| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.8 This is a very straightforward question requiring only basic probability axioms: (i) use the fact that probabilities sum to 1 to find the missing value by simple arithmetic, (ii) apply the standard expectation formula E(X) = Σx·P(X=x). Both parts are direct recall and calculation with no problem-solving or conceptual challenge. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 3 } { 10 }\) | \(\frac { 1 } { 5 }\) | \(\frac { 2 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - (\frac{1}{10} + \frac{1}{5} + \frac{2}{5})\) | M1 | or \((\frac{1}{10} + \frac{1}{5} + \frac{2}{5}) + p = 1\) |
| \(\sqrt{10} + 2 \times \frac{1}{5} + 3 \times \frac{2}{5}\) | M1 | |
| \(\frac{19}{10}\) oe | A1 | \(\div 4 \text{ or } 6 \Rightarrow \text{M0A0}\) |
**i)**
$1 - (\frac{1}{10} + \frac{1}{5} + \frac{2}{5})$ | M1 | or $(\frac{1}{10} + \frac{1}{5} + \frac{2}{5}) + p = 1$
$\sqrt{10} + 2 \times \frac{1}{5} + 3 \times \frac{2}{5}$ | M1 |
$\frac{19}{10}$ oe | A1 | $\div 4 \text{ or } 6 \Rightarrow \text{M0A0}$
**Total for Question 1:** 4 marks1 Part of the probability distribution of a variable, $X$, is given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & & $\frac { 3 } { 10 }$ & $\frac { 1 } { 5 }$ & $\frac { 2 } { 5 }$ \\
\hline
\end{tabular}
\end{center}
(i) Find $\mathrm { P } ( X = 0 )$.\\
(ii) Find $\mathrm { E } ( X )$.
\hfill \mbox{\textit{OCR S1 2007 Q1 [4]}}