Answers should be supported by working and/or explanations.
All numerical answers must be given exactly or correct to three significant figures, unless otherwise stated in the question.
You may use Excel, GeoGebra, WolframAlpha and other programs on your computer as well.
Questions are here.
1. The surveyor's triangular garden problem based on geometric and trigonometric analysis.
a) Diagram sketch
Line \(AB\) goes \(80\text{ m}\) straight up (due North, \(000^{\circ }\)).
Line \(BC\) goes \(70\text{ m}\) down and to the right at an angle of \(125^{\circ }\) clockwise from North.
Line \(CA\) connects point \(C\) back to the origin \(A\).
b) Length of side AC
To find the length of side \(AC\), we use the Law of Cosines on \(\triangle ABC\) with the interior angle \(\angle B = 55^\circ\):
\(AC^{2}=AB^{2}+BC^{2}-2\times AB\times BC\times \cos B\)
\(=80^{2}+70^{2}-2\times80\times70\times\cos 55^{\circ }\)
\(=6400+4900-11200\times 0.5736\) (using the trigonometric table)
\(=4875.94\)
\(AC=\sqrt{4875.94}=69.82\cdots\fallingdotseq \mathbf{69.8}\ \text{m}\) (three significant figures)
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| "mode" and calculation by GDC |
c) Area of the triangular garden
Using the sine area formula Area$= \frac{1}{2}ab\sin C$ for the two known sides and their included angle:
Area$=\frac{1}{2}\times AB\times BC\times \sin 55^{\circ }$
\(=0.5\times 80\times 70\times 0.8192\)
\(=2293.76 \fallingdotseq \mathbf{2290}\ \text{m}^{\mathbf{2}}\) (three significant figures)
d) Bearing from C to A
First, find the interior angle at \(A\) (\(\angle BAC\)) using the Law of Sines:
\(\frac{\sin A}{BC}=\frac{\sin B}{AC}\)
\(\sin A=\frac{70\times \sin 55^{\circ }}{69.83}\fallingdotseq 0.8212\)
Using linear interpolation from the trigonometric table or "$\sin^{-1}(\ )$" by GDC/GeoGebra or "=DEGREES(ASIN( ))" by Excel,
\(A=\sin^{-1}(0.8212)=\arcsin (0.8212)\fallingdotseq 55.2^{\circ }\)
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| linear interpolation with the trigonometric table |
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| arcsine by GDC |
Bearing from $C$ to $A$ $=55.2^{\circ }+180^{\circ }=235.2^{\circ}\fallingdotseq \mathbf{235^{\circ}}$
2. We are given that \(T\) is normally distributed with a mean \(\mu = 50\) minutes and standard deviation \(\sigma = 5\) minutes: \(T\sim N(50,5^{2})\)
a) Diagram representation
b) Probability between 45 and 55 minutes
We want to find \(P(45 \leqq T \leqq 55)\).
Converting to the standard normal variable \(Z = \frac{T - \mu}{\sigma}\):
\(Z_{1}=\frac{45-50}{5}=-1\)， \(Z_{2}=\frac{55-50}{5}=1\)
Using the standard normal distribution table:
\(P(-1\leqq Z\leqq 1)=2\times P(0\leqq Z\leqq 1)=2\times 0.341345=0.68269 \fallingdotseq \mathbf{0.683}\text{ (or 68.3%)}\)
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| $P(45 \leqq T \leqq 55), P(-1\leqq Z\leqq 1)$ by GDC |
c) Probability less than 40 minutes
We want to find \(P(T < 40)\).
Standardizing the value: \(Z=\frac{40-50}{5}=-2\)
Using standard normal distribution tables:
\(P(Z<-2)=0.5-P(0\leqq Z\leqq 2)=0.5-0.47725=0.02275\fallingdotseq\mathbf{0.0228}\text{ (or 2.28%)}\)
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| \(P(T < 40)\) by GDC |
d) Time limit for a medal
"Fastest 75%" refers to the competitors who achieved the top 75% of race completion times.
