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2016 Higher Level - Paper Two

Section A

Question 1

The points \(A(6,-2)\), \(B(5,3)\) and \(C(-3,4)\) are shown on the diagram.

xyCBA–2–15643217654321–1–2–3

(a) Find the equation of the line through \(b\) which is perpendicular to \(AC\).

Answer

\(3x-2y-9=0\)

Solution

\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-(-2)}{-3-6}\\&=-\frac{2}{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m_{\perp}=\frac{3}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-y_1=m(x-x_1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-3=\frac{3}{2}(x-5)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2y-6=3x-15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x-2y-9=0\end{align}

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(b) Use your answer to part (a) above to find the co-ordinates of the orthocentre of the triangle \(ABC\).

Answer

\((7,6)\)

Solution

\begin{align}m_{AB}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-(-2)}{5-6}\\&=-5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m_{\perp}=\frac{1}{5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-y_1=m(x-x_1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-4=\frac{1}{5}(x-(-3))\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5y-20=x+3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x-5y+23=0\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}3x-2y-9=0\end{align}

\begin{align}x-5y+23=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x-2y-9=0\end{align}

\begin{align}3x-15y+69=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}13y-78=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{78}{13}\\&=6\end{align}

and

\begin{align}x-5y+23=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=5y-23\\&=5(6)-23\\&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x,y)=(7,6)\end{align}

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Question 2

A point \(X\) has co-ordinates \((-1,6)\) and the slope of the line \(XC\) is \(\dfrac{1}{7}\).

(a) Find the equation of \(XC\). Give your answer in the form \(ax+by+c=0\), where \(a,b,c,\in\mathbb{Z}\).

XC(–g,–f)(–1, 6)l5 cms
Answer

\(x-7y+43=0\)

Solution

\begin{align}y-y_1=m(x-x_1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-6=\frac{1}{7}(x-(-1))\end{align}

\begin{align}\downarrow\end{align}

\begin{align}7y-42=x+1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x-7y+43=0\end{align}

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(b) \(C\) is the centre of a circle \(S\), of radius \(5\mbox{ cm}\). The line \(l:3x+4y-21=0\) is a tangent to \(s\) and passes through \(X\), as shown. Find the equation of one such circle \(s\).

Answer

\((x-6)^2+(y-7)^2=25\) or \((x+8)^2+(y-5)^2=25\)

Solution

\begin{align}x-7y+43=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-g-7(-f)+43=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-g+7f+43=0\end{align}

and

\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{|(3)(-g)+4(-f)-21|}{\sqrt{3^2+4^2}}=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{|-3g-4f-21|}{5}=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|-3g-4f-21|=25\end{align}

Choosing the positive value:

\begin{align}-3g-4f-21=25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3g+4f+46=0\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}-g+7f+43=0\end{align}

\begin{align}3g+4f+46=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-3g+21f+129=0\end{align}

\begin{align}3g+4f+46=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}25f+175=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f&=-\frac{175}{25}\\&=-7\end{align}

and

\begin{align}-g+7f+43=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}g&=7f+43\\&=7(-7)+43\\&=-6\end{align}

\begin{align}\downarrow\end{align}

Centre: \((6,7)\)

\begin{align}\downarrow\end{align}

\begin{align}(x-6)^2+(y-7)^2=25\end{align}

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Question 3

(a) Show that \(\dfrac{\cos 7A+\cos A}{\sin 7A-\sin A}=\cot 3A\).

Answer

The answer is already in the question!

Solution

\begin{align}\frac{\cos7A+\cos A}{\sin7A-\sin A}&=\frac{\cos(4A+3A)+\cos(4A-3A)}{\sin(4A+3A)-\sin(4A-3A)}\\&=\frac{2\cos4A\cos3A}{2\cos4A\sin3A}\\&=\frac{\cos3A}{\sin3A}\\&=\cot3A\end{align}

as required.

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(b) Given that \(\cos2\theta=\dfrac{1}{9}\), find \(\cos\theta\) in the form \(\pm\dfrac{\sqrt{a}}{b}\), where \(a,b\in\mathbb{N}\).

Answer

\(\pm\dfrac{\sqrt{5}}{3}\)

Solution

\begin{align}\cos^2\theta=\frac{1}{2}(1+\cos2\theta)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos\theta&=\pm\sqrt{\frac{1}{2}(1+\cos2\theta)}\\&=\pm\sqrt{\frac{1}{2}\left(1+\frac{1}{9}\right)}\\&=\pm\sqrt{\frac{5}{9}}\\&=\pm\frac{\sqrt{5}}{3}\end{align}

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Question 4

The diagram shows a semi-circle standing on a diameter \([AC]\), and \([BD]\perp[AC]\).

