| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine integration to find k, then systematic integration of each piece to build the CDF, followed by straightforward percentile calculation. While multi-part with several steps, each component uses well-practiced techniques with no novel insight required—slightly easier than average due to the symmetric, piecewise-linear structure making integration particularly straightforward. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
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| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch: horizontal line at height \(k\) from \(x=2\) to \(x=6\), with triangular ends | B1 | Correct shape with end points on the \(x\)-axis |
| Values \(k\), 2, 3, 5, 6 marked in correct places | B1 | Correct shape with \(k\), 2, 3, 5, 6 marked in correct places; allow \(\frac{1}{3}\) for \(k\) (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times k + 2 \times k + \frac{1}{2} \times k = 1\) | M1 | An attempt to find area using any correct method and putting equal to 1 |
| \(3k = 1\) | ||
| \(k = \frac{1}{3}\) | A1 cso | AG; method must be shown with no incorrect working; need these 3 lines as minimum |
| Alternative: \(\int_2^3 k(x-2)\,dx + \int_3^5 k\,dx + \int_5^6 k(6-x)\,dx = 1\) | M1 | Correct integration to find whole area, put \(= 1\) and attempt to integrate; ignore limits for \(x^n \to x^{n+1}\) |
| \(\left[\frac{kx^2}{2} - 2kx\right]_2^3 + [kx]_3^5 + k\left[6x - \frac{x^2}{2}\right]_5^6 = 1\) | ||
| \(\frac{k}{2} + 2k\frac{\ddot{o}}{\ddot{o}} + (5k - 3k) + \frac{\ddot{o}}{\ddot{o}}8k - \frac{35}{2}k\frac{\ddot{o}}{\ddot{o}} = 1\) | ||
| \(3k = 1\); \(k = \frac{1}{3}\) | A1 cso | Method must be shown – at least one step between integration and \(k = \frac{1}{3}\); no incorrect working; SC for using verification: M1 A0 if no errors (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(x) = 0,\quad x < 2\) | ||
| \(F(x) = \frac{x^2}{6} - \frac{2x}{3} + \frac{2}{3},\quad 2 \leq x \leq 3\) | M1A1 | |
| \(F(x) = \frac{x}{3} - \frac{5}{6},\quad 3 < x < 5\) | M1A1 | |
| \(F(x) = 2x - \frac{x^2}{6} - 5,\quad 5 \leq x \leq 6\) | M1A1 | |
| \(F(x) = 1,\quad x > 6\) | B1 | |
| Alternative forms: \(\frac{1}{6}(x-2)^2\); \(\frac{x}{3} - \frac{5}{6}\); \(1 - \frac{1}{6}(6-x)^2\); \(1\) | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x - \frac{x^2}{6} - 5 = 0.9\) | M1 | Using their cdf for \(5 \leq x \leq 6 = 0.9\) |
| \(\frac{x^2}{6} - 2x + 5.9 = 0\) | M1 | Using either the quadratic formula or completing the square or factorising or any correct method to solve their 3 term quadratic which must have been correctly rearranged. If they write the formula down then allow a slip. If no formula written down then it must be correct for their equation. May be implied by awrt 5.23 or 6.77 |
| \(x = \frac{2 \pm \sqrt{4 - 4 \times \frac{1}{6} \times 5.9}}{\frac{1}{3}}\) | ||
| \(x =\) awrt \(5.23\) | A1 | awrt \(5.23\) — (allow \(\frac{30 - \sqrt{15}}{5}\)). If they have \(6.77\ldots\) this must be eliminated |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = 4\) | ||
| \(F(5.5) - F(4) = \frac{11}{24}\) | M1 | For writing or attempting to find \(F(5.5) - F(4)\) or \(P(X \leq 5.5) - P(x \leq 4)\) or \(P(X < 5.5) - P(x < 4)\) or \(F(5.5) - 0.5\) or \(\int_4^5 k\, dx + \int_5^{5.5} k(6-x)\, dx\) with correct limits and \(x^n \to x^{n+1}\). May be implied by a correct answer. |
| A1 | \(\frac{11}{24}\) oe or awrt \(0.458\) | |
| (2) | ||
| (Total 16) |
