| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Moderate -0.3 This is a standard S2 CDF question requiring routine techniques: using boundary conditions to find k, differentiating to get pdf, reading the mode, solving F(t)=0.5 for median, and applying conditional probability. All steps are algorithmic with no novel insight required, making it slightly easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(8)=0\) or \(F(12)=1\); \(\frac{1}{96}(74(8)-\frac{5}{2}(8^2)+k)=0\); \(432+k=0\); \(k=-432\) | M1, A1cso | Using \(k=-432\) to verify \(F(8)=0\) or \(F(12)=1\); correct solution with at least one intermediate line of working |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(t)=\frac{d}{dt}F(t)\); \([f(t)=]\begin{cases}\frac{1}{96}(74-5t) & 8\leq t\leq 12\\ 0 & \text{otherwise}\end{cases}\) | M1, A1 | Attempting to differentiate \(F(t)\) (at least one \(t^n \to t^{n-1}\)); for both lines of \(f(t)\) with correct limits |
| Answer | Marks |
|---|---|
| \(8\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(m)=0.5\) or \(\int_8^m \frac{1}{96}(74-5t)dt=0.5\); \(5m^2-148m+960=0\); \(m=\frac{148\pm\sqrt{148^2-4(5)(960)}}{10}\); \(m=9.6\) | M1, dM1, A1 | Use of \(F(m)=0.5\); arranging to quadratic and solving (dep on 1st M1); A1 for 9.6 only (must reject \(m=20\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(9)\) or \(\int_8^9 \frac{1}{96}(74-5t)dt=\frac{21}{64}\) | M1 A1 | Writing or using \(F(9)\) or correct integral; \(\frac{21}{64}\) or awrt 0.328 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(11)-\text{'F(9)'}\) or \(\int_9^{11}\frac{1}{96}(74-5t)dt=\frac{1}{2}\); \([P(T<11 | T>9)=]\frac{P(9 |
M1 A1ft, M1, A1 |
# Question 2:
## Part (a)
| $F(8)=0$ or $F(12)=1$; $\frac{1}{96}(74(8)-\frac{5}{2}(8^2)+k)=0$; $432+k=0$; $k=-432$ | M1, A1cso | Using $k=-432$ to verify $F(8)=0$ or $F(12)=1$; correct solution with at least one intermediate line of working |
## Part (b)
| $f(t)=\frac{d}{dt}F(t)$; $[f(t)=]\begin{cases}\frac{1}{96}(74-5t) & 8\leq t\leq 12\\ 0 & \text{otherwise}\end{cases}$ | M1, A1 | Attempting to differentiate $F(t)$ (at least one $t^n \to t^{n-1}$); for both lines of $f(t)$ with correct limits |
## Part (c)
| $8$ | B1 | |
## Part (d)
| $F(m)=0.5$ or $\int_8^m \frac{1}{96}(74-5t)dt=0.5$; $5m^2-148m+960=0$; $m=\frac{148\pm\sqrt{148^2-4(5)(960)}}{10}$; $m=9.6$ | M1, dM1, A1 | Use of $F(m)=0.5$; arranging to quadratic and solving (dep on 1st M1); A1 for 9.6 only (must reject $m=20$) |
## Part (e)
| $F(9)$ or $\int_8^9 \frac{1}{96}(74-5t)dt=\frac{21}{64}$ | M1 A1 | Writing or using $F(9)$ or correct integral; $\frac{21}{64}$ or awrt **0.328** |
## Part (f)
| $F(11)-\text{'F(9)'}$ or $\int_9^{11}\frac{1}{96}(74-5t)dt=\frac{1}{2}$; $[P(T<11|T>9)=]\frac{P(9<T<11)}{P(T>9)}=\frac{F(11)-\text{'F(9)'}} {1-\text{'F(9)'}}$; $=\frac{\frac{1}{2}}{1-\frac{21}{64}}=\frac{32}{43}$ | M1 A1ft, M1, A1 | 1st M1: $F(11)-F(9)$ or correct integral limits 9 and 11; 1st A1ft: 0.5 allow ft through $F(11)$; 2nd M1: correct ratio of probabilities (num $<$ denom); $\frac{32}{43}$ or awrt **0.744** |
---\begin{enumerate}
\item The lifetime of a particular battery, $T$ hours, is modelled using the cumulative distribution function
\end{enumerate}
$$\mathrm { F } ( t ) = \left\{ \begin{array} { l r }
0 & t < 8 \\
\frac { 1 } { 96 } \left( 74 t - \frac { 5 } { 2 } t ^ { 2 } + k \right) & 8 \leqslant t \leqslant 12 \\
1 & t > 12
\end{array} \right.$$
(a) Show that $k = - 432$\\
(b) Find the probability density function of $T$, for all values of $t$.\\
(c) Write down the mode of $T$.\\
(d) Find the median of $T$.\\
(e) Find the probability that a randomly selected battery has a lifetime of less than 9 hours.
A battery is selected at random. Given that its lifetime is at least 9 hours,\\
(f) find the probability that its lifetime is no more than 11 hours.\\
\hfill \mbox{\textit{Edexcel S2 2016 Q2 [14]}}