(a)

(i) Prove that the triangles \(ABD\) and \(DBC\) are similar.

ABCD

(ii) If \(|AB|=x\), \(|BC|=1\), and \(|BD|=y\), write \(y\) in terms of \(x\).

Answer

(i) The answer is already in the question!

(ii) \(y=\sqrt{x}\)

Solution

(i)

\(|\angle BDC|+|\angle DCB|=90^{\circ}\) (angles in triangle)

and

\(|\angle BDC|+|\angle ADB|=90^{\circ}\) (angle in semicircle)

\begin{align}\downarrow\end{align}

\(|\angle DCB|=|\angle ADB|\)

Since both triangle also contain a right angles, both triangles contain the same three angles and are therefore similar.

(ii)

\begin{align}\frac{y}{x}=\frac{1}{y}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y^2=x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y=\sqrt{x}\end{align}

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(b) Use your result from part (a)(ii) to construct a line segment equal in length (in centimetres) to the square root of the length of the line segment \([TU]\) which is drawn below.

TU
Answer
TU√|T U|1 cm
Solution
TU√|T U|1 cm
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Question 5

(a)

(i) In an archery competition, the team consisting of John, David, and Mike will win \(1\)st prize if at least two of them hit the bullseye with their last arrows. From past experience, they know that the probability that John, David, and Mike will hit the
bullseye on their last arrow is \(\dfrac{1}{5}\), \(\dfrac{1}{6}\) and \(\dfrac{1}{4}\) respectively.
Complete the table below to show all the ways in which they could win \(1\)st prize.

\(t\) Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii) Hence or otherwise find the probability that they will win the competition.

Answer

(i)

\(t\) Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii) \(\dfrac{13}{120}\)

Solution

(i)

\(t\) Way 1 Way 2 Way 3 Way 4

John

David

Mike

(ii)

\begin{align}P&=\frac{1}{5}\times\frac{1}{6}\times\frac{3}{4}+\frac{1}{5}\times\frac{5}{6}\times\frac{1}{4}+\frac{4}{5}\times\frac{1}{6}\times\frac{1}{4}+\frac{1}{5}\times\frac{1}{6}\times\frac{1}{4}\\&=\frac{13}{120}\end{align}

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(b) Two events, A and B, are represented in the diagram.
\(P(A\cap B)=0.1\), \(P(B\backslash A)=0.3\) and \(P(A\backslash B)=x\).
Write \(P(A)\) in terms of \(x\) and hence, or otherwise, find the value of \(x\) for which the events \(A\) and \(B\) are independent.

AxB0.10.3
Answer

\(P(A)=x+0.1\) and \(x=0.15\)

Solution

\(P(A)=x+0.1\).

\begin{align}P(A\cap B)=P(A)\times P(B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.1=(x+0.1)\times(0.3+0.1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{0.1}{0.3+0.1}-0.1\\&=0.15\end{align}

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Question 6

A local sports club is planning to run a weekly lotto. To win the Jackpot of €\(1000\), contestants must match one letter chosen from the \(26\) letters in the alphabet and two numbers chosen, in the correct order, from the numbers \(0\) to \(9\). In this lotto, repetition of numbers is allowed (e.g. \(M\), \(3\), \(3\) is an outcome).

(a) Calculate the probability that \(M,3,3\) would be the winning outcome in a particular week.

Answer

\(\dfrac{1}{2600}\)

Solution

\begin{align}P&=\frac{1}{26}\times\frac{1}{10}\times\frac{1}{10}\\&=\frac{1}{2600}\end{align}

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(b) If a contestant matches the letter only, or the letter and one number (but not both numbers), they will win €\(50\). Using the table below, or otherwise, find how much the club should expect to make or lose on each play, correct to the nearest cent, if they charge €\(2\) per play.