## Question 6:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch: horizontal line at height $k$ from $x=2$ to $x=6$, with triangular ends | B1 | Correct shape with end points on the $x$-axis |
| Values $k$, 2, 3, 5, 6 marked in correct places | B1 | Correct shape with $k$, 2, 3, 5, 6 marked in correct places; allow $\frac{1}{3}$ for $k$ **(2 marks)** |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times k + 2 \times k + \frac{1}{2} \times k = 1$ | M1 | An attempt to find area using any correct method and putting equal to 1 |
| $3k = 1$ | | |
| $k = \frac{1}{3}$ | A1 cso | AG; method must be shown with no incorrect working; need these 3 lines as minimum |
| **Alternative:** $\int_2^3 k(x-2)\,dx + \int_3^5 k\,dx + \int_5^6 k(6-x)\,dx = 1$ | M1 | Correct integration to find whole area, put $= 1$ and attempt to integrate; ignore limits for $x^n \to x^{n+1}$ |
| $\left[\frac{kx^2}{2} - 2kx\right]_2^3 + [kx]_3^5 + k\left[6x - \frac{x^2}{2}\right]_5^6 = 1$ | | |
| $\frac{k}{2} + 2k\frac{\ddot{o}}{\ddot{o}} + (5k - 3k) + \frac{\ddot{o}}{\ddot{o}}8k - \frac{35}{2}k\frac{\ddot{o}}{\ddot{o}} = 1$ | | |
| $3k = 1$; $k = \frac{1}{3}$ | A1 cso | Method must be shown – at least one step between integration and $k = \frac{1}{3}$; no incorrect working; **SC** for using verification: M1 A0 if no errors **(2 marks)** |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(x) = 0,\quad x < 2$ | | |
| $F(x) = \frac{x^2}{6} - \frac{2x}{3} + \frac{2}{3},\quad 2 \leq x \leq 3$ | M1A1 | |
| $F(x) = \frac{x}{3} - \frac{5}{6},\quad 3 < x < 5$ | M1A1 | |
| $F(x) = 2x - \frac{x^2}{6} - 5,\quad 5 \leq x \leq 6$ | M1A1 | |
| $F(x) = 1,\quad x > 6$ | B1 | |
| **Alternative forms:** $\frac{1}{6}(x-2)^2$; $\frac{x}{3} - \frac{5}{6}$; $1 - \frac{1}{6}(6-x)^2$; $1$ | | **(7 marks)** |
## CDF / Probability Distribution Question
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x - \frac{x^2}{6} - 5 = 0.9$ | **M1** | Using their cdf for $5 \leq x \leq 6 = 0.9$ |
| $\frac{x^2}{6} - 2x + 5.9 = 0$ | **M1** | Using either the quadratic formula or completing the square or factorising or any correct method to solve their **3 term quadratic** which must have been correctly rearranged. If they write the formula down then allow a slip. If no formula written down then it must be correct for their equation. May be implied by awrt 5.23 or 6.77 |
| $x = \frac{2 \pm \sqrt{4 - 4 \times \frac{1}{6} \times 5.9}}{\frac{1}{3}}$ | | |
| $x =$ awrt $5.23$ | **A1** | awrt $5.23$ — (allow $\frac{30 - \sqrt{15}}{5}$). If they have $6.77\ldots$ this must be eliminated |
| | **(3)** | |
---
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = 4$ | | |
| $F(5.5) - F(4) = \frac{11}{24}$ | **M1** | For writing or attempting to find $F(5.5) - F(4)$ or $P(X \leq 5.5) - P(x \leq 4)$ or $P(X < 5.5) - P(x < 4)$ or $F(5.5) - 0.5$ or $\int_4^5 k\, dx + \int_5^{5.5} k(6-x)\, dx$ with correct limits and $x^n \to x^{n+1}$. May be implied by a correct answer. |
| | **A1** | $\frac{11}{24}$ **oe** or awrt $0.458$ |
| | **(2)** | |
| | **(Total 16)** | |6. The continuous random variable $X$ has a probability density function
$$\mathrm { f } ( x ) = \left\{ \begin{array} { c c }
k ( x - 2 ) & 2 \leqslant x \leqslant 3 \\
k & 3 < x < 5 \\
k ( 6 - x ) & 5 \leqslant x \leqslant 6 \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $\mathrm { f } ( x )$.
\item Show that the value of $k$ is $\frac { 1 } { 3 }$
\item Define fully the cumulative distribution function $\mathrm { F } ( x )$.
\item Hence find the 90th percentile of the distribution.
\item Find $\mathrm { P } [ \mathrm { E } ( X ) < X < 5.5 ]$
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\hfill \mbox{\textit{Edexcel S2 2017 Q6 [16]}}