| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Piecewise or conditional probability function |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability distribution question requiring only standard procedures: summing probabilities to find k, calculating E(X) using the definition, computing Var(X) from E(X²) - [E(X)]², and applying the variance transformation rule Var(aX+b) = a²Var(X). All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(k + 2k + 3k + 5k + 6k = 1\) | M1 | Use of \(\sum P(X=x) = 1\) |
| Answer | Marks |
|---|---|
| \(k = \frac{1}{17} = 0.0588\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1 \times \frac{1}{17} + 2 \times \frac{2}{17} + .... + 5 \times \frac{6}{17} = \frac{64}{17}\) | M1 | Use of \(\sum xP(X=x)\) and at least 2 prob correct |
| \(= 3\frac{13}{17}\) | A1 | Do not ignore subsequent working |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 1^2 \times \frac{1}{17} + 2^2 \times \frac{2}{17} + .... + 5^2 \times \frac{6}{17} = \left(\frac{266}{17} = 15.6\right)\) | M1 A1 | Use of \(\sum x^2 P(X=x)\) and at least 2 prob correct |
| \(\text{Var}(X) = \frac{266}{17} - \left(\frac{64}{17}\right)^2\) | M1 | Use of \(\sum x^2 P(X=x) - (E(X))^2\) |
| \(= 1.4740...\) | A1 | awrt 1.47 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(4 - 3X) = 9\ \text{Var}(X) = 9 \times 1.47 = 13.23 \Rightarrow 13.2\), or \(9 \times 1.4740... = 13.266 \Rightarrow 13.3\) | M1 A1 | cao; \(9\ \text{Var}\ X\) |
# Question 5:
## Part (a)
$k + 2k + 3k + 5k + 6k = 1$ | M1 | Use of $\sum P(X=x) = 1$
$17k = 1$
$k = \frac{1}{17} = 0.0588$ | A1 |
## Part (b)
$E(X) = 1 \times \frac{1}{17} + 2 \times \frac{2}{17} + .... + 5 \times \frac{6}{17} = \frac{64}{17}$ | M1 | Use of $\sum xP(X=x)$ and at least 2 prob correct
$= 3\frac{13}{17}$ | A1 | Do not ignore subsequent working
## Part (c)
$E(X^2) = 1^2 \times \frac{1}{17} + 2^2 \times \frac{2}{17} + .... + 5^2 \times \frac{6}{17} = \left(\frac{266}{17} = 15.6\right)$ | M1 A1 | Use of $\sum x^2 P(X=x)$ and at least 2 prob correct
$\text{Var}(X) = \frac{266}{17} - \left(\frac{64}{17}\right)^2$ | M1 | Use of $\sum x^2 P(X=x) - (E(X))^2$
$= 1.4740...$ | A1 | awrt 1.47
## Part (d)
$\text{Var}(4 - 3X) = 9\ \text{Var}(X) = 9 \times 1.47 = 13.23 \Rightarrow 13.2$, or $9 \times 1.4740... = 13.266 \Rightarrow 13.3$ | M1 A1 | cao; $9\ \text{Var}\ X$
---5. The random variable $X$ has probability function
$$P ( X = x ) = \begin{cases} k x , & x = 1,2,3 \\ k ( x + 1 ) , & x = 4,5 \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the exact value of $\mathrm { E } ( X )$.
\item Show that, to 3 significant figures, $\operatorname { Var } ( X ) = 1.47$.
\item Find, to 1 decimal place, $\operatorname { Var } ( 4 - 3 X )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2005 Q5 [10]}}