| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Moderate -0.3 This is a straightforward S2 question requiring standard techniques: finding k by integrating the PDF to equal 1, then calculating E(X) and Var(X) using standard formulas. The linear PDF and simple limits make the integration routine. Slightly easier than average due to the geometric hint and mechanical nature of the calculations. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks |
|---|---|
| (a) Graph: straight line from \((1, k)\) to \((4, 4k)\); on \(x\)-axis elsewhere | B2 |
| Area of trapezium \(= \frac{1}{2} \times 3 \times (k + 4k) = 1\), so \(k = \frac{2}{15}\) | M1 A1 |
| (b) \(E(X) = \int_1^4 \frac{2}{15}x^2 \, dx = 2.8\); \(E(X^2) = \int_1^4 \frac{2}{15}x^3 \, dx = 8.5\) | M1 A1 M1 A1 |
| \(\text{Var}(X) = 8.5 - 2.8^2 = 0.66\) | M1 A1 |
(a) Graph: straight line from $(1, k)$ to $(4, 4k)$; on $x$-axis elsewhere | B2 |
Area of trapezium $= \frac{1}{2} \times 3 \times (k + 4k) = 1$, so $k = \frac{2}{15}$ | M1 A1 |
(b) $E(X) = \int_1^4 \frac{2}{15}x^2 \, dx = 2.8$; $E(X^2) = \int_1^4 \frac{2}{15}x^3 \, dx = 8.5$ | M1 A1 M1 A1 |
$\text{Var}(X) = 8.5 - 2.8^2 = 0.66$ | M1 A1 |
**Total: 10 marks**
---4. A continuous random variable $X$ has probability density function
$$\begin{array} { l l }
\mathrm { f } ( x ) = 0 & x < 1 , \\
\mathrm { f } ( x ) = k x & 1 \leq x \leq 4 , \\
\mathrm { f } ( x ) = 0 & x > 4 .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch a graph of $\mathrm { f } ( x )$, and hence find the value of $k$.
\item Calculate the mean and the variance of $X$.
\section*{STATISTICS 2 (A)TEST PAPER 4 Page 2}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q4 [10]}}