| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Moderate -0.3 This is a standard S2 question testing routine CDF techniques: integrating a polynomial PDF to find the constant, deriving the CDF by integration, and applying basic probability rules including conditional probability. All parts follow textbook procedures with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^9 c(81-t^2)\,dt = 1\) | M1 | Attempting to integrate; \(x^n \to x^{n+1}\), \(c\) must remain as \(c\) or \(\frac{1}{486}\). Ignore limits |
| \(c\left[81t - \frac{t^3}{3}\right]_0^9 = 1\) | A1 | Correct integration, ignore limits |
| \(c\left[81\times9 - \frac{9^3}{3}\right] = 1\) | M1d | Dependent on previous M1. Putting \(=1\) and substituting \(9\) as limit. Need at least one intermediate step before getting 486 |
| \(486c = 1\) | ||
| \(c = \frac{1}{486}\) | A1cso | cso, or if verifying, the statement \(c = \frac{1}{486}\) leads to answer of 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(t) = \frac{1}{486}\int_0^t 81 - x^2\,dx\) | M1 | Attempting to integrate with correct limits or \(\int f(t)\,dt + C\) and \(F(0)=0\) or \(F(9)=1\). Subst in \(c\) at some point |
| \(= \frac{1}{486}\left[81x - \frac{x^3}{3}\right]_0^t\) | ||
| \(= \frac{t}{6} - \frac{t^3}{1458}\) | ||
| \(F(t) = \begin{cases} 0 & t < 0 \\ \frac{t}{6} - \frac{t^3}{1458} & 0 \leq t \leq 9 \\ 1 & t > 9 \end{cases}\) | A1cso | \(F(t)\) must be stated and cso. Condone use of \(<\) instead of \(\leq\) etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T>3) = 1 - \left(\frac{3}{6} - \frac{3^3}{1458}\right)\) | M1 | Using or writing \(1 - F(3)\) or \(\frac{1}{486}\int_3^9 81-x^2\,dx\) or \(1 - P(X \leq 3)\) |
| \(= \frac{14}{27}\) or awrt \(0.519\) | A1 | awrt 0.519 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T>7 \mid T>3) = \frac{0.068587}{0.5185}\) | M1A1ft | M1: \(\frac{\text{a probability}}{\text{their (c)}}\) where \(0 < \text{a probability} < \text{their (c)} < 1\). If a probability \(\geq\) their (c), give M0. A1ft: \(\frac{50/729}{\text{their (c)}}\) or \(\frac{\text{awrt}\,0.0686}{\text{their (c)}}\) |
| \(= \frac{25}{189}\) or awrt \(0.132\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(^3C_2(0.5185)^2(1-0.5185) = \frac{2548}{6561}\) or awrt \(0.388/0.387\) | M1A1ftA1 | M1: Allow \((\text{their}\,0.5185)^2(1-\text{their}\,0.5185)\). A1ft: Allow \(^3C_2(\text{their}\,0.5185)^2(1-\text{their}\,0.5185)\). A1: awrt 0.388 or 0.387 |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^9 c(81-t^2)\,dt = 1$ | M1 | Attempting to integrate; $x^n \to x^{n+1}$, $c$ must remain as $c$ or $\frac{1}{486}$. Ignore limits |
| $c\left[81t - \frac{t^3}{3}\right]_0^9 = 1$ | A1 | Correct integration, ignore limits |
| $c\left[81\times9 - \frac{9^3}{3}\right] = 1$ | M1d | Dependent on previous M1. Putting $=1$ and substituting $9$ as limit. Need at least one intermediate step before getting 486 |
| $486c = 1$ | | |
| $c = \frac{1}{486}$ | A1cso | cso, or if verifying, the statement $c = \frac{1}{486}$ leads to answer of 1 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(t) = \frac{1}{486}\int_0^t 81 - x^2\,dx$ | M1 | Attempting to integrate with correct limits **or** $\int f(t)\,dt + C$ and $F(0)=0$ or $F(9)=1$. Subst in $c$ at some point |
| $= \frac{1}{486}\left[81x - \frac{x^3}{3}\right]_0^t$ | | |
| $= \frac{t}{6} - \frac{t^3}{1458}$ | | |
| $F(t) = \begin{cases} 0 & t < 0 \\ \frac{t}{6} - \frac{t^3}{1458} & 0 \leq t \leq 9 \\ 1 & t > 9 \end{cases}$ | A1cso | $F(t)$ must be stated and cso. Condone use of $<$ instead of $\leq$ etc. |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T>3) = 1 - \left(\frac{3}{6} - \frac{3^3}{1458}\right)$ | M1 | Using or writing $1 - F(3)$ or $\frac{1}{486}\int_3^9 81-x^2\,dx$ or $1 - P(X \leq 3)$ |
| $= \frac{14}{27}$ or awrt $0.519$ | A1 | awrt 0.519 |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T>7 \mid T>3) = \frac{0.068587}{0.5185}$ | M1A1ft | M1: $\frac{\text{a probability}}{\text{their (c)}}$ where $0 < \text{a probability} < \text{their (c)} < 1$. If a probability $\geq$ their (c), give M0. A1ft: $\frac{50/729}{\text{their (c)}}$ or $\frac{\text{awrt}\,0.0686}{\text{their (c)}}$ |
| $= \frac{25}{189}$ or awrt $0.132$ | A1 | |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $^3C_2(0.5185)^2(1-0.5185) = \frac{2548}{6561}$ or awrt $0.388/0.387$ | M1A1ftA1 | M1: Allow $(\text{their}\,0.5185)^2(1-\text{their}\,0.5185)$. A1ft: Allow $^3C_2(\text{their}\,0.5185)^2(1-\text{their}\,0.5185)$. A1: awrt 0.388 or 0.387 |
---2. The length of time, in minutes, that a customer queues in a Post Office is a random variable, $T$, with probability density function
$$\mathrm { f } ( t ) = \left\{ \begin{array} { c c }
c \left( 81 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 9 \\
0 & \text { otherwise }
\end{array} \right.$$
where $c$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $c$ is $\frac { 1 } { 486 }$
\item Show that the cumulative distribution function $\mathrm { F } ( t )$ is given by
$$\mathrm { F } ( t ) = \left\{ \begin{array} { c c }
0 & t < 0 \\
\frac { t } { 6 } - \frac { t ^ { 3 } } { 1458 } & 0 \leqslant t \leqslant 9 \\
1 & t > 9
\end{array} \right.$$
\item Find the probability that a customer will queue for longer than 3 minutes.
A customer has been queueing for 3 minutes.
\item Find the probability that this customer will be queueing for at least 7 minutes.
Three customers are selected at random.
\item Find the probability that exactly 2 of them had to queue for longer than 3 minutes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2014 Q2 [14]}}