| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Find parameter n from mean |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring recall of the discrete uniform distribution formula E(X) = (n+1)/2, then simple algebraic manipulation to find n=9. Parts (b) and (c) involve basic probability counting and applying the variance formula. All steps are routine with no problem-solving insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02e Discrete uniform distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | ||
| \(E[X] = \sum x \cdot P[X = x] = \frac{1}{n} + \frac{2}{n} + \ldots + \frac{n}{n}\) | M1 | Use of \(E(X)\) |
| \(= \frac{1}{n}[1 + 2 + \ldots + n]\) | ||
| \(= \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}\) | M1 | Use of \(\sum\) formula |
| \(\therefore \frac{n+1}{2} = 5 \Rightarrow n = 9\) | A1 | c.s.o |
| Part (b) | ||
| \(P(X < T) = \frac{1}{9} \times 6 = \frac{2}{3}\) | M1, A1 | (2 marks) |
| Part (c) | ||
| \(\text{Var}[X] = E[X^2] - [E[X]]^2\) | M1 | Use of Var\((X)\) |
| \(= \frac{1^2}{9} + \frac{2^2}{9} + \ldots + \frac{9^2}{9} - 5^2\) | M1 | Use of \(\sum n^3\) |
| \(= \frac{1}{9} \cdot \frac{1}{6} \times 9 \times 10 \times 19 - 5^2\) | ||
| Correct | A1 | |
| \(= \frac{20}{3}\) | A1 | (4 marks) |
| Answer | Marks |
|---|---|
| \(\text{Var}(X) = \frac{n^2 - 1}{12} = \frac{80}{12} = \frac{20}{3}\) | M2, A1, A1 |
**Part (a)** | |
$E[X] = \sum x \cdot P[X = x] = \frac{1}{n} + \frac{2}{n} + \ldots + \frac{n}{n}$ | M1 | Use of $E(X)$ |
$= \frac{1}{n}[1 + 2 + \ldots + n]$ | |
$= \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$ | M1 | Use of $\sum$ formula |
$\therefore \frac{n+1}{2} = 5 \Rightarrow n = 9$ | A1 | c.s.o | (3 marks)
**Part (b)** | |
$P(X < T) = \frac{1}{9} \times 6 = \frac{2}{3}$ | M1, A1 | (2 marks)
**Part (c)** | |
$\text{Var}[X] = E[X^2] - [E[X]]^2$ | M1 | Use of Var$(X)$ |
$= \frac{1^2}{9} + \frac{2^2}{9} + \ldots + \frac{9^2}{9} - 5^2$ | M1 | Use of $\sum n^3$ |
$= \frac{1}{9} \cdot \frac{1}{6} \times 9 \times 10 \times 19 - 5^2$ | |
Correct | A1 |
$= \frac{20}{3}$ | A1 | (4 marks)
OR
$\text{Var}(X) = \frac{n^2 - 1}{12} = \frac{80}{12} = \frac{20}{3}$ | M2, A1, A1 |
---5. The random variable $X$ has the discrete uniform distribution
$$\mathrm { P } ( X = x ) = \frac { 1 } { n } , \quad x = 1,2 , \ldots , n$$
Given that $\mathrm { E } ( X ) = 5$,
\begin{enumerate}[label=(\alph*)]
\item show that $n = 9$.
Find
\item $\mathrm { P } ( X < 7 )$,
\item $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q5 [9]}}