Question 4 - A-Level Maths

Edexcel S2 2016 June — Question 4 13 marks

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2016
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Uniform Random Variables
Type	Waiting time applications
Difficulty	Moderate -0.3 This is a straightforward S2 continuous uniform distribution question requiring only standard techniques: identifying the distribution, calculating probabilities using the PDF/CDF, applying binomial probability for repeated trials, deriving and sketching the CDF, and using linear transformation properties for mean and variance. All parts are routine textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration 5.03e Find cdf: by integration 5.03f Relate pdf-cdf: medians and percentiles

The waiting times, in minutes, between flight take-offs at an airport are modelled by the continuous random variable $X$ with probability density function

$$f ( x ) = \begin{cases} \frac { 1 } { 5 } & 2 \leqslant x \leqslant 7 \\ 0 & \text { otherwise } \end{cases}$$

Write down the name of this distribution. A randomly selected flight takes off at 9am
Find the probability that the next flight takes off before 9.05 am
Find the probability that at least 1 of the next 5 flights has a waiting time of more than 6 minutes.
Find the cumulative distribution function of $X$, for all $x$
Sketch the cumulative distribution function of $X$ for $2 \leqslant x \leqslant 7$ On foggy days, an extra 2 minutes is added to each waiting time.
Find the mean and variance of the waiting times between flight take-offs on foggy days.

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks
(Continuous) Uniform/Rectangular	B1

Part (b):

Answer	Marks
$\left[\frac{1}{5}(5-2)\right] = \frac{3}{5}$	B1

Part (c):

$P(X > 6) = p$ where $p = \frac{1}{5}$ or $\frac{7-6}{7-2}$

$Y =$ number of flights with waiting time more than 6 minutes

Answer	Marks	Guidance
$[P(Y \geqslant 1) = 1 - P(Y=0)] = 1-(1-p)^5 = 1-\left(\frac{4}{5}\right)^5 =\ 0.67232$ awrt 0.672	M1, M1, A1	$1^{st}$ M1 for $P(X>6) = \frac{1}{5}$. $2^{nd}$ M1 correct expression of form $1-(1-p)^5$ ft their $p = P(X>6)$ provided $0 < p < 1$

Part (d):

Answer	Marks	Guidance
$\int_2^x \frac{1}{5}\, dt = \left[\frac{t}{5}\right]_2^x$ or $\frac{x}{5} + c$ and $\frac{7}{5}+c=1$ or $\frac{2}{5}+c=0$	M1	Correct integration and sight of correct limits, or integrating with $+c$ and attempt to use $F(7)=1$ or $F(2)=0$

\[F(x) = \begin{cases} 0 & x < 2 \\ \frac{x-2}{5} & 2 \leqslant x \leqslant 7 \\ 1 & x > 7 \end{cases}\]

Answer	Marks	Guidance
Second line correct with correct limits	A1	Allow $<$ instead of $\leqslant$
First and third lines correct with correct limits	B1	Allow $\leqslant$ and $\geqslant$ instead of $<$ and $>$

Part (e):

Answer	Marks	Guidance
Shape (single straight line of positive gradient wholly above $x$-axis), with or without horizontal line ("lid")	B1
Correct sketch with labels 2, 7 on $x$-axis and 1 on $y$-axis	dB1	$2^{nd}$ dB1 dependent on first B1. 2, 7 and 1 in correct places.

Part (f):

Answer	Marks	Guidance
Mean $= \frac{2+7}{2} = 4.5$, so on foggy days Mean $= 6.5$	B1
Variance $= \frac{(7-2)^2}{12} = \frac{25}{12}$ or awrt 2.08	M1 A1	M1 for correct expression $\frac{(7-2)^2}{12}$ or $\frac{(9-4)^2}{12}$, or $\int_\alpha^\beta \frac{1}{5}x^2 dx - \mu^2$. If $\mu=4.5$ use [2,7]; if $\mu=6.5$ use [4,9]. A1 for $\frac{25}{12}$ or awrt 2.08, do not isw.

## Question 4:

### Part (a):
(Continuous) Uniform/Rectangular | B1 |

### Part (b):
$\left[\frac{1}{5}(5-2)\right] = \frac{3}{5}$ | B1 |

### Part (c):
$P(X > 6) = p$ where $p = \frac{1}{5}$ or $\frac{7-6}{7-2}$

$Y =$ number of flights with waiting time more than 6 minutes

$[P(Y \geqslant 1) = 1 - P(Y=0)] = 1-(1-p)^5 = 1-\left(\frac{4}{5}\right)^5 =\ 0.67232$ awrt **0.672** | M1, M1, A1 | $1^{st}$ M1 for $P(X>6) = \frac{1}{5}$. $2^{nd}$ M1 correct expression of form $1-(1-p)^5$ ft their $p = P(X>6)$ provided $0 < p < 1$

