1. The surveyor's triangular garden problem based on geometric and trigonometric analysis.
a) Diagram Sketch
To visualize the information, imagine a coordinate plane where point \(A\) is the starting origin:
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. Since the line straight down from \(B\) represents South (\(180^{\circ }\)), the internal angle \(\angle ABC\) is \(180^\circ - 125^\circ = 55^\circ\).
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 \(\angle B = 55^\circ\):
\(AC^{2}=AB^{2}+BC^{2}-2(AB)(BC)\cos (B)\)
\(AC^{2}=80^{2}+70^{2}-2(80)(70)\cos (55^{\circ })\)
\(AC^{2}=6400+4900-11200(0.5736)\)
\(AC^{2}=11300-6424.06=4875.94\)
\(AC=\sqrt{4875.94}\approx \mathbf{69.83}\ \text{m}\)
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:
\(\text{Area}=\frac{1}{2}\times AB\times BC\times \sin (55^{\circ })\)
\(\text{Area}=0.5\times 80\times 70\times 0.8192\)
\(\text{Area}=2800\times 0.8192\approx \mathbf{2293.63} \ \text{m}^{\mathbf{2}}\)
d) Bearing from C to A
First, find the interior angle at \(A\) (\(\angle BAC\)) using the Law of Sines:
\(\frac{\sin (A)}{BC}=\frac{\sin (B)}{AC}\)
\(\sin (A)=\frac{70\times \sin (55^{\circ })}{69.83}\approx 0.8212\)
\(A=\arcsin (0.8212)\approx 55.2^{\circ }\)
Since the line \(AB\) points due North (\(000^{\circ }\)), the bearing from A to C is simply the interior angle: \(55.2^{\circ }\).
Because the surveyor walks in the exact opposite direction (from C to A), we add \(180^{\circ }\) to find the back-bearing:
\(\text{Bearing from C to A}=55.2^{\circ }+180^{\circ }=\mathbf{235.2}^{\mathbf{\circ }}\)
2. Let the continuous random variable \(T\) be the time taken by a competitor to complete the 10,000 m race. 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
The graph of this normal distribution is a symmetrical bell-shaped curve centered at the mean \(\mu = 50\). The standard deviation \(\sigma = 5\) determines the spread, with inflection points located one standard deviation away from the mean at \(45\) and \(55\).
b) Probability between 45 and 55 minutes
We want to find \(P(45 \le T \le 55)\). 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 empirical rule (68-95-99.7 rule) or a standard normal table:
\(P(-1\le Z\le 1)\approx \mathbf{0.6827}\text{ (or 68.3%)}\)
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:
\(P(Z<-2)=1-P(Z\le 2)=1-0.9772=\mathbf{0.0228}\text{(or 2.28%)}\)
d) Time limit for a medal
"Fastest 75%" refers to the competitors who achieved the shortest 75% of race completion times. Therefore, we need to find the 75th percentile time \(k\) such that \(P(T \le k) = 0.75\).
Using a standard normal inverse lookup table, the \(Z\)-score for a cumulative area of \(0.75\) is approximately \(Z \approx 0.674\).
\(k=\mu +Z\cdot \sigma =50+0.674\times 5=\mathbf{53.37}\ \text{minutes}\)
A competitor must finish under 53.37 minutes (or roughly 53 minutes and 22 seconds) to receive a medal.
e) Conditional probability
We are asked to find the probability that a competitor finished under 40 minutes given that they won a medal:
\(P(T<40\mid T\le 53.37)\)
Using the conditional probability formula:
\(P(T<40\mid T\le 53.37)=\frac{P(T<40\cap T\le 53.37)}{P(T\le 53.37)}\)
Since any time under 40 minutes is automatically under 53.37 minutes, the intersection is simply \(P(T < 40)\):
\(=\frac{P(T<40)}{0.75}=\frac{0.02275}{0.75}\approx \mathbf{0.0303}\ \text{(or 3.03%)}\)
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 a standard normal table:
\(P(Z<-3.4)\approx 0.000337\)
Multiplying by the total number of competitors (\(10,000\)):
\(\text{Expected number}=10000\times 0.000337=3.37\)
Rounding to the nearest whole integer (as "nearest minute" in the prompt's context implies a typo for headcount rounding unit):
\(\approx \mathbf{3}\ \text{competitors}\)
3. The problems based on the depth equation \(d = 11 + 5\sin(30t^\circ)\), where \(t\) is the number of hours after midnight (\(0 \le t \le 24\)).
a) Depth of water at midnight
At midnight, the time elapsed since midnight is \(t = 0\).
\(d=11+5\sin (30\times 0)=11+5(0)=\mathbf{11}\text{ m}\)
b) Period of this function
The period of a sine function in degrees, given as \(\sin(Bt)\), is calculated using \(\frac{360^{\circ }}{B}\). Here, \(B = 30^\circ\).
\(\text{Period}=\frac{360^{\circ }}{30^{\circ }}=\mathbf{12}\text{ hours}\)
c) Maximum and Minimum depths and their times
i) Maximum depth:
The maximum value of the sine function is \(1\).
Maximum depth = \(11 + 5(1) = \mathbf{16\text{ m}}\)
This occurs when \(\sin(30t^\circ) = 1 \implies 30t = 90^\circ, 450^\circ\)
\(t = \mathbf{3\text{ hours}}\) (03:00) and \(t = \mathbf{15\text{ hours}}\) (15:00)
ii) Minimum depth:
The minimum value of the sine function is \(-1\).
Minimum depth = \(11 + 5(-1) = \mathbf{6\text{ m}}\)
This occurs when \(\sin(30t^\circ) = -1 \implies 30t = 270^\circ, 630^\circ\)
\(t = \mathbf{9\text{ hours}}\) (09:00) and \(t = \mathbf{21\text{ hours}}\) (21:00)
d) Sketch of the graph (\(0 \le t \le 24\))
The graph completes two full cycles within the 24-hour day, centered along the midline \(d = 11\text{ m}\).
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):
\(30t=\sin ^{-1}(0.6)\approx 36.87^{\circ }\)
\(t_{1}=\frac{36.87}{30}\approx 1.229\text{ hours}\)
Converting \(0.229\text{ hours}\) to minutes: \(0.229 \times 60 \approx 14\text{ minutes}\).
Time: 01:14
Second boundary point (exit time):
Using the symmetric property of sine waves in the second quadrant:
\(30t=180^{\circ }-36.87^{\circ }=143.13^{\circ }\)
\(t_{2}=\frac{143.13}{30}\approx 4.771\text{ hours}\)
Converting \(0.771\text{ hours}\) to minutes: \(0.771 \times 60 \approx 46\text{ minutes}\).
Time: 04:46
Conclusion:
The morning time interval during which the boat can safely enter the harbour is between 01:14 and 04:46 (or for a duration of 3 hours and 32 minutes).
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:
\(\text{Mean\ }(\mu )=n\times p=9\times \frac{2}{3}=\mathbf{6}\)
iii) Probability that Mich scores exactly seven times (\(X = 7\)):
\(P(X=7)={9 \choose 7}\left(\frac{2}{3}\right)^{7}\left(\frac{1}{3}\right)^{2}=36\times \frac{128}{2187}=\frac{512}{2187}\approx \mathbf{0.2341}\)
iv) Probability that he scores four or more times (\(X \ge 4\)):
\(P(X\ge 4)=1-P(X\le 3)=1-\sum _{k=0}^{3}{9 \choose k}\left(\frac{2}{3}\right)^{k}\left(\frac{1}{3}\right)^{9-k}\approx \mathbf{0.9576}\)
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:
\(1-\left(\frac{2}{3}\right)^{n}\ge 0.99\)
\(\left(\frac{2}{3}\right)^{n}\le 0.01\)
3. Solve for \(n\) using logarithms or trial and error:
Taking the natural logarithm (\(\ln \)) on both sides:
\(n\cdot \ln \left(\frac{2}{3}\right)\le \ln (0.01)\)Since \(\ln\left(\frac{2}{3}\right)\) is a negative value (\(\approx -0.4055\)), the inequality sign flips when dividing:
\(n\ge \frac{\ln (0.01)}{\ln (2/3)}\)
\(n\ge \frac{-4.6052}{-0.4055}\)
\(n\ge 11.36\)
Thus, Ken must take at least 12 free throws.
5. Based on the data provided, the full reconstructed data set for the eleven students is as follows:
Student A B C D E F G H I J K
Hours (\(h\)) 10 9.5 9 8.5 8 7.5 7 6.5 6 5 0
Score (\(s\)) 100 91 93 90 80 85 79 70 69 60 65
a) Identify any outliers in the values of \(h\).
Justification: To mathematically determine outliers, we calculate the boundaries using the Interquartile Range (\(\text{IQR}\)):
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 \(=9\)
\(\text{IQR} = Q_3 - Q_1 = 9 - 6 = 3\)
Lower Boundary \(= Q_1 - 1.5(\text{IQR}) = 6 - 1.5(3) = \mathbf{1.5}\)
Upper Boundary \(= Q_3 + 1.5(\text{IQR}) = 9 + 1.5(3) = \mathbf{13.5}\)
Since Student K's revision time (\(h = 0\)) is less than the lower boundary of \(1.5\), it is a confirmed outlier.
b) Calculate the Pearson's product-moment correlation coefficient, \(r\).
The Pearson's product-moment correlation coefficient is \(r \approx 0.816\) (or \(0.8163\) to 4 decimal places), indicating a strong positive linear relationship between the hours spent revising and the test scores.
c) Write down the equation of the line of best fit of \(s\) on \(h\).
The least-squares regression line of \(s\) on \(h\) is:
\(\mathbf{s=3.79h+53.6}\)
(values given to 3 significant figures)
d) Estimate the score of a twelfth student who spent 5.5 hours revising.
Using the regression line equation:
\(s=3.7922(5.5)+53.6364=74.4935\)
Rounding to the nearest integer, the estimated score is 74.
e) Complete the table of rankings.
Rank Type A B C D E F G H I J K
Hour Rank 1 2 3 4 5 6 7 8 9 10 11
Score Rank 1 3 2 4 6 5 7 8 9 11 10
f) Calculate the Spearman's rank correlation coefficient, \(r_{s}\).
Using the rankings above, the Spearman's rank correlation coefficient is \(r_s \approx 0.973\) (or \(0.9727\) to 4 decimal places), showing a very strong positive monotonic relationship.
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g) Explain why \(r_s > r\).
The Spearman's coefficient (\(r_{s}\)) is notably higher than the Pearson's coefficient (\(r\)) because Pearson's \(r\) is highly sensitive to outliers, such as Student K (\(h=0, s=65\)), which severely deviates from a straight linear path and artificially pulls down the linear correlation score.
Conversely, Spearman's \(r_{s}\) operates purely on relative ranks rather than raw numeric values. Because Student K ranks near the bottom for both metrics (11th in hours, 10th in score), they perfectly fit the overall monotonic trend (more hours \(\approx \) higher score), leaving \(r_{s}\) unaffected by the numeric extreme of the outlier.
6. Answers to the Rollercoaster Modelling Problem
Here are the step-by-step solutions to each part of the problem based on the rollercoaster height equation:
\(h(x)=\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x+5\quad \text{for }0<x<7\)
a) Find \(h'(x)\)
To find the derivative, apply the power rule (\(\frac{d}{dx}x^n = n x^{n-1}\)) to each term:
\(h^{\prime }(x)=3\cdot \frac{x^{2}}{3}-2\cdot \frac{5x}{2}+4\)
\({h}^{\prime }{(x)=x}^{{2}}{-5x+4}\)
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 \(x = 1\):
\(h(1)=\frac{1}{3}-\frac{5}{2}+4+5=\frac{41}{6}\approx 6.83\)
For \(x = 4\):
\(h(4)=\frac{64}{3}-\frac{80}{2}+16+5=\frac{7}{3}\approx 2.33\)
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:
\((1, 6.83)\) is a Local Maximum
\((4, 2.33)\) is a Local Minimum
d) Sketch the graph of \(h(x)\) against \(x\)
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 or a GDC to solve for \(0 < x < 7\) yields:
\(x\approx 5.90\text{ m}\)
Since the coaster is moving upwards after its local minimum at \(x = 4\), Alison will experience acrophobia when:
\(\mathbf{5.90<x<7}\)
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 $x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(5)}}{2}$:
\(x_1 = \frac{5 - \sqrt{5}}{2} \approx 1.38\text{ m}\)
\(x_2 = \frac{5 + \sqrt{5}}{2} \approx 3.62\text{ m}\)
Since the quadratic opens upwards, the gradient is less than \(-1\) between these two values:
\(\mathbf{1.38<x<3.62}\)
