| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| 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 probability distribution question requiring basic recall and application of standard formulas. Part (i) involves summing probabilities to find k (routine algebra), and part (ii) applies standard E(X) and Var(X) formulas with no conceptual challenges. The calculations are mechanical with no problem-solving insight needed, making it easier than average for A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(2k + 4k + 6k + 8k = 1\) with \(k = \frac{1}{20}\) AND \(6 \times \frac{1}{20} = \frac{3}{10}\) | M1, A1 | Must see correct wk'g for \(k=\frac{1}{20}\), otherwise M0A0. NB \(k×6 = \frac{3}{10} \Rightarrow k = \frac{1}{20}\) M0A0. Even if tested by showing that \(k = \frac{1}{20}\) gives \(\Sigma p=1\). Just showing \(\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = 1\) M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2× \frac{1}{10} + 4× \frac{2}{10} + 6 × \frac{3}{10} + 8 × \frac{4}{10}\) oe = 6 | M1, A1, [5] | ≥ 3 terms correct; ft their values of \(p\), dep \(\Sigma p = 1\). cao. \(2^3× \frac{1}{10} +4^2× \frac{2}{10} + 6^2× \frac{3}{10} + 8^2× \frac{4}{10}\) oe (= 10) \(-'6'^2\) |
## (i)
$2k + 4k + 6k + 8k = 1$ with $k = \frac{1}{20}$ AND $6 \times \frac{1}{20} = \frac{3}{10}$ | M1, A1 | Must see correct wk'g for $k=\frac{1}{20}$, otherwise M0A0. NB $k×6 = \frac{3}{10} \Rightarrow k = \frac{1}{20}$ M0A0. Even if tested by showing that $k = \frac{1}{20}$ gives $\Sigma p=1$. Just showing $\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = 1$ M0A0 | or $2k + 4k + 6k + 8k = 20k$ with M1 or P(X = 6) = $\frac{6k}{20k} = \frac{3}{10}$ | M1, A1 | Allow i.t.o. $k$ for M1
## (ii)
$2× \frac{1}{10} + 4× \frac{2}{10} + 6 × \frac{3}{10} + 8 × \frac{4}{10}$ oe = 6 | M1, A1, [5] | ≥ 3 terms correct; ft their values of $p$, dep $\Sigma p = 1$. cao. $2^3× \frac{1}{10} +4^2× \frac{2}{10} + 6^2× \frac{3}{10} + 8^2× \frac{4}{10}$ oe (= 10) $-'6'^2$ | M1, A1 | ≥ 3 terms correct; ft their values of $p$; dep $\Sigma p = 1$. ft their values of $p$; dep +ve result & $\Sigma p = 1$ cao | M1, A1 | Allow ito $k$ for M1M1; NOT $-m^2 ÷ 4$ ✓ 4 = 2 lose final A1, not ISW, unless labelled sd | Allow i.t.o. $k$ for M1M1 ÷ 4 M0
---When a four-sided spinner is spun, the number on which it lands is denoted by $X$, where $X$ is a random variable taking values 2, 4, 6 and 8. The spinner is biased so that P($X = x$) = $kx$, where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that P($X = 6$) = $\frac{3}{10}$. [2]
\item Find E($X$) and Var($X$). [5]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q1 [7]}}