### Part (d):
$\int_2^x \frac{1}{5}\, dt = \left[\frac{t}{5}\right]_2^x$ or $\frac{x}{5} + c$ and $\frac{7}{5}+c=1$ or $\frac{2}{5}+c=0$ | M1 | Correct integration and sight of correct limits, or integrating with $+c$ and attempt to use $F(7)=1$ or $F(2)=0$

$$F(x) = \begin{cases} 0 & x < 2 \\ \frac{x-2}{5} & 2 \leqslant x \leqslant 7 \\ 1 & x > 7 \end{cases}$$

Second line correct with correct limits | A1 | Allow $<$ instead of $\leqslant$

First and third lines correct with correct limits | B1 | Allow $\leqslant$ and $\geqslant$ instead of $<$ and $>$

### Part (e):
Shape (single straight line of positive gradient wholly above $x$-axis), with or without horizontal line ("lid") | B1 |

Correct sketch with labels 2, 7 on $x$-axis and 1 on $y$-axis | dB1 | $2^{nd}$ dB1 dependent on first B1. 2, 7 and 1 in correct places.

### Part (f):
Mean $= \frac{2+7}{2} = 4.5$, so on foggy days Mean $= 6.5$ | B1 |

Variance $= \frac{(7-2)^2}{12} = \frac{25}{12}$ or awrt **2.08** | M1 A1 | M1 for correct expression $\frac{(7-2)^2}{12}$ or $\frac{(9-4)^2}{12}$, or $\int_\alpha^\beta \frac{1}{5}x^2 dx - \mu^2$. If $\mu=4.5$ use [2,7]; if $\mu=6.5$ use [4,9]. A1 for $\frac{25}{12}$ or awrt **2.08**, do not isw.

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Show LaTeX source

\begin{enumerate}
  \item The waiting times, in minutes, between flight take-offs at an airport are modelled by the continuous random variable $X$ with probability density function
\end{enumerate}

$$f ( x ) = \begin{cases} \frac { 1 } { 5 } & 2 \leqslant x \leqslant 7 \\ 0 & \text { otherwise } \end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Write down the name of this distribution.

A randomly selected flight takes off at 9am
\item Find the probability that the next flight takes off before 9.05 am
\item Find the probability that at least 1 of the next 5 flights has a waiting time of more than 6 minutes.
\item Find the cumulative distribution function of $X$, for all $x$
\item Sketch the cumulative distribution function of $X$ for $2 \leqslant x \leqslant 7$

On foggy days, an extra 2 minutes is added to each waiting time.
\item Find the mean and variance of the waiting times between flight take-offs on foggy days.

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2016 Q4 [13]}}

This paper (7 questions)

View full paper

Q1 11 Q2 10 Q3 6 Q4 13 Q5 9 Q6 11 Q7 15

Edexcel S2 2016 June — Question 4 13 marks

Question 4:

Part (a):

Part (b):

Part (c):

\(P(X > 6) = p\) where \(p = \frac{1}{5}\) or \(\frac{7-6}{7-2}\)

\(Y =\) number of flights with waiting time more than 6 minutes

Part (d):

\[F(x) = \begin{cases} 0 & x < 2 \\ \frac{x-2}{5} & 2 \leqslant x \leqslant 7 \\ 1 & x > 7 \end{cases}\]

Part (e):

Part (f):

This paper (7 questions)

Answer	Marks	Guidance
Second line correct with correct limits	A1	Allow \(<\) instead of \(\leqslant\)
First and third lines correct with correct limits	B1	Allow \(\leqslant\) and \(\geqslant\) instead of \(<\) and \(>\)

Answer	Marks	Guidance
Shape (single straight line of positive gradient wholly above \(x\)-axis), with or without horizontal line ("lid")	B1
Correct sketch with labels 2, 7 on \(x\)-axis and 1 on \(y\)-axis	dB1	\(2^{nd}\) dB1 dependent on first B1. 2, 7 and 1 in correct places.

Answer	Marks	Guidance
Mean \(= \frac{2+7}{2} = 4.5\), so on foggy days Mean \(= 6.5\)	B1
Variance \(= \frac{(7-2)^2}{12} = \frac{25}{12}\) or awrt 2.08	M1 A1	M1 for correct expression \(\frac{(7-2)^2}{12}\) or \(\frac{(9-4)^2}{12}\), or \(\int_\alpha^\beta \frac{1}{5}x^2 dx - \mu^2\). If \(\mu=4.5\) use [2,7]; if \(\mu=6.5\) use [4,9]. A1 for \(\frac{25}{12}\) or awrt 2.08, do not isw.