| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Moderate -0.8 This is a straightforward geometric distribution problem with a given probability table. Part (i) requires routine calculation of E(X) and Var(X) using standard formulas with provided probabilities. Part (ii) is a simple linear transformation (multiply by constant). Part (iii) asks for a basic bar chart. All steps are mechanical with no problem-solving or insight required—easier than average A-level. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04a Discrete probability distributions |
| \(r\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | 0.2 | 0.16 | 0.128 | 0.512 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1 \times 0.2 + 2 \times 0.16 + 3 \times 0.128 + 4 \times 0.512 = 2.952\) | M1 for \(\sum rp\) (at least 3 terms correct); A1 cao | Division by 4 or other spurious value at end loses A mark |
| \(E(X^2) = 1 \times 0.2 + 4 \times 0.16 + 9 \times 0.128 + 16 \times 0.512 = 10.184\) | M1 for \(\sum x^2p\) at least 3 terms correct | |
| \(\text{Var}(X) = 10.184 - 2.952^2 = 1.47\) (to 3 s.f.) | M1 for \(E(X^2) - [E(X)]^2\); Provided ans > 0; A1 FT their \(E(X)\) but not a wrong \(E(X^2)\) | |
| TOTAL: 5 |
| Answer | Marks |
|---|---|
| Expected cost \(= 2.952 \times £45000 = £133000\) (3sf) | B1 FT (no extra multiples/divisors introduced at this stage) |
| TOTAL: 1 |
| Answer | Marks |
|---|---|
| [Bar chart showing P(X = r) vs r, with bars at r = 1, 2, 3, 4 with heights approximately 0.2, 0.1, 0.1, 0.4] | G1 labelled linear scales; G1 height of lines |
| TOTAL: 2 |
## Q4(i)
$E(X) = 1 \times 0.2 + 2 \times 0.16 + 3 \times 0.128 + 4 \times 0.512 = 2.952$ | M1 for $\sum rp$ (at least 3 terms correct); A1 cao | Division by 4 or other spurious value at end loses A mark
$E(X^2) = 1 \times 0.2 + 4 \times 0.16 + 9 \times 0.128 + 16 \times 0.512 = 10.184$ | M1 for $\sum x^2p$ at least 3 terms correct |
$\text{Var}(X) = 10.184 - 2.952^2 = 1.47$ (to 3 s.f.) | M1 for $E(X^2) - [E(X)]^2$; Provided ans > 0; A1 FT their $E(X)$ but not a wrong $E(X^2)$ |
| TOTAL: 5 |
## Q4(ii)
Expected cost $= 2.952 \times £45000 = £133000$ (3sf) | B1 FT (no extra multiples/divisors introduced at this stage) |
| TOTAL: 1 |
## Q4(iii)
[Bar chart showing P(X = r) vs r, with bars at r = 1, 2, 3, 4 with heights approximately 0.2, 0.1, 0.1, 0.4] | G1 labelled linear scales; G1 height of lines |
| TOTAL: 2 |
---4 A company is searching for oil reserves. The company has purchased the rights to make test drillings at four sites. It investigates these sites one at a time but, if oil is found, it does not proceed to any further sites. At each site, there is probability 0.2 of finding oil, independently of all other sites.
The random variable $X$ represents the number of sites investigated. The probability distribution of $X$ is shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & 0.2 & 0.16 & 0.128 & 0.512 \\
\hline
\end{tabular}
\end{center}
(i) Find the expectation and variance of $X$.\\
(ii) It costs $\pounds 45000$ to investigate each site. Find the expected total cost of the investigation.\\
(iii) Draw a suitable diagram to illustrate the distribution of $X$.
\hfill \mbox{\textit{OCR MEI S1 2008 Q4 [8]}}