Event Payout (\(x\)) Probability (\(P(x)\)) \(xP(x)\)

Win Jackpot

Match letter and first number only

Match letter and second number only

Match letter and neither number

Fail to win

Answer
Event Payout (\(x\)) Probability (\(P(x)\)) \(xP(x)\)

Win Jackpot

\(1{,}000\)

\(\dfrac{1}{2{,}600}\)

\(\dfrac{1{,}000}{2{,}600}\)

Match letter and first number only

\(50\)

\(\dfrac{9}{2{,}600}\)

\(\dfrac{450}{2{,}600}\)

Match letter and second number only

\(50\)

\(\dfrac{9}{2{,}600}\)

\(\dfrac{450}{2{,}600}\)

Match letter and neither number

\(50\)

\(\dfrac{81}{2{,}600}\)

\(\dfrac{4{,}050}{2{,}600}\)

Fail to win

\(0\)

\(0\)

The club will lose \(0.29\mbox{ euro}\) per play.

Solution
Event Payout (\(x\)) Probability (\(P(x)\)) \(xP(x)\)

Win Jackpot

\(1{,}000\)

\(\dfrac{1}{2{,}600}\)

\(\dfrac{1{,}000}{2{,}600}\)

Match letter and first number only

\(50\)

\(\dfrac{9}{2{,}600}\)

\(\dfrac{450}{2{,}600}\)

Match letter and second number only

\(50\)

\(\dfrac{9}{2{,}600}\)

\(\dfrac{450}{2{,}600}\)

Match letter and neither number

\(50\)

\(\dfrac{81}{2{,}600}\)

\(\dfrac{4{,}050}{2{,}600}\)

Fail to win

\(0\)

\(0\)

\begin{align}E(x)&=\Sigma xP(x)\\&=\frac{1000}{2600}+\frac{450}{2600}+\frac{450}{2600}+\frac{4050}{2600}\\&\approx2.29\mbox{ euro}\end{align}

Therefore, the club will lose \(2.29-2=0.29\mbox{ euro}\) per play.

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(c) The club estimates that the average number of plays per week will be \(845\). If the club wants to make an average profit of €\(600\) per week from the lotto, how much should the club charge per play, correct to the nearest cent?

Answer

\(3\mbox{ euro}\)

Solution

\begin{align}845(x-2.29)=600\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=2.29+\frac{600}{845}\\&\approx3\mbox{ euro}\end{align}

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Section B

Question 7

A glass Roof Lantern in the shape of a pyramid has a rectangular base \(CDEF\) and its apex is at \(B\) as
shown. The vertical height of the pyramid is \(|AB|\), where \(A\) is the point of intersection of the
diagonals of the base as shown in the diagram.
Also \(|CD|=2.5\mbox{ m}\) and \(|CF|=3\mbox{ m}\).

(a)

(i) Show that \(|AC|=1.95\mbox{ m}\), correct to two decimal places.

CAFEDB

(ii) The angle of elevation of \(B\) from \(C\) is \(50^{\circ}\) (i.e. \(|\angle BCA|=50^{\circ}\)).
Show that \(|AB|=2.3\mbox{ m}\), correct to one decimal place.

(iii) Find \(|BC|\), correct to the nearest metre.

(iv) Find \(|\angle BCD|\), correct to the nearest degree.

(v) Find the area of glass required to glaze all four triangular sides of the pyramid.
Give your answer correct to the nearest \(\mbox{m}^2\).

Answer

(i) The answer is already in the question!

(ii) The answer is already in the question!

(iii) \(3\mbox{ m}\)

(iv) \(65^{\circ}\)

(v) \(15\mbox{ m}^2\)

Solution

(i)

\begin{align}|EC|^2=|ED|^2+|CD|^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(2|AC|)^2=|ED|^2+|CD|^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|AC|&=\frac{\sqrt{|ED|^2+|CD|^2}}{2}\\&=\frac{\sqrt{3^2+2.5^2}}{2}\\&\approx1.95\mbox{ m}\end{align}

(ii)

\begin{align}\tan50^{\circ}=\frac{|AB|}{1.95}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|AB|&=1.95\tan50^{\circ}\\&\approx2.3\mbox{ m}\end{align}

as required.

(iii)

\begin{align}|BC|&=\sqrt{1.95^2+2.3^2}\\&\approx3\mbox{ m}\end{align}

(iv)

\begin{align}|BD|^2=|BC|^2+|CD|^2-2|BC||CD|\cos|\angle BCD|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|\angle BCD|&=\cos^{-1}\left(\frac{|BC|^2+|CD|^2-|BD|^2}{2|BC||CD|}\right)\\&=\cos^{-1}\left(\frac{3^2+2.5^2-3^2}{2(3)(2.5)}\right)\\&\approx65^{\circ}\end{align}

(v)

\begin{align}A&=2\times\left(\frac{1}{2}(3)(3)\sin60^{\circ}\right)+2\times\left(\frac{1}{2}(2.5)(3)\sin65^{\circ}\right)\\&\approx15\mbox{ m}^2\end{align}

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(b) Another Roof Lantern, in the shape of a pyramid, has a square base \(CDEF\). The vertical height \(|AB|=3\mbox{ m}\), where
\(A\) is the point of intersection of the diagonals of the base as shown.
The angle of elevation of \(B\) to \(C\) is \(60^{\circ}\)
(i.e. \(|\angle BCA|=60^{\circ}\)).
Find the length of the side of the square base of the lantern.
Give your answer in the form \(\sqrt{a}\mbox{ m}\), where \(a\in\mathbb{N}\).

