| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Moderate -0.3 This is a standard FS2 CDF question requiring integration of a simple polynomial pdf, finding a constant, and applying the CDF to solve probability problems. All techniques are routine for Further Statistics students, though part (d) requires setting up an equation using the complement rule. The integration and algebra are straightforward with no conceptual challenges. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \dfrac{t}{120}\,dt = \dfrac{t^2}{240}\) and use of \(F(4)=0\) or \(F(16)=1\) or limits of \(t\) and 4; or attempt at area of trapezium (allow 1 mistake): \(\frac{1}{2}\times(t-4)\!\left(\dfrac{4}{120}+\dfrac{t}{120}\right)\) | M1 | 2.1 — For attempting to integrate and a correct method |
| \(= \dfrac{t^2}{240} - \dfrac{1}{15}\) | A1 | 1.1b — or \(\dfrac{t^2}{240} - 0.0667\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(10) - F(5) = \dfrac{100}{240} - "c" - \dfrac{25}{240} + "c"\) | M1 | 1.1b — Writing or using \(F(10)-F(5)\) |
| \(= \dfrac{5}{16}\) | A1 | 1.1b — awrt \(\dfrac{5}{16}\) or \(0.3125\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{m^2}{240} - \dfrac{1}{15} = 0.5\) | M1 | 1.1b — Setting their \(F(t) = 0.5\) |
| \(m = 11.66...\) | A1 | 1.1b — awrt 11.7 or \(2\sqrt{34}\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(k) = \dfrac{2}{3}(1 - F(k))\) or \(\displaystyle\int_4^k \dfrac{t}{120}\,dt = \dfrac{2}{3}\int_k^{16}\dfrac{t}{120}\,dt\) | M1 | 3.1a — Setting up a correct equation; or setting up correct equation to find \(p\) and an attempt to solve |
| \(\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{3}\!\left(1 - \left(\dfrac{k^2}{240} - \dfrac{1}{15}\right)\right)\) or \(\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{3}\times\left(\dfrac{16}{15} - \dfrac{k^2}{240}\right)\) | dM1 | 1.1b — Attempted to integrate and limits substituted, or using "Their \(F(k)\)" = "their \(p\)" |
| \(\dfrac{k^2}{144} = \dfrac{7}{9}\) | ||
| \(k = \sqrt{112}\) or awrt \(10.6\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
Let \(P(T| (M1) |
| |
| \(\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{5}\) | (dM1) | |
| \(k = \sqrt{112}\) or awrt \(10.6\) | (A1) |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \dfrac{t}{120}\,dt = \dfrac{t^2}{240}$ and use of $F(4)=0$ or $F(16)=1$ or limits of $t$ and 4; **or** attempt at area of trapezium (allow 1 mistake): $\frac{1}{2}\times(t-4)\!\left(\dfrac{4}{120}+\dfrac{t}{120}\right)$ | M1 | 2.1 — For attempting to integrate and a correct method |
| $= \dfrac{t^2}{240} - \dfrac{1}{15}$ | A1 | 1.1b — or $\dfrac{t^2}{240} - 0.0667$ |
**(2 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(10) - F(5) = \dfrac{100}{240} - "c" - \dfrac{25}{240} + "c"$ | M1 | 1.1b — Writing or using $F(10)-F(5)$ |
| $= \dfrac{5}{16}$ | A1 | 1.1b — awrt $\dfrac{5}{16}$ or $0.3125$ or exact equivalent |
**(2 marks)**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{m^2}{240} - \dfrac{1}{15} = 0.5$ | M1 | 1.1b — Setting their $F(t) = 0.5$ |
| $m = 11.66...$ | A1 | 1.1b — awrt 11.7 or $2\sqrt{34}$ or exact equivalent |
**(2 marks)**
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(k) = \dfrac{2}{3}(1 - F(k))$ **or** $\displaystyle\int_4^k \dfrac{t}{120}\,dt = \dfrac{2}{3}\int_k^{16}\dfrac{t}{120}\,dt$ | M1 | 3.1a — Setting up a correct equation; or setting up correct equation to find $p$ and an attempt to solve |
| $\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{3}\!\left(1 - \left(\dfrac{k^2}{240} - \dfrac{1}{15}\right)\right)$ **or** $\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{3}\times\left(\dfrac{16}{15} - \dfrac{k^2}{240}\right)$ | dM1 | 1.1b — Attempted to integrate and limits substituted, or using "Their $F(k)$" = "their $p$" |
| $\dfrac{k^2}{144} = \dfrac{7}{9}$ | | |
| $k = \sqrt{112}$ or awrt $10.6$ | A1 | 1.1b |
**Alternative:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $P(T<k)=p$, then $p = \dfrac{2}{3}(1-p) \;\therefore p = \dfrac{2}{5}$ | (M1) | |
| $\dfrac{k^2}{240} - \dfrac{1}{15} = \dfrac{2}{5}$ | (dM1) | |
| $k = \sqrt{112}$ or awrt $10.6$ | (A1) | |
**(3 marks)**\begin{enumerate}
\item Lloyd regularly takes a break from work to go to the local cafe. The amount of time Lloyd waits to be served, in minutes, is modelled by the continuous random variable $T$, having probability density function
\end{enumerate}
$$f ( t ) = \left\{ \begin{array} { c c }
\frac { t } { 120 } & 4 \leqslant t \leqslant 16 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Show that the cumulative distribution function is given by
$$\mathrm { F } ( t ) = \left\{ \begin{array} { c r }
0 & t < 4 \\
\frac { t ^ { 2 } } { 240 } - c & 4 \leqslant t \leqslant 16 \\
1 & t > 16
\end{array} \right.$$
where the value of $c$ is to be found.\\
(b) Find the exact probability that the amount of time Lloyd waits to be served is between 5 and 10 minutes.\\
(c) Find the median of $T$.\\
(d) Find the value of $k$ such that
$$\mathrm { P } ( T < k ) = \frac { 2 } { 3 } \mathrm { P } ( T > k )$$
giving your answer to 3 significant figures.
\hfill \mbox{\textit{Edexcel FS2 AS 2019 Q2 [9]}}