| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(aX+b) or Var(aX+b) given distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability distribution properties: finding k using ΣP(X=x)=1, then applying standard formulas for E(aX+b) and Var(aX+b). All steps are routine calculations with no problem-solving insight required, making it easier than average but not trivial due to the arithmetic involved. |
| 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 |
|---|---|---|
| \(k(1 + 2 + 3 + 4 + 5) = 1 \Rightarrow k = \frac{1}{15}\) | M1 A1 A1 | Use of \(\sum P(X = x) = 1\) (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \frac{1}{15}(1 \times 2 + 2 \times 2 + \ldots + 5 \times 5) = 15\) | M1 A1 A1 | Use of \(E(X) = \sum xP(X = x)\) |
| \(E(2X + 3) = 2E(X) + 3 = 31/3\) | M1 A1 ft | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = \frac{1}{15}(1 \times 2^2 \times 2 + \ldots + 5^2 \times 5) = 15\) | M1 A1 | Use of \(E(X^2) = \sum x^2 P(X = x)\) |
| \(\text{Var}(X) = 15 - \left(\frac{11}{3}\right)^2 = \frac{14}{9}\) | M1 A1 | Use of \(\text{Var}(X) = E(X^2) - [E(X)]^2\) |
| \(\text{Var}(2X - 4) = 4\text{Var}(X) = \frac{56}{9}\) | M1 A1 ft | Use of \(\text{Var}(aX) = a^2\text{Var}(X)\) (6 marks) |
## (a)
$k(1 + 2 + 3 + 4 + 5) = 1 \Rightarrow k = \frac{1}{15}$ | M1 A1 A1 | Use of $\sum P(X = x) = 1$ (3 marks)
## (b)
$E(X) = \frac{1}{15}(1 \times 2 + 2 \times 2 + \ldots + 5 \times 5) = 15$ | M1 A1 A1 | Use of $E(X) = \sum xP(X = x)$
$E(2X + 3) = 2E(X) + 3 = 31/3$ | M1 A1 ft | (5 marks)
## (c)
$E(X^2) = \frac{1}{15}(1 \times 2^2 \times 2 + \ldots + 5^2 \times 5) = 15$ | M1 A1 | Use of $E(X^2) = \sum x^2 P(X = x)$
$\text{Var}(X) = 15 - \left(\frac{11}{3}\right)^2 = \frac{14}{9}$ | M1 A1 | Use of $\text{Var}(X) = E(X^2) - [E(X)]^2$
$\text{Var}(2X - 4) = 4\text{Var}(X) = \frac{56}{9}$ | M1 A1 ft | Use of $\text{Var}(aX) = a^2\text{Var}(X)$ (6 marks)
**(14 marks)**
---The discrete random variable $X$ has probability function
$$P(X = x) = \begin{cases} kx, & x = 1, 2, 3, 4, 5, \\ 0, & \text{otherwise.} \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{1}{15}$. [3]
\end{enumerate}
Find the value of
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item E$(2X + 3)$, [5]
\item Var$(2X - 4)$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [14]}}