Therefore, we need to find the 75th percentile time \(t\) such that \(P(T \leqq t) = 0.75\)
Using linear interpolation with the normal distribution table or Using "invNorm" by GDC or "InverseNormal" by GeoGebra or "=NORM.S.INV(0.75)" by Excel ,
$P(T \leqq t) = 0.75\Longleftrightarrow P(50\leqq T \leqq t) = 0.25$
$\Longleftrightarrow P(0\leqq Z \leqq u) = 0.25\Longleftrightarrow u\fallingdotseq 0.6745$
\(t=\mu +u\cdot \sigma =50+0.6745\times 5=53.37...\fallingdotseq\mathbf{53.4}\ \text{minutes}\)
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| linear interpolation with the normal distribution table |
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| "Inverse Normal" by GDC |
e) Conditional probability
We are asked to find the probability that a competitor finished under 40 minutes given that they won a medal:
\(P(T<40\mid T\leqq 53.4)\)
Using the conditional probability formula:
\(P(T<40\mid T\leqq 53.4)=\frac{P(T<40\cap T\leqq 53.4)}{P(T\leqq 53.4)}\)
Since any time under 40 minutes is automatically under 53.4 minutes, the intersection is simply \(P(T < 40)\):
\(=\frac{P(T<40)}{0.75}=\frac{0.0228}{0.75}\fallingdotseq \mathbf{0.0304}\ \text{(or 3.04%)}\)
f) Competitors finishing in less than 33 minutes
First, calculate the probability of finishing a race in under 33 minutes by standardizing:
\(Z=\frac{33-50}{5}=-3.4\)
From the standard normal distribution table from 0.00 to 4.09
\(P(Z<-3.4)=0.5-P(Z\leqq 3.4)=0.5-0.499663\fallingdotseq 0.000337\)
Multiplying by the total number of competitors (\(10,000\)):
Expected number $=10000\times 0.000337=3.37$
Round down 3.37 to the nearest integer, \( \mathbf{3}\ \text{competitors}\)
| $P(T<33)$, "E-4" means "$\times 10^{-4}$" by GDC |
3. The problems based on the depth equation \(d = 11 + 5\sin 30t^\circ\), where \(t\) is the number of hours after midnight (\(0 \leqq t \leqq 24\)).
a) Depth of water at midnight
At midnight, the time elapsed since midnight is \(t = 0\).
\(d=11+5\sin (30\times 0)=11+5\times 0=\mathbf{11}\text{ m}\)
b) Period of this function
The period of a sine function in degrees, given as \(\sin Bt^\circ\), is calculated using \(\frac{360^{\circ }}{B}\). Here, \(B = 30^\circ\).
Period $=\frac{360^{\circ }}{30^{\circ }}=\mathbf{12}\text{ hours}$
c) Maximum and Minimum depths and their times
i) Maximum depth:
The maximum value of the sine function is \(1\).
Maximum depth = \(11 + 5\times 1= 16\text{ m}\)
This occurs when \(\sin 30t^\circ= 1 \implies 30t = 90^\circ, 450^\circ\)
\(t =3\)，\(t =15\)
$\mathbf{16\text{m at }} \mathbf{03:00} \text{ and } \mathbf{15:00}$
ii) Minimum depth:
The minimum value of the sine function is \(-1\).
Minimum depth = \(11 + 5\times (-1) =6\text{ m}\)
This occurs when \(\sin 30t^\circ = -1 \implies 30t = 270^\circ, 630^\circ\)
\(t =9\)，\(t =21\)
$\mathbf{6\text{m at }} \mathbf{09:00} \text{ and } \mathbf{21:00}$
d) Sketch of the graph (\(0 \leqq t \leqq 24\))
(see figure below)
e) Morning time interval for the boat to enter (\(d > 14\text{ m}\))
To find when the boat can enter, set \(d > 14\):
\(11+5\sin 30t^{\circ }>14\)
\(5\sin 30t^{\circ }>3\)
\(\sin 30t^{\circ }>0.6\)
Find the boundary times by solving \(\sin 30t^\circ = 0.6\):
First boundary point (entry time):
\(30t=\sin ^{-1}(0.6)\fallingdotseq 36.87^{\circ }\) (You can now find the value)
\(t_{1}=\frac{36.87}{30}\fallingdotseq 1.229\text{ hours}\)
Converting \(0.229\text{ hours}\) to minutes: \(0.229 \times 60 \fallingdotseq 14\text{ minutes}\).
