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Part (a): We want to find \(P(W < 242)\) where \(W \sim N(250, 4^2)\). Using a graphic display calculator, we find \(P(W < 242) = 0.022750... \approx 0.0228\). Part (b): Since 95% of the bags are accepted, the probability of a bag being within the acceptable range is \(P(242 \le W \le k) = 0.95\). We can express the total probability as \(P(W \le k) = P(W < 242) + P(242 \le W \le k) = 0.022750 + 0.95 = 0.972750\). Using the inverse normal function on a GDC with an area of 0.972750, a mean of 250, and a standard deviation of 4, we obtain \(k \approx 257.69\). To 3 significant figures, \(k = 258\).

(a) (M1) for attempting to use the normal distribution with mean 250 and standard deviation 4. (A1) for 0.0228. (b) (M1) for setting up the relation \(P(242 \le W \le k) = 0.95\). (M1) for finding \(P(W \le k) = 0.97275\) (or equivalent). (M1) for an attempt to use inverse normal. (A1) for 258.

Part (a): The platform height is the height at \(t = 0\), which is \(h(0) = 10\) metres. Part (b)(i): The maximum height occurs at the vertex of the quadratic function, where \(t = -\frac{b}{2a}\). Substituting the values, we get \(1.5 = -\frac{b}{2(-4.9)}\), which simplifies to \(b = 1.5 \times 9.8 = 14.7\). Part (b)(ii): Substituting \(t = 1.5\) and \(b = 14.7\) back into \(h(t)\) gives \(h(1.5) = -4.9(1.5)^2 + 14.7(1.5) + 10 = 21.025 \approx 21.0\) metres. Part (c): The ball hits the ground when \(h(t) = 0\), so \(-4.9t^2 + 14.7t + 10 = 0\). Using the quadratic formula or GDC to solve for \(t > 0\) yields \(t \approx 3.57\) seconds.

(a) (A1) for 10. (b)(i) (M1) for using the vertex formula \(t = -b/(2a)\). (A1) for \(b = 14.7\). (b)(ii) (A1) for 21.0. (c) (M1) for setting \(h(t) = 0\). (A1) for 3.57.

Part (a): Let \(O\) be the center of the square base. The distance from \(O\) to the midpoint \(M\) is half the side length of the base, so \(OM = 4\) cm. In the right-angled triangle \(VOM\), we apply Pythagoras' theorem: \(VM^2 = VO^2 + OM^2 = 12^2 + 4^2 = 160\). Thus, \(VM = \sqrt{160} \approx 12.6\) cm. Part (b): The angle between the face \(VAB\) and the base is the angle \(\angle VMO\). In triangle \(VOM\), \(\tan(\angle VMO) = \frac{VO}{OM} = \frac{12}{4} = 3\). Hence, \(\angle VMO = \arctan(3) \approx 71.6^\circ\). Part (c): The total surface area is the sum of the square base and four identical triangular faces: \(\text{Area} = 8^2 + 4 \times \left(\frac{1}{2} \times 8 \times \sqrt{160}\right) = 64 + 16\sqrt{160} \approx 266\) cm\(^2\).

(a) (M1) for identifying the correct right-angled triangle and using Pythagoras. (A1) for 12.6. (b) (M1) for using trigonometric ratio on triangle VOM. (A1) for 71.6. (c) (M1) for attempting to calculate base area plus four triangular faces. (A1) for 266.

Part (a): Using the arithmetic sequence formula: \(u_{10} = u_1 + 9d = 15 + 9(4) = 51\). Part (b): Using the geometric series sum formula: \(S_{10} = \frac{v_1(r^{10} - 1)}{r - 1} = \frac{3(1.2^{10} - 1)}{1.2 - 1} \approx 77.9\). Part (c): We want to find the smallest integer \(n\) such that \(3 \times 1.2^{n-1} > 15 + (n-1)4\). Evaluating this using a table of values on a GDC: for \(n = 19\), \(v_{19} \approx 79.9\) and \(u_{19} = 87\) (so \(v_{19} < u_{19}\)); for \(n = 20\), \(v_{20} \approx 95.8\) and \(u_{20} = 91\) (so \(v_{20} > u_{20}\)). Thus, the smallest value is \(n = 20\).

(a) (M1) for substituting into arithmetic term formula. (A1) for 51. (b) (M1) for substituting into geometric sum formula. (A1) for 77.9. (c) (M1) for writing or setting up the inequality. (A1) for 20.

Part (a): If the width perpendicular to the wall is \(x\), the side parallel to the wall is \(80 - 2x\). The area is \(A(x) = x(80 - 2x) = 80x - 2x^2\). Part (b): Differentiating \(A\) with respect to \(x\) gives \(\frac{\mathrm{d}A}{\mathrm{d}x} = 80 - 4x\). Part (c): Setting \(\frac{\mathrm{d}A}{\mathrm{d}x} = 0\) to locate the stationary point: \(80 - 4x = 0 \implies x = 20\). To verify that it is a maximum, we find the second derivative: \(\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = -4\). Since the second derivative is negative, \(x = 20\) gives a maximum area.

(a) (M1) for finding side parallel to wall is \(80-2x\). (A1) for \(A = 80x - 2x^2\). (b) (M1)(A1) for \(80 - 4x\). (c) (M1) for setting derivative to 0 to get \(x = 20\). (R1) for verifying it is a maximum using the second derivative test.

Part (a): Using a GDC to find the correlation coefficient for the given data, we get \(r \approx 0.998\). Part (b): Using linear regression on a GDC, the equation of the regