5.
A buggy is pulling a roller-skater, in a straight line along a horizontal road, by means of a connecting rope as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-06_243_1004_390_497}
The combined mass of the buggy and driver is 410 kg
A driving force of 300 N and a total resistance force of 140 N act on the buggy.
The mass of the roller-skater is 72 kg
A total resistance force of \(R\) newtons acts on the roller-skater.
The buggy and the roller-skater have an acceleration of \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\)
- Find \(R\).
- Find the tension in the rope.
Name: □
\section*{U8th A LEVEL Single Mathematics Assessment
Statistics
18 \({ } ^ { \text {th } }\) January 2021 }
\section*{Instructions}
- Answer all the questions
- Write your answer to each question in the space provided under each question. The question number(s) must be clearly shown.
- Use black or blue ink. Pencil may be used for graphs and diagrams only.
- You should clearly write your name at the top of this page and on any additional sheets that you use.
- You are permitted to use a scientific or graphical calculator in this paper.
- Final answers should be given to a degree of accuracy appropriate to the context.
Information
- The total mark for this paper is \(\mathbf { 3 0 }\) marks.
- The marks for each question are shown in brackets.
- You are reminded of the need for clear presentation in your answers.
- You should allow approximately 30 minutes for this section of the test
\section*{Formulae}
\section*{A Level Mathematics A (H240)}
\section*{Arithmetic series}
\(S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}\)
\section*{Geometric series}
\(S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }\)
\(S _ { \infty } = \frac { a } { 1 - r }\) for \(| r | < 1\)
\section*{Binomial series}
\(( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )\), where \({ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }\)
\(( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )\)
\section*{Differentiation}
| \(\mathrm { f } ( x )\) | \(\mathrm { f } ^ { \prime } ( x )\) |
| \(\tan k x\) | \(k \sec ^ { 2 } k x\) |
| \(\sec x\) | \(\sec x \tan x\) |
| \(\cot x\) | \(- \operatorname { cosec } ^ { 2 } x\) |
| \(\operatorname { cosec } x\) | \(- \operatorname { cosec } x \cot x\) |
\(\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c\)
Integration by parts \(\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x\)
Small angle approximations
\(\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta\) where \(\theta\) is measured in radians
\section*{Trigonometric identities}
\(\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B\)
\(\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B\)
\(\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)\)
\section*{Numerical methods}
Trapezium rule: \(\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}\), where \(h = \frac { b - a } { n }\)
The Newton-Raphson iteration for solving \(\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }\)
\section*{Probability}
\(\mathrm { P } ( A \cup B ) = \mathrm { P } ( A ) + \mathrm { P } ( B ) - \mathrm { P } ( A \cap B )\)
\(\mathrm { P } ( A \cap B ) = \mathrm { P } ( A ) \mathrm { P } ( B \mid A ) = \mathrm { P } ( B ) \mathrm { P } ( A \mid B )\) or \(\mathrm { P } ( A \mid B ) = \frac { \mathrm { P } ( A \cap B ) } { \mathrm { P } ( B ) }\)
\section*{Standard deviation}
\(\sqrt { \frac { \sum ( x - \bar { x } ) ^ { 2 } } { n } } = \sqrt { \frac { \sum x ^ { 2 } } { n } - \bar { x } ^ { 2 } }\) or \(\sqrt { \frac { \sum f ( x - \bar { x } ) ^ { 2 } } { \sum f } } = \sqrt { \frac { \sum f x ^ { 2 } } { \sum f } - \bar { x } ^ { 2 } }\)
\section*{The binomial distribution}
If \(X \sim \mathbf { B } ( n , p )\) then \(\mathbf { P } ( X = x ) = \binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }\), Mean of \(X\) is \(n p\), Variance of \(X\) is \(n p ( 1 - p )\)
\section*{Hypothesis test for the mean of a normal distribution}
If \(X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)\) then \(\bar { X } \sim \mathrm {~N} \left( \mu , \frac { \sigma ^ { 2 } } { n } \right)\) and \(\frac { \bar { X } - \mu } { \sigma / \sqrt { n } } \sim \mathrm {~N} ( 0,1 )\)
\section*{Percentage points of the normal distribution}
If \(Z\) has a normal distribution with mean 0 and variance 1 then, for each value of \(p\), the table gives the value of \(z\) such that \(\mathrm { P } ( Z \leqslant z ) = p\).