Time: 01:14
Second boundary point (exit time):
Using the symmetric property of sine waves in the second quadrant:
\(30t=180^{\circ }-36.87^{\circ }=143.13^{\circ }\)
\(t_{2}=\frac{143.13}{30}\fallingdotseq 4.771\text{ hours}\)
Converting \(0.771\text{ hours}\) to minutes: \(0.771 \times 60 \fallingdotseq 46\text{ minutes}\).
Time: 04:46
Conclusion:
The morning time interval during which the boat can safely enter the harbour is between $\mathbf{01:14}$ and $\mathbf{04:46}$.
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| Graphs and Intersections by GeoGebra |
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| "Intersection" and "Degree, Minute, Second" by GDC |
4.
a) Mich takes \(n = 9\) free throws with a constant success probability of \(p = \frac{2}{3}\).
i) Distribution and parameters:
\(X\) satisfies a Binomial distribution, written as \(X \sim B\left(9, \frac{2}{3}\right)\).
Number of trials \(n\) = \(9\)
Probability of success \(p\) = \(\frac{2}{3}\)
ii) Mean of the distribution:
Mean \(\mu=n\times p=9\times \frac{2}{3}=\mathbf{6}\)
iii) Let ${}_n \mathrm{ C }_k={n \choose k}$, probability that Mich scores exactly seven times (\(X = 7\)):
\(P(X=7)={9 \choose 7}\left(\frac{2}{3}\right)^{7}\left(\frac{1}{3}\right)^{2}=36\times \frac{128}{2187}=\frac{512}{2187}\fallingdotseq \mathbf{0.234}\)
iv) Probability that he scores four or more times (\(X \geqq 4\)):
\(P(X\geqq 4)=1-P(X\leqq 3)=1-\sum _{k=0}^{3}{9 \choose k}\left(\frac{2}{3}\right)^{k}\left(\frac{1}{3}\right)^{9-k}\fallingdotseq \mathbf{0.958}\)
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| "Binomial Probability Density Function" and "Binomial Cumulative Distribution Function" by GDC |
b) Let \(n\) be the number of free throws Ken takes. We want the probability of scoring at least once to be at least \(99\%\) (\(0.99\)).The probability of scoring at least once is the complement of missing all \(n\) free throws:
\(\text{P(scoring at least once)}=1-\text{P(missing all shots)}=1-\left(\frac{2}{3}\right)^{n}\)
We need to solve for \(n\) where:
\(1-\left(\frac{2}{3}\right)^{n}\geqq 0.99\)
\(\left(\frac{2}{3}\right)^{n}\leqq 0.01\)
Taking the common logarithm on both sides:
\(n\cdot \log \left(\frac{2}{3}\right)\leqq \log (0.01)\)
Since \(\log\left(\frac{2}{3}\right)\) is a negative value , the inequality sign flips when dividing:
\(n\geqq \frac{\log (0.01)}{\log (2/3)}\)
\(n\geqq \frac{-2}{-0.1761}\) (Using Log Table)
\(n\geqq 11.36\)
Thus, Ken must take at least $\mathbf{12}$ free throws.
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| "equation solver" by GDC |
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| Graphs and Intersection by GeoGebra |
5. Based on the data provided, the full reconstructed data set for the eleven students is as follows:
a) Identify any outliers in the values of \(h\).
Student K's revision time $h = 0$ is an outlier.
Justification:
To determine outliers, we calculate the boundaries using the Interquartile Range, $IQR$.
First, sort the \(h\) values: \(0, 5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10\)
Median \(=7.5\)
Lower Quartile (\Q_{1}\) = Median of lower half \(=6\)
Upper Quartile (\Q_{3}\) = Median of upper half \(=9\)
\(IQR = Q_3 - Q_1 = 9 - 6 = 3\)
Lower Boundary \(= Q_1 - 1.5\times IQR = 6 - 1.5\times3=1.5\)
Upper Boundary \(= Q_3 + 1.5\times IQR= 9 + 1.5\times3 =13.5\)
So the value of $h=0$ for student K is an outlier.
b) Calculate the Pearson's product-moment correlation coefficient, \(r\)
Letting \(s_{xy}\) be the covariance and \(s_{x}\), \(s_{y}\) be the standard deviations,
\(r=\frac{s_{xy}}{s_x s_y}=0.8163... \fallingdotseq \mathbf{0.816}\)
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| correlation coefficient by Excel |
c) Write down the equation of the line of best fit of \(s\) on \(h\).
You can get the answer by Excel or GDC like figure below.
Visit here if you'd like to know how to get the regression line.
\(s\mathbf{=3.79}h\mathbf{+53.6}\)
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| scatter plot and regression line by Excel |
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| "List" and "Linear Regression" by GDC |
d) Estimate the score of a twelfth student who spent 5.5 hours revising.
Using the regression line equation:
\(s=3.79\times 5.5+53.6=74.445\)
Rounding to the nearest integer, the estimated score is $\mathbf{74}$
e) Complete the table of rankings.
f) Calculate the Spearman's rank correlation coefficient, \(r_{s}\)
Using the rankings above, the Spearman's rank correlation coefficient is \(r_s =0.9727...\fallingdotseq \mathbf{0.973}\)
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| correlation coefficient by Excel |
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| scatter plot and regression line by Excel |
g) Explain why \(r_s > r\).
This is because, while outliers lower the Pearson correlation coefficient calculated from actual data, Spearman’s rank correlation coefficient—which is calculated from ranks—is largely unaffected by them.
6. Answers to the Rollercoaster Modelling Problem
\(h(x)=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x+5\quad \text{for }0<x<7\)
a) Find \(h'(x)\)
To find the derivative, apply the power rule (\(\frac{d}{dx}x^n = n x^{n-1}\)) to each term:
$h^{\prime }(x)=3\cdot \frac{x^{2}}{3}-2\cdot \frac{5x}{2}+4=x^2-5x+4$
b) Solve \(h'(x) = 0\)
Set the derivative equal to zero and factor the quadratic equation:
\(x^{2}-5x+4=0\)
\((x-1)(x-4)=0\)
\({x=1}\quad \text{or}\quad {x=4}\)
c) Find the stationary points and determine their nature
Substitute the \(x\)-values back into the original function \(h(x)\) to find the \(y\)-coordinates:
For \(x = 1\):
\(h(1)=\frac{1}{3}-\frac{5}{2}+4+5=\frac{41}{6}\fallingdotseq 6.83\)
For \(x = 4\):
\(h(4)=\frac{64}{3}-\frac{80}{2}+16+5=\frac{7}{3}\fallingdotseq 2.33\)
To find their nature, use the second derivative: \(h''(x) = 2x - 5\).
\(h''(1) = 2(1) - 5 = -3 < 0 \rightarrow\) Local Maximum
\(h''(4) = 2(4) - 5 = 3 > 0 \rightarrow\) Local Minimum
Stationary Points:
\(\mathbf{(1, 6.83)}\) is a Local Maximum
\(\mathbf{(4, 2.33)}\) is a Local Minimum
d) Sketch the graph of \(h(x)\) against \(x\)
(see figure below)
e) Find the values of \(x\) for which Alison suffers from acrophobia (\(h(x) > 10\))
Set the height equation equal to \(10\):
\(\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x+5=10\)
\(\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x-5=0\)
Using numerical methods (ex. Newton's method) or GeoGbra or GDC to solve for \(0 < x < 7\) yields:
\(x\fallingdotseq 5.90\text{ m}\)
Since the coaster is moving upwards after its local minimum at \(x = 4\), Alison will experience acrophobia when:
$\mathbf{5.90<} x\mathbf{\leqq7}$
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| Graphs and Intersection by GeoGebra |
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| Graphs and Intersection by GDC |
f) Find the values of \(x\) for which Alison screams (\(h'(x) < -1\))
Set the gradient function to be less than \(-1\):
\(x^{2}-5x+4<-1\)
\(x^{2}-5x+5<0\)
Using the quadratic formula to find the boundary roots $x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(5)}}{2}$:
\(x_1 = \frac{5 - \sqrt{5}}{2} \fallingdotseq 1.38\text{ m}\)
\(x_2 = \frac{5 + \sqrt{5}}{2} \fallingdotseq 3.62\text{ m}\)
Since the quadratic opens upwards, the gradient is less than \(-1\) between these two values:
$\mathbf{1.38<}x\mathbf{<3.62}$
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