| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | October |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Moderate -0.3 This is a standard S2 piecewise PDF question requiring routine techniques: sketching, using ∫f(x)dx=1 to find k, calculating E(X) by integration, and finding quartiles. All steps are textbook exercises with no novel problem-solving required, making it slightly easier than average A-level maths. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Trapezium shape with height \(2k - 0.75\), base from \(-0.5\) to \(k\), flat section height \(0.25\) from \(-0.5\) to \(0.5\) | M1 | A correct shape. Must not go below zero |
| Correct shape including labels | A1 | Allow 1.25 for \(k\) and 1.75 for \(2k - 0.75\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Area} = \frac{1}{4} + \frac{1}{2}\left(\frac{1}{4} + 2k - \frac{3}{4}\right)\left(k - \frac{1}{2}\right) = 1\) | M1 | Equating area to 1. A correct method to find the area - allow 1 sign error. May be implied by a correct 3 term quadratic |
| \(8k^2 - 6k - 5 = 0\) or \(k^2 - \frac{3}{4}k - \frac{5}{8} = 0\) | A1 | For a correct 3 term quadratic |
| \((4k-5)(2k+1) = 0\) or \(k = \dfrac{\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 + 4 \times \frac{5}{8}}}{2}\) | M1 | A correct method to solve a 3 term quadratic |
| \(k = 1.25\) | A1* | 1.25 must be the only answer given. All previous marks must be awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[\int_{-0.5}^{0.5} \frac{1}{4}x\, dx\right] + \int_{0.5}^{1.25} 2x^2 - \frac{3}{4}x\, dx = [0] + \left[\frac{2x^3}{3} - \frac{3}{8}x^2\right]_{0.5}^{1.25}\) | M1A1 | \(\int_{0.5}^{1.25} 2x^2 - \frac{3}{4}x\, dx\) on its own also acceptable; correct integration of \(2x^2 - \frac{3}{4}x\) |
| \(= \left(\frac{2 \times 1.25^3}{3} - \frac{3}{8} \times 1.25^2\right) - \left(\frac{2 \times 0.5^3}{3} - \frac{3}{8} \times 0.5^2\right)\) | dM1 | Dep on previous M. Substituting in the correct limits |
| \(= \dfrac{93}{128}\) | A1 | awrt 0.727 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Q_1 = 0.5\) | B1 | For 0.5. May be seen in their IQR |
| \(Q_3^2 - \frac{3}{4}Q_3 + 0.125 = 0.5\) or \(Q_3^2 - \frac{3}{4}Q_3 - 0.375 = 0\) | M1 | A correct equation for finding \(Q_3\) |
| \(Q_3 = \dfrac{\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 + 4 \times \frac{1}{8}}}{2}\) | M1 | A correct method to solve a 3 term quadratic. Implied by \(Q_3 = \frac{3+\sqrt{33}}{8}\) or awrt 1.093 |
| \(\text{IQR} = \text{"1.093"} - 0.5\) | M1 | Correct method to find the IQR |
| \(= 0.593\ldots\) | A1 | awrt 0.59. Allow \(\frac{-1+\sqrt{33}}{8}\) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trapezium shape with height $2k - 0.75$, base from $-0.5$ to $k$, flat section height $0.25$ from $-0.5$ to $0.5$ | M1 | A correct shape. Must not go below zero |
| Correct shape including labels | A1 | Allow 1.25 for $k$ and 1.75 for $2k - 0.75$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{4} + \frac{1}{2}\left(\frac{1}{4} + 2k - \frac{3}{4}\right)\left(k - \frac{1}{2}\right) = 1$ | M1 | Equating area to 1. A correct method to find the area - allow 1 sign error. May be implied by a correct 3 term quadratic |
| $8k^2 - 6k - 5 = 0$ or $k^2 - \frac{3}{4}k - \frac{5}{8} = 0$ | A1 | For a correct 3 term quadratic |
| $(4k-5)(2k+1) = 0$ or $k = \dfrac{\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 + 4 \times \frac{5}{8}}}{2}$ | M1 | A correct method to solve a 3 term quadratic |
| $k = 1.25$ | A1* | 1.25 must be the only answer given. All previous marks must be awarded |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\int_{-0.5}^{0.5} \frac{1}{4}x\, dx\right] + \int_{0.5}^{1.25} 2x^2 - \frac{3}{4}x\, dx = [0] + \left[\frac{2x^3}{3} - \frac{3}{8}x^2\right]_{0.5}^{1.25}$ | M1A1 | $\int_{0.5}^{1.25} 2x^2 - \frac{3}{4}x\, dx$ on its own also acceptable; correct integration of $2x^2 - \frac{3}{4}x$ |
| $= \left(\frac{2 \times 1.25^3}{3} - \frac{3}{8} \times 1.25^2\right) - \left(\frac{2 \times 0.5^3}{3} - \frac{3}{8} \times 0.5^2\right)$ | dM1 | Dep on previous M. Substituting in the correct limits |
| $= \dfrac{93}{128}$ | A1 | awrt 0.727 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q_1 = 0.5$ | B1 | For 0.5. May be seen in their IQR |
| $Q_3^2 - \frac{3}{4}Q_3 + 0.125 = 0.5$ or $Q_3^2 - \frac{3}{4}Q_3 - 0.375 = 0$ | M1 | A correct equation for finding $Q_3$ |
| $Q_3 = \dfrac{\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 + 4 \times \frac{1}{8}}}{2}$ | M1 | A correct method to solve a 3 term quadratic. Implied by $Q_3 = \frac{3+\sqrt{33}}{8}$ or awrt 1.093 |
| $\text{IQR} = \text{"1.093"} - 0.5$ | M1 | Correct method to find the IQR |
| $= 0.593\ldots$ | A1 | awrt 0.59. Allow $\frac{-1+\sqrt{33}}{8}$ |
---\begin{enumerate}
\item A random variable $X$ has probability density function given by
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 1 } { 4 } & - \frac { 1 } { 2 } \leqslant x < \frac { 1 } { 2 } \\
2 x - \frac { 3 } { 4 } & \frac { 1 } { 2 } \leqslant x \leqslant k \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ is a positive constant.\\
(a) Sketch the graph of $\mathrm { f } ( x )$\\
(b) By forming and solving an equation in $k$, show that $k = 1.25$\\
(c) Use calculus to find $\mathrm { E } ( X )$\\
(d) Calculate the interquartile range of $X$
\hfill \mbox{\textit{Edexcel S2 2022 Q2 [15]}}