| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability distribution calculations. Part (a) requires routine application of the variance formula with simple arithmetic (E(W)=2, E(W²)=6, Var(W)=2). Part (b)(i) uses ∑P=1 to find k, and (b)(ii) applies the expectation formula. All steps are standard textbook exercises with no problem-solving or insight required, making it easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(w\) | 0 | 2 | 4 |
| \(\mathrm { P } ( W = w )\) | 0.3 | 0.4 | 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0^2 \times 0.3) + 2^2 \times 0.4 + 4^2 \times 0.3\) | M1 | last two terms correct. NOT \(\div 6\) or \(\div 3\) |
| \(-2^2\) or \(-4\) | M1 | allow \(-(\text{any number})^2\), dep +ve result |
| \(= 2.4\) | A1 | \(\div 3\) M0M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2k + 3k + 4k + 5k = 1\) oe; \(\left(k = \frac{1}{14}\right)\) | B1 | or \(14k = 1\) oe; "\(= 1\)" is essential. NOT just \(2+3+4+5=14\) so \(k=\frac{1}{14}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2}{14}, \frac{3}{14}, \frac{4}{14}, \frac{5}{14}\) or \(\frac{2}{14}, \frac{6}{14}, \frac{12}{14}, \frac{20}{14}\) | B1 | \(\geq 3\) correct |
| \(\Sigma xp\) | M1 | \(\geq 3\) correct terms added |
| \(= \frac{20}{7}\) or \(2\frac{6}{7}\) or 2.86 (3 sf) | A1 | SC \(1 \times \frac{1}{14} + 2 \times \frac{2}{14} + 3 \times \frac{3}{14} + 4 \times \frac{4}{14}\) \((= 2.143)\) B0M1A0 |
# Question 2:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0^2 \times 0.3) + 2^2 \times 0.4 + 4^2 \times 0.3$ | M1 | last two terms correct. NOT $\div 6$ or $\div 3$ |
| $-2^2$ or $-4$ | M1 | allow $-(\text{any number})^2$, dep +ve result |
| $= 2.4$ | A1 | $\div 3$ M0M0A0 |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2k + 3k + 4k + 5k = 1$ oe; $\left(k = \frac{1}{14}\right)$ | B1 | or $14k = 1$ oe; "$= 1$" is essential. NOT just $2+3+4+5=14$ so $k=\frac{1}{14}$ |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2}{14}, \frac{3}{14}, \frac{4}{14}, \frac{5}{14}$ or $\frac{2}{14}, \frac{6}{14}, \frac{12}{14}, \frac{20}{14}$ | B1 | $\geq 3$ correct |
| $\Sigma xp$ | M1 | $\geq 3$ correct terms added |
| $= \frac{20}{7}$ or $2\frac{6}{7}$ or 2.86 (3 sf) | A1 | SC $1 \times \frac{1}{14} + 2 \times \frac{2}{14} + 3 \times \frac{3}{14} + 4 \times \frac{4}{14}$ $(= 2.143)$ B0M1A0 |
---2
\begin{enumerate}[label=(\alph*)]
\item The probability distribution of a random variable $W$ is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$w$ & 0 & 2 & 4 \\
\hline
$\mathrm { P } ( W = w )$ & 0.3 & 0.4 & 0.3 \\
\hline
\end{tabular}
\end{center}
Calculate $\operatorname { Var } ( W )$.
\item The random variable $X$ has probability distribution given by
$$\mathrm { P } ( X = x ) = k ( x + 1 ) \quad \text { for } x = 1,2,3,4 .$$
\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac { 1 } { 14 }$.
\item Calculate $\mathrm { E } ( X )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR S1 2014 Q2 [7]}}