| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Find parameter n from mean |
| Difficulty | Standard +0.3 This is a straightforward S1 question testing standard discrete distribution properties. Part (a) uses the uniform distribution mean formula (routine algebra), part (b) is basic probability calculation, part (c) applies standard linear transformation rules for mean/variance (direct recall), and parts (d)-(e) involve summing a simple arithmetic series and calculating probabilities. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02e Discrete uniform distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((n + 1)/2 = 10\) → \(n = 19\) | M1 A1 | |
| \(\text{Var}(X) = (19^2 - 1)/12 = 30\) | M1 A1 | |
| (b) \(P(X = 4, 5 \text{ or } 6) = \frac{3}{19}\) | M1 A1 | |
| (c) \(E(Y) = 0\), \(\text{Var}(Y) = 270\) | B1 M1 A1 | |
| (d) \(k(2 + 3 + \ldots + 20) = 1\) → \(209k = 1\) → \(k = \frac{1}{209}\) | M1 A1 A1 | |
| (e) \(P(3 < X \leq 6) = f(4) + f(5) + f(6) = \frac{1}{209}(5 + 6 + 7) = \frac{18}{209}\) | M1 A1 A1 | 15 marks total |
(a) $(n + 1)/2 = 10$ → $n = 19$ | M1 A1 |
$\text{Var}(X) = (19^2 - 1)/12 = 30$ | M1 A1 |
(b) $P(X = 4, 5 \text{ or } 6) = \frac{3}{19}$ | M1 A1 |
(c) $E(Y) = 0$, $\text{Var}(Y) = 270$ | B1 M1 A1 |
(d) $k(2 + 3 + \ldots + 20) = 1$ → $209k = 1$ → $k = \frac{1}{209}$ | M1 A1 A1 |
(e) $P(3 < X \leq 6) = f(4) + f(5) + f(6) = \frac{1}{209}(5 + 6 + 7) = \frac{18}{209}$ | M1 A1 A1 | 15 marks totalThe random variable $X$, which can take any value from $\{1, 2, \ldots, n\}$, is modelled by the discrete uniform distribution with mean 10.
\begin{enumerate}[label=(\alph*)]
\item Show that $n = 19$ and find the variance of $X$. [4 marks]
\item Find $\text{P}(3 < X \leq 6)$. [2 marks]
\end{enumerate}
The random variable $Y$ is defined by $Y = 3(X - 10)$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item State the mean and the variance of $Y$. [3 marks]
\end{enumerate}
The model for the distribution of $X$ is found to be unsatisfactory, and in a refined model the probability distribution of $X$ is taken to be
$$\text{f}(x) = \begin{cases} k(x + 1) & x = 1, 2, \ldots, 19, \\ 0 & \text{otherwise}. \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Show that $k = \frac{1}{209}$. [3 marks]
\item Find $\text{P}(3 < X \leq 6)$ using this model. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}