabilities \(\geq 0\) and \(<1\)) |
| \(E(X^2) = (4\times0.6)+(9\times0.2)+(16\times0.1)+(25\times0.06)+(36\times0.04)\) | M1* | For \(\Sigma r^2 p\) (at least 3 terms correct); Division by 5 or other spurious value at end and/or rooting final answer gives max M1A1M1A1A0 |
| \(= 8.74\) or \(\frac{437}{50}\) | ||
| \(Var(X) = 8.74 - 2.74^2 = 1.23\) or \(1.232\) or \(\frac{3081}{2500}\) | M1* dep A1 [5] | for \(-\) their \(E(X)^2\); FT their \(E(X)\) provided \(Var(X)>0\); Condone 1.2324 despite the fact that this is over-specified since it is the exact answer |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{k}{2} + \frac{k}{6} + \frac{k}{12} + \frac{k}{20} + \frac{k}{30} = 1$ | M1 | For correct equation including $=1$; Allow substitution of $k=1.2$ to show probabilities add to 1 with convincing working |
| $(30+10+5+3+2)\frac{k}{60} = 1$ | A1 | Need one further intermediate step after equation |
| $50k = 60$ | | **NB Answer Given** |
| $k = 1.2$ | | |
| Table: $r$: 2, 3, 4, 5, 6; $P(X=r)$: $0.6, 0.2, 0.1, 0.06, 0.04$ (i.e. $\frac{3}{5}, \frac{1}{5}, \frac{1}{10}, \frac{3}{50}, \frac{1}{25}$) | B1 [3] | Complete correct table in fraction or decimal form NOT in terms of $k$; Must tabulate probabilities; If fractions any denominator is ok provided numerators are integers |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = (2\times0.6)+(3\times0.2)+(4\times0.1)+(5\times0.06)+(6\times0.04)$ | M1 | For $\Sigma rp$ (at least 3 terms correct) provided 5 reasonable probabilities seen; CAO; Use of $E(X-\mu)^2$ gets M1 for attempt at $(x-\mu)^2$ |
| $E(X) = 2.74$ or $\frac{137}{50}$ | A1 | If probs wrong but sum $=1$ allow max M1A0M1A1; If sum $\neq 1$ allow max M1A0M1A0 (provided all probabilities $\geq 0$ and $<1$) |
| $E(X^2) = (4\times0.6)+(9\times0.2)+(16\times0.1)+(25\times0.06)+(36\times0.04)$ | M1* | For $\Sigma r^2 p$ (at least 3 terms correct); Division by 5 or other spurious value at end and/or rooting final answer gives max M1A1M1A1A0 |
| $= 8.74$ or $\frac{437}{50}$ | | |
| $Var(X) = 8.74 - 2.74^2 = 1.23$ or $1.232$ or $\frac{3081}{2500}$ | M1* dep A1 [5] | for $-$ their $E(X)^2$; FT their $E(X)$ provided $Var(X)>0$; Condone 1.2324 despite the fact that this is over-specified since it is the exact answer |
---4 The probability distribution of the random variable $X$ is given by the formula
$$\mathrm { P } ( X = r ) = \frac { k } { r ( r - 1 ) } \text { for } r = 2,3,4,5,6 .$$
(i) Show that the value of $k$ is 1.2 . Using this value of $k$, show the probability distribution of $X$ in a table.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2016 Q4 [8]}}