Edexcel S2 2011 January — Question 7 13 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2011
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeMultiple independent observations
DifficultyStandard +0.3 This is a standard S2 continuous probability distribution question requiring routine integration techniques. Part (a) uses the fundamental property that pdf integrates to 1, parts (b) and (c) apply standard E(X) and P(X>a) formulas, and part (d) is straightforward binomial probability. While it involves multiple steps and careful algebra, all techniques are textbook applications with no novel problem-solving required, making it slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration
The queuing time in minutes, \(X\), of a customer at a post office is modelled by the probability density function $$\text{f}(x) = \begin{cases} kx(81 - x^2) & 0 \leqslant x \leqslant 9 \\ 0 & \text{otherwise} \end{cases}$$
  1. Show that \(k = \frac{4}{6561}\). [3]
Using integration, find
  1. the mean queuing time of a customer, [4]
  2. the probability that a customer will queue for more than 5 minutes. [3]
Three independent customers shop at the post office.
  1. Find the probability that at least 2 of the customers queue for more than 5 minutes. [3]
AnswerMarks Guidance
(a) \(\int_0^9 k(81x - x^3)dx = 1\); \(k\left[\frac{81}{2}x^2 - \frac{1}{4}x^4\right]_0^9 = 1\); \(k\left(\frac{6561}{2} - \frac{6561}{4}\right) = 1\); \(k = \frac{4}{6561}\) agM1, M1, A1cso (3 marks)
(b) \(E(X) = \int_0^9 kx^2(81 - x^2)dx\); \(E(X) = k\left[\frac{81}{3}x^3 - \frac{x^5}{5}\right]_0^9 = k(19683 - 11809.8) = 4.8\)M1A1, dM1, A1cao (4 marks)
(c) \(P(X > 5) = \int_5^9 k(81x - x^3)\); \(P(X > 5) = k\left[\frac{81}{2}x^2 - \frac{1}{4}x^4\right]_5^9 = k\left(\frac{6561}{4} - 856.25\right) = \text{awrt } 0.478 \text{ or } \frac{3136}{6561}\)M1, M1d, A1 (3 marks)
(d) \(P(\text{At least 2 queue for more than 5 mins}) = 3(1-0.478)(0.478)^2 + 0.478^3 = 0.467\)M1A1ft, A1 (3 marks)
Notes:
- (a) M1 putting integral = 1 ignore limits, = 1 must appear at least once in the working. M1 attempting to integrate at least one part must have correct power of \(x\) (ignore limits). A1 cso for subsst of at least 9. Allow 1/1640.25.
- (b) M1 attempt to use \(xf(x)\) and attempt to multiply out bracket and attempt integration – must have \(x^3\) and \(x^5\) terms (ignore limits). A1 correct integration (ignore limits). dM1 substituting correct limits (need not explicitly see 0). Dependent on having been awarded the first M1. A1 correct z value ±2.1 or ± \(\frac{14.5-25}{5}\); A1 awrt 0.01786.
- (c) M1 attempting to integrate at least one part must have correct power of \(x\) (ignore limits). M1 dep on previous M being awarded, substituting correct limits [may use \(1 - \int_0^5 k(81x - x^3)\) with limits 0 and 5].
- (d) M1 \(3(1-p) p^2 + p^3\) or \(1 - (1-p)^3 - 3(1-p)^2p\); 3 not needed. A1 for \(3(1-p)p^2 + p^3\) where \(p\) is their solution to part (c). A1 awrt 0.467.
[13 marks total]
**(a)** $\int_0^9 k(81x - x^3)dx = 1$; $k\left[\frac{81}{2}x^2 - \frac{1}{4}x^4\right]_0^9 = 1$; $k\left(\frac{6561}{2} - \frac{6561}{4}\right) = 1$; $k = \frac{4}{6561}$ **ag** | M1, M1, A1cso | (3 marks)

**(b)** $E(X) = \int_0^9 kx^2(81 - x^2)dx$; $E(X) = k\left[\frac{81}{3}x^3 - \frac{x^5}{5}\right]_0^9 = k(19683 - 11809.8) = 4.8$ | M1A1, dM1, A1cao | (4 marks)

**(c)** $P(X > 5) = \int_5^9 k(81x - x^3)$; $P(X > 5) = k\left[\frac{81}{2}x^2 - \frac{1}{4}x^4\right]_5^9 = k\left(\frac{6561}{4} - 856.25\right) = \text{awrt } 0.478 \text{ or } \frac{3136}{6561}$ | M1, M1d, A1 | (3 marks)

**(d)** $P(\text{At least 2 queue for more than 5 mins}) = 3(1-0.478)(0.478)^2 + 0.478^3 = 0.467$ | M1A1ft, A1 | (3 marks)

**Notes:**
- (a) M1 putting integral = 1 ignore limits, = 1 must appear at least once in the working. M1 attempting to integrate at least one part must have correct power of $x$ (ignore limits). A1 cso for subsst of at least 9. Allow 1/1640.25.
- (b) M1 attempt to use $xf(x)$ and attempt to multiply out bracket and attempt integration – must have $x^3$ and $x^5$ terms (ignore limits). A1 correct integration (ignore limits). dM1 substituting correct limits (need not explicitly see 0). Dependent on having been awarded the first M1. A1 correct z value ±2.1 or ± $\frac{14.5-25}{5}$; A1 awrt 0.01786.
- (c) M1 attempting to integrate at least one part must have correct power of $x$ (ignore limits). M1 dep on previous M being awarded, substituting correct limits [may use $1 - \int_0^5 k(81x - x^3)$ with limits 0 and 5].
- (d) M1 $3(1-p) p^2 + p^3$ or $1 - (1-p)^3 - 3(1-p)^2p$; 3 not needed. A1 for $3(1-p)p^2 + p^3$ where $p$ is their solution to part (c). A1 awrt 0.467.

**[13 marks total]**
The queuing time in minutes, $X$, of a customer at a post office is modelled by the probability density function

$$\text{f}(x) = \begin{cases}
kx(81 - x^2) & 0 \leqslant x \leqslant 9 \\
0 & \text{otherwise}
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{4}{6561}$.
[3]
\end{enumerate}

Using integration, find

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the mean queuing time of a customer,
[4]

\item the probability that a customer will queue for more than 5 minutes.
[3]
\end{enumerate}

Three independent customers shop at the post office.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the probability that at least 2 of the customers queue for more than 5 minutes.
[3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2011 Q7 [13]}}
This paper (7 questions)
View full paper
Q1 10 Q2 6 Q3 11 Q4 6 Q5 13 Q6 16 Q7 13