CAFEDB
Answer

\(\sqrt{3}\mbox{ m}\)

Solution

\begin{align}\tan60^{\circ}=\frac{3}{|CA|}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sqrt{3}=\frac{3}{|CA|}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|CA|&=\frac{3}{\sqrt{3}}\\&=\sqrt{3}\mbox{ m}\end{align}

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Question 8

The height of the water in a port was measured over a period of time. The average height was found to be \(1.6\mbox{ m}\). The height measured in metres, \(h(t)\), was modelled using the function

\begin{align}h(t)=1.6+1.5\cos\left(\frac{\pi}{6}t\right)\end{align}

where \(t\) represents the number of hours since the last recorded high tide and \(\left(\dfrac{\pi}{6}t\right)\) is expressed in
radians.

(a) Find the period and range of \(h(t)\).

Answer

Period: \(12\mbox{ hr}\)

Range: \([0.1\mbox{ m},3.1\mbox{ m}]\)

Solution

\begin{align}T&=\frac{2\pi}{\omega}\\&=\frac{2\pi}{\frac{\pi}{6}}\\&=12\mbox{ hr}\end{align}

and

\begin{align}\mbox{Range}&=[1.6-1.5,1.6+1.5]\\&=[0.1\mbox{ m},3.1\mbox{ m}]\end{align}

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(b) Find the maximum height of the water in the port.

Answer

\(3.1\mbox{ m}\)

Solution

\begin{align}1.6+1.5(1)=3.1\mbox{ m}\end{align}

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(c) Find the rate at which the height of the water is changing when \(t=2\), correct to two decimal places. Explain your answer in the context of the question.

Answer

Two hours since the last recorded high tide, the tide is retreating at a rate of \(0.68\mbox{ m/hr}\).

Solution

\begin{align}h'(t)&=1.5\left(\frac{\pi}{6}\right)\left(-\sin\frac{\pi t}{6}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h'(2)&=1.5\left(\frac{\pi}{6}\right)\left(-\sin\frac{\pi (2)}{6}\right)\\&\approx-0.68\mbox{ m/hr}\end{align}

Two hours since the last recorded high tide, the tide is retreating at a rate of \(0.68\mbox{ m/hr}\).

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(d)

(i) On a particular day the high tide occurred at midnight (i.e. \(t=0\)). Use the function to complete the table and show the height, \(h(t)\), of the water between midnight and the following midnight.

Time Midnight 3 a.m. 6 a.m. 9 a.m. 12 noon 3 p.m. 6 p.m. 9 p.m. Midnight

\(\mathbf{t}\)

\(0\)

\(3\)

\(\mathbf{h(t)}\)

(ii) Sketch the graph of \(h(t)\) between midnight and the following midnight.

3219 p.m.Midnight6 p.m.3 p.m.MiddayTime9 a.m.6 a.m.3 a.m.MidnightHeight (m)
Answer

(i)

Time Midnight 3 a.m. 6 a.m. 9 a.m. 12 noon 3 p.m. 6 p.m. 9 p.m. Midnight

\(\mathbf{t}\)

\(0\)

\(3\)

\(6\)

\(9\)

\(12\)

\(15\)

\(18\)

\(21\)

\(24\)

\(\mathbf{h(t)}\)

\(3.1\)

\(1.6\)

\(0.1\)

\(1.6\)

\(3.1\)

\(1.6\)

\(0.1\)

\(1.6\)

\(3.1\)

(ii)

3219 p.m.Midnight6 p.m.3 p.m.MiddayTime9 a.m.6 a.m.3 a.m.MidnightHeight (m)
Solution

(i)

Time Midnight 3 a.m. 6 a.m. 9 a.m. 12 noon 3 p.m. 6 p.m. 9 p.m. Midnight

\(\mathbf{t}\)

\(0\)

\(3\)

\(6\)

\(9\)

\(12\)

\(15\)

\(18\)

\(21\)

\(24\)

\(\mathbf{h(t)}\)

\(3.1\)

\(1.6\)

\(0.1\)

\(1.6\)

\(3.1\)

\(1.6\)

\(0.1\)

\(1.6\)

\(3.1\)

(ii)

3219 p.m.Midnight6 p.m.3 p.m.MiddayTime9 a.m.6 a.m.3 a.m.MidnightHeight (m)
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(e) Find, from your sketch, the difference in water height between low tide and high tide.

