Question 2 - A-Level Maths

WJEC Further Unit 2 2018 June — Question 2 15 marks

Exam Board	WJEC
Module	Further Unit 2 (Further Unit 2)
Year	2018
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Cumulative distribution functions
Type	Expectation from CDF/PDF
Difficulty	Standard +0.8 This is a Further Maths statistics question requiring multiple techniques: evaluating a CDF, working with independence of random variables, finding quartiles, differentiating to find a PDF, and computing expected values via integration. While each individual part uses standard methods, the multi-step nature, the cubic polynomial requiring careful differentiation and integration, and the conceptual understanding needed for part (f) place this above average difficulty but not at the extreme end for Further Maths students.
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 length of time a battery works, in tens of hours, is modelled by a random variable $X$ with cumulative distribution function $$F(x) = \begin{cases} 0 & \text{for } x < 0, \\ \frac{x^3}{432}(8-x) & \text{for } 0 \leq x \leq 6, \\ 1 & \text{for } x > 6. \end{cases}$$

Find $P(X > 5)$. [2]
A head torch uses three of these batteries. All three batteries must work for the torch to operate. Find the probability that the head torch will operate for more than 50 hours. [2]
Show that the upper quartile of the distribution lies between 4·5 and 4·6. [3]
Find $f(x)$, the probability density function for $X$. [3]
Find the mean lifetime of the batteries in hours. [4]
The graph of $f(x)$ is given below. \includegraphics{figure_1} Give a reason why the model may not be appropriate. [1]

Show mark scheme Show mark scheme source

2(a)

Answer	Marks
$P(X > 5) = 1 - F(5) = 0.1319(44 \ldots)$ or $\frac{19}{144}$ awrt 0.132	M1 A1

2(b)

Answer	Marks	Guidance
$P(\text{torch will operate for more than 50 hours}) = 0.01394444\ldots^3 = 0.00229(70\ldots)$ awrt 0.0023	M1 A1	'Their (a)'$^3$

2(c)

Answer	Marks	Guidance
$F(4.5) = 0.7382 \ldots$	M1	M1 for attempt to find $F(4.5)$ and $F(4.6)$
$F(4.6) = 0.7660 \ldots$	A1	A1 for both answers. If rearranged to $q^4 - 8q^3 + 324 = 0$ A1 is for 5.0625 and -6.9424. Accept oe
Since $F(4.6)$ is greater than 0.75 and $F(4.5)$ is less than 0.75 the solution to $F(q) = 0.75$ is between 4.5 and 4.6	E1

2(d)

Answer	Marks	Guidance
$f(x) = F'(x)$	M1	M1 Attempt at differentiating with at least one power of $x$ decreasing
$f(x) = \frac{8 \times 3x^2}{432} - \frac{4x^3}{432}$	A1	A1 Correct expression for $f(x)$ for $x$ between 0 and 6.
$f(x) = \begin{cases} \frac{x^2}{108}(6-x) & 0 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}$	B1	B1 for "0 otherwise" and range $0 \le x \le 6$

2(e)

Answer	Marks	Guidance
$E(X) = \int_0^6 \frac{x^3}{108}(6-x)dx$	M1	M1 Attempt at integrating $xf(x)$ with at least one power of $x$ increasing. FT their $f(x)$ of equivalent difficulty (ignore limits here)
$E(X) = \frac{1}{108}\int_0^6 (6x^3 - x^4)dx$	A1
$E(X) = \frac{1}{108}\left[\frac{6x^4}{4} - \frac{x^5}{5}\right]_0^6$	A1	A1 correct integration with correct limits FT cao
$E(X) = 3.6$ Mean = 36 hours	A1 B1	FT their derived $E(X)$

2(f)

Answer	Marks	Guidance
Valid explanation e.g. It is possible for a battery to last more than 60 hours. e.g. X could be greater than 6.	E1	[15]

## 2(a)
$P(X > 5) = 1 - F(5) = 0.1319(44 \ldots)$ or $\frac{19}{144}$ awrt 0.132 | M1 A1 |

## 2(b)
$P(\text{torch will operate for more than 50 hours}) = 0.01394444\ldots^3 = 0.00229(70\ldots)$ awrt 0.0023 | M1 A1 | 'Their (a)'$^3$

## 2(c)
$F(4.5) = 0.7382 \ldots$ | M1 | M1 for attempt to find $F(4.5)$ and $F(4.6)$

$F(4.6) = 0.7660 \ldots$ | A1 | A1 for both answers. If rearranged to $q^4 - 8q^3 + 324 = 0$ A1 is for 5.0625 and -6.9424. Accept oe

Since $F(4.6)$ is greater than 0.75 and $F(4.5)$ is less than 0.75 the solution to $F(q) = 0.75$ is between 4.5 and 4.6 | E1 |

## 2(d)
$f(x) = F'(x)$ | M1 | M1 Attempt at differentiating with at least one power of $x$ decreasing

$f(x) = \frac{8 \times 3x^2}{432} - \frac{4x^3}{432}$ | A1 | A1 Correct expression for $f(x)$ for $x$ between 0 and 6.

$f(x) = \begin{cases} \frac{x^2}{108}(6-x) & 0 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}$ | B1 | B1 for "0 otherwise" and range $0 \le x \le 6$

## 2(e)
$E(X) = \int_0^6 \frac{x^3}{108}(6-x)dx$ | M1 | M1 Attempt at integrating $xf(x)$ with at least one power of $x$ increasing. FT their $f(x)$ of equivalent difficulty (ignore limits here)

$E(X) = \frac{1}{108}\int_0^6 (6x^3 - x^4)dx$ | A1 |

$E(X) = \frac{1}{108}\left[\frac{6x^4}{4} - \frac{x^5}{5}\right]_0^6$ | A1 | A1 correct integration with correct limits FT cao

$E(X) = 3.6$ Mean = 36 hours | A1 B1 | FT their derived $E(X)$

## 2(f)
Valid explanation e.g. It is possible for a battery to last more than 60 hours. e.g. X could be greater than 6. | E1 | [15]

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Show LaTeX source

The length of time a battery works, in tens of hours, is modelled by a random variable $X$ with cumulative distribution function

$$F(x) = \begin{cases}
0 & \text{for } x < 0, \\
\frac{x^3}{432}(8-x) & \text{for } 0 \leq x \leq 6, \\
1 & \text{for } x > 6.
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\item Find $P(X > 5)$. [2]
\item A head torch uses three of these batteries. All three batteries must work for the torch to operate. Find the probability that the head torch will operate for more than 50 hours. [2]
\item Show that the upper quartile of the distribution lies between 4·5 and 4·6. [3]
\item Find $f(x)$, the probability density function for $X$. [3]
\item Find the mean lifetime of the batteries in hours. [4]
\item The graph of $f(x)$ is given below.

\includegraphics{figure_1}

Give a reason why the model may not be appropriate. [1]
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 2 2018 Q2 [15]}}

This paper (7 questions)

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Q1 8 Q2 15 Q3 11 Q4 9 Q5 12 Q6 10 Q7 7

Answer	Marks	Guidance
\(F(4.5) = 0.7382 \ldots\)	M1	M1 for attempt to find \(F(4.5)\) and \(F(4.6)\)
\(F(4.6) = 0.7660 \ldots\)	A1	A1 for both answers. If rearranged to \(q^4 - 8q^3 + 324 = 0\) A1 is for 5.0625 and -6.9424. Accept oe
Since \(F(4.6)\) is greater than 0.75 and \(F(4.5)\) is less than 0.75 the solution to \(F(q) = 0.75\) is between 4.5 and 4.6	E1

Answer	Marks	Guidance
\(f(x) = F'(x)\)	M1	M1 Attempt at differentiating with at least one power of \(x\) decreasing
\(f(x) = \frac{8 \times 3x^2}{432} - \frac{4x^3}{432}\)	A1	A1 Correct expression for \(f(x)\) for \(x\) between 0 and 6.
\(f(x) = \begin{cases} \frac{x^2}{108}(6-x) & 0 \le x \le 6 \\ 0 & \text{otherwise} \end{cases}\)	B1	B1 for "0 otherwise" and range \(0 \le x \le 6\)

Answer	Marks	Guidance
\(E(X) = \int_0^6 \frac{x^3}{108}(6-x)dx\)	M1	M1 Attempt at integrating \(xf(x)\) with at least one power of \(x\) increasing. FT their \(f(x)\) of equivalent difficulty (ignore limits here)
\(E(X) = \frac{1}{108}\int_0^6 (6x^3 - x^4)dx\)	A1
\(E(X) = \frac{1}{108}\left[\frac{6x^4}{4} - \frac{x^5}{5}\right]_0^6\)	A1	A1 correct integration with correct limits FT cao
\(E(X) = 3.6\) Mean = 36 hours	A1 B1	FT their derived \(E(X)\)