| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Derive or verify variance formula |
| Difficulty | Moderate -0.3 Parts (a)-(c) are straightforward applications of uniform distribution formulas requiring minimal calculation. Parts (d)-(e) involve standard integration and variance derivation that are textbook exercises for S2, though the algebraic manipulation in part (e) requires some care. Overall slightly easier than average due to being a routine multi-part question with clear structure and standard techniques. |
| Spec | 5.02e Discrete uniform distribution5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(6.5\) | A1 |
| (b) | \(2.4 \times \frac{5}{9} = \frac{4}{15}\) or \(0.2667 \text{ (4sf)}\) | M1 A1 |
| (c) | \(= P(3 < X < 7) = 4 \times \frac{5}{9} = \frac{20}{9}\) or \(0.4444 \text{ (4sf)}\) | M1 A1 |
| (d) | \(f(y) = \frac{1}{b-a}\) | B1 |
| \(E(Y^2) = \int_a^b \frac{1}{b-a}y^2 \, dy\) | M1 | |
| \(= \frac{1}{b-a}\left[\frac{1}{3}y^3\right]_a^b\) | A1 | |
| \(= \frac{b^3 - a^3}{3(b-a)}\) | M1 | |
| \(= \frac{1}{3}(b^2 + ab + a^2)\) | A1 | |
| (e) | \(\text{Var}(Y) = E(Y^2) - [E(Y)]^2\) | M1 |
| \(= \frac{1}{3}(b^2 + ab + a^2) - \frac{1}{4}(a^2 + 2ab + b^2)\) | M1 | |
| \(= \frac{1}{12}(4b^2 + 4ab + 4a^2 - 3a^2 - 6ab - 3b^2)\) | M1 | |
| \(= \frac{1}{12}(b^2 - 2ab + a^2) = \frac{1}{12}(b - a)^2\) | A1 | (14) |
(a) | $6.5$ | A1 |
(b) | $2.4 \times \frac{5}{9} = \frac{4}{15}$ or $0.2667 \text{ (4sf)}$ | M1 A1 |
(c) | $= P(3 < X < 7) = 4 \times \frac{5}{9} = \frac{20}{9}$ or $0.4444 \text{ (4sf)}$ | M1 A1 |
(d) | $f(y) = \frac{1}{b-a}$ | B1 |
| $E(Y^2) = \int_a^b \frac{1}{b-a}y^2 \, dy$ | M1 |
| $= \frac{1}{b-a}\left[\frac{1}{3}y^3\right]_a^b$ | A1 |
| $= \frac{b^3 - a^3}{3(b-a)}$ | M1 |
| $= \frac{1}{3}(b^2 + ab + a^2)$ | A1 |
(e) | $\text{Var}(Y) = E(Y^2) - [E(Y)]^2$ | M1 |
| $= \frac{1}{3}(b^2 + ab + a^2) - \frac{1}{4}(a^2 + 2ab + b^2)$ | M1 |
| $= \frac{1}{12}(4b^2 + 4ab + 4a^2 - 3a^2 - 6ab - 3b^2)$ | M1 |
| $= \frac{1}{12}(b^2 - 2ab + a^2) = \frac{1}{12}(b - a)^2$ | A1 | (14)
**Total: (75)**The random variable $X$ follows a continuous uniform distribution over the interval $[2, 11]$.
\begin{enumerate}[label=(\alph*)]
\item Write down the mean of $X$. [1 mark]
\item Find P($X \geq 8.6$). [2 marks]
\item Find P($|X - 5| < 2$). [2 marks]
\end{enumerate}
The random variable $Y$ follows a continuous uniform distribution over the interval $[a, b]$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Show by integration that
$$\text{E}(Y^2) = \frac{1}{3}(b^2 + ab + a^2).$$ [5 marks]
\item Hence, prove that
$$\text{Var}(Y) = \frac{1}{12}(b - a)^2.$$
You may assume that E($Y$) = $\frac{1}{2}(a + b)$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [14]}}