Answer

\(3\mbox{ m}\)

Solution

\begin{align}3.1-1=3\mbox{ m}\end{align}

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(f) A fully loaded barge enters the port, unloads its cargo and departs some time later.
The fully loaded barge requires a minimum water level of \(2\mbox{ m}\).
When the barge is unloaded it only requires \(1.5\mbox{ m}\).
Use your graph to estimate the maximum amount of time that the barge can spend in port, without resting on the sea-bed.

Answer

\(5\mbox{ hr }45\mbox{ min}\)

Solution

\begin{align}t&=15\colon15-9\colon30\\&=5\mbox{ hr }45\mbox{ min}\end{align}

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Question 9

Data on earnings were published for a particular country. The data showed that the annual income of people in full-time employment was normally distributed with a mean of €\(39{,}400\) and a standard deviation of €\(12{,}920\).

(a)

(i) The government intends to impose a new tax on incomes over €\(60{,}000\).
Find the percentage of full-time workers who will be liable for this tax, correct to one decimal place.

(ii) The government will also provide a subsidy to the lowest \(10\%\) of income earners.
Find the level of income at which the government will stop paying the subsidy, correct to the nearest euro.

(iii) Some time later a research institute surveyed a sample of \(1000\) full-time workers, randomly selected, and found that the mean annual income of the sample was €\(38{,}280\).
Test the hypothesis, at the \(5\%\) level of significance, that the mean annual income of full-time workers has changed since the national data were published.
State the null hypothesis and the alternative hypothesis.
Give your conclusion in the context of the question.

Answer

(i) \(5.6\%\)

(ii) \(22{,}862\mbox{ euro}\)

(iii) The mean income has changed.

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{60{,}000-39{,}400}{12{,}920}\\&\approx1.59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(z>1.59)&=1-P(z<1.59)\\&=1-0.9441\\&=0.0559\\&\approx5.6\%\end{align}

(ii)

\begin{align}z=-1.28\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{x-39{,}400}{12{,}920}=-1.28\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=39{,}400-1.28(12{,}920)\\&\approx22{,}862\mbox{ euro}\end{align}

(iii)

Null hypothesis: The mean income has not changed.

Alternative hypothesis: The mean income has changed

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{38{,}280-39{,}400}{12{,}920/\sqrt{1{,}000}}\\&\approx-2.74\end{align}

Conclusion: As \(-2.74<-1.96\), the mean income has changed.

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(b) The research institute surveyed \(400\) full-time farmers, randomly selected from all the full-time farmers in the country, and found that the mean income for the sample was €\(26{,}974\) and the standard deviation was €\(5120\).
Assuming that annual farm income is normally distributed in this country, create a \(95\%\) confidence interval for the mean income of full-time farmers.

Answer

\(26{,}472.24\leq\mu\leq27{,}475.76\)

Solution

\begin{align}\bar{x}-1.96\left(\frac{\sigma}{\sqrt{n}}\right)\leq\mu\leq\bar{x}+1.96\left(\frac{\sigma}{\sqrt{n}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}26{,}974-1.96\left(\frac{5{,}120}{\sqrt{400}}\right)\leq\mu\leq26{,}974+1.96\left(\frac{5{,}120}{\sqrt{400}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}26{,}472.24\leq\mu\leq27{,}475.76\end{align}

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(c) It is known that data on farm size are not normally distributed.
The research institute could take many large random samples of farm size and create a sampling distribution of the means of all these samples.
Give one reason why they might do this.

Answer

The sampling distribution will be a normal distribution.

Solution

The sampling distribution will be a normal distribution.

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(d) The research institute also carried out a survey into the use of agricultural land.
\(n\) farmers were surveyed.
If the margin of error of the survey was \(4.5\%\), find the value of \(n\).

Answer

\(494\)

Solution

\begin{align}\frac{1}{\sqrt{n}}=0.045\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n&=\frac{1}{0.045^2}\\&\approx494\end{align}

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