| \(p\) | 0.75 | 0.90 | 0.95 | 0.975 | 0.99 | 0.995 | 0.9975 | .0 .999 | 0.9995 |
| \(z\) | 0.674 | 1.282 | 1.645 | 1.960 | 2.326 | 2.576 | 2.807 | 3.090 | 3.291 |
\(v = u + a t\)
\(s = u t + \frac { 1 } { 2 } a t ^ { 2 }\)
\(s = \frac { 1 } { 2 } ( u + v ) t\)
\(v ^ { 2 } = u ^ { 2 } + 2 a s\)
\(s = v t - \frac { 1 } { 2 } a t ^ { 2 }\)
Motion in two dimensions
\(\mathbf { v } = \mathbf { u } + \mathbf { a } t\)
\(\mathbf { s } = \mathbf { u } t + \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }\)
\(\mathbf { s } = \frac { 1 } { 2 } ( \mathbf { u } + \mathbf { v } ) \boldsymbol { t }\)
\(\mathbf { s } = \mathbf { v } t - \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }\)
1.
The histogram below shows the heights, in cm, of male A-level students at a particular school.
\includegraphics[max width=\textwidth, alt={}, center]{82e828ab-0812-4de1-b066-720f3632046f-10_853_1095_402_429}
Which class interval contains the median height?
2.
A teacher in a college asks her mathematics students what other subjects they are studying.
She finds that, of her 24 students:
12 study physics
8 study geography
4 study geography and physics
A student is chosen at random from the class.
Determine whether the event 'the student studies physics' and the event 'the student studies geography' are independent.
[0pt]
[2 marks]
3.
Abu visits his local hardware store to buy six light bulbs.
He knows that \(15 \%\) of all bulbs at this store are faulty.
(a) State a distribution which can be used to model the number of faulty bulbs he buys.
(b) Find the probability that all of the bulbs he buys are faulty.
(c) Find the probability that at least two of the bulbs he buys are faulty.
(d) Find the mean of the distribution stated in part (a).
(e) State two necessary assumptions in context so that the distribution stated in part (a) is valid.
4.
A survey of 120 adults found that the volume, \(X\) litres per person, of carbonated drinks they consumed in a week had the following results:
$$\sum x = 165.6 \quad \sum x ^ { 2 } = 261.8$$
(a) (i) Calculate the mean of \(X\).
(a) (ii) Calculate the standard deviation of \(X\).
(b) Assuming that \(X\) can be modelled by a normal distribution find
(b) (i) \(\mathrm { P } ( 0.5 < X < 1.5 )\)
(b) (ii) \(\mathrm { P } ( X = 1 )\)
(c) Determine with a reason, whether a normal distribution is suitable to model this data.
[0pt]
[2 marks]
(d) It is known that the volume, \(Y\) litres per person, of energy drinks consumed in a week may be modelled by a normal distribution with standard deviation 0.21
Given that \(\mathrm { P } ( Y > 0.75 ) = 0.10\), find the value of \(\mu\), correct to three significant figures.
[0pt]
[4 marks]
5.
In a region of England, the government decides to use an advertising campaign to encourage people to eat more healthily.
Before the campaign, the mean consumption of chocolate per person per week was known to be 66.5 g , with a standard deviation of 21.2 g
(a) After the campaign, the first 750 available people from this region were surveyed to find out their average consumption of chocolate.
(a) (i) State the sampling method used to collect the survey.
(a) (ii) Explain why this sample should not be used to conduct a hypothesis test.
(b) A second sample of 750 people revealed that the mean consumption of chocolate per person per week was 65.4 g
Investigate, at the \(10 \%\) level of significance, whether the advertising campaign has decreased the mean consumption of chocolate per person per week.
Assume that an appropriate sampling method was used and that the consumption of chocolate is normally distributed with an unchanged standard deviation.