| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Interquartile range calculation |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question with routine calculations. Parts (a)-(d) involve straightforward sketching, integration for E(X) and Var(X), and finding the CDF. Part (e) requires solving quadratic equations from the CDF to find quartiles, which is a standard technique. All steps follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks |
|---|---|
| Graph with correct shape and key points labeled: \(\frac{3}{8}\) on y-axis, points at \((0, O)\), \((2, \frac{1}{8})\), \((6, \frac{3}{8})\) | B2 |
| \(E(X) = \int_2^6 x \cdot \frac{1}{16}x \, dx = \frac{1}{16}\int_2^6 x^2 \, dx\) | M1 |
| \(= \frac{1}{48}[x^3]_2^6 = \frac{1}{48}(216 - 8) = \frac{13}{3}\) | M1 A1 |
| \(E(X^2) = \int_2^6 x^2 \cdot \frac{1}{16}x \, dx = \frac{1}{16}\int_2^6 x^3 \, dx\) | M1 |
| \(= \frac{1}{64}[x^4]_2^6 = \frac{1}{64}(1296 - 16) = 20\) | M1 A1 |
| \(\therefore \text{Var}(X) = 20 - (\frac{13}{3})^2 = \frac{11}{9}\) | M1 A1 |
| \(F(t) = \int_2^t \frac{1}{16}x \, dx\) | M1 |
| \(= \frac{1}{32}[x^2]_2^t = \frac{1}{32}(t^2 - 4)\) | M1 A1 |
| \[F(x) = \begin{cases} 0, & x < 2, \\ \frac{1}{32}(x^2 - 4) & 2 \leq x \leq 6, \\ 1, & x > 6. \end{cases}\] | A1 |
| \(F(Q_1) = \frac{1}{4} \therefore \frac{1}{32}(x^2 - 4) = \frac{1}{4}\) | M1 |
| \(x^2 - 4 = 8; x^2 = 12; x = \pm 2\sqrt{3}; 2 \leq x \leq 6\) so \(Q_1 = 2\sqrt{3}\) | M1 A1 |
| \(F(Q_3) = \frac{3}{4} \therefore \frac{1}{32}(x^2 - 4) = \frac{3}{4}\) | M1 |
| \(x^2 - 4 = 24; x^2 = 28; x = \pm 2\sqrt{7}; 2 \leq x \leq 6\) so \(Q_3 = 2\sqrt{7}\) | A1 |
| \(\therefore \text{IQR} = 2\sqrt{7} - 2\sqrt{3} = 2(\sqrt{7} - \sqrt{3})\) | A1 |
Graph with correct shape and key points labeled: $\frac{3}{8}$ on y-axis, points at $(0, O)$, $(2, \frac{1}{8})$, $(6, \frac{3}{8})$ | B2 |
$E(X) = \int_2^6 x \cdot \frac{1}{16}x \, dx = \frac{1}{16}\int_2^6 x^2 \, dx$ | M1 |
$= \frac{1}{48}[x^3]_2^6 = \frac{1}{48}(216 - 8) = \frac{13}{3}$ | M1 A1 |
$E(X^2) = \int_2^6 x^2 \cdot \frac{1}{16}x \, dx = \frac{1}{16}\int_2^6 x^3 \, dx$ | M1 |
$= \frac{1}{64}[x^4]_2^6 = \frac{1}{64}(1296 - 16) = 20$ | M1 A1 |
$\therefore \text{Var}(X) = 20 - (\frac{13}{3})^2 = \frac{11}{9}$ | M1 A1 |
$F(t) = \int_2^t \frac{1}{16}x \, dx$ | M1 |
$= \frac{1}{32}[x^2]_2^t = \frac{1}{32}(t^2 - 4)$ | M1 A1 |
$$F(x) = \begin{cases} 0, & x < 2, \\ \frac{1}{32}(x^2 - 4) & 2 \leq x \leq 6, \\ 1, & x > 6. \end{cases}$$ | A1 |
$F(Q_1) = \frac{1}{4} \therefore \frac{1}{32}(x^2 - 4) = \frac{1}{4}$ | M1 |
$x^2 - 4 = 8; x^2 = 12; x = \pm 2\sqrt{3}; 2 \leq x \leq 6$ so $Q_1 = 2\sqrt{3}$ | M1 A1 |
$F(Q_3) = \frac{3}{4} \therefore \frac{1}{32}(x^2 - 4) = \frac{3}{4}$ | M1 |
$x^2 - 4 = 24; x^2 = 28; x = \pm 2\sqrt{7}; 2 \leq x \leq 6$ so $Q_3 = 2\sqrt{7}$ | A1 |
$\therefore \text{IQR} = 2\sqrt{7} - 2\sqrt{3} = 2(\sqrt{7} - \sqrt{3})$ | A1 |6. The continuous random variable $X$ has the following probability density function:
$$f ( x ) = \begin{cases} \frac { 1 } { 16 } x , & 2 \leq x \leq 6 \\ 0 , & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch $\mathrm { f } ( x )$ for all values of $x$.
\item Find $\mathrm { E } ( X )$.
\item Show that $\operatorname { Var } ( X ) = \frac { 11 } { 9 }$.
\item Define fully the cumulative distribution function $\mathrm { F } ( x )$ of $X$.
\item Show that the interquartile range of $X$ is $2 ( \sqrt { } 7 - \sqrt { 3 } )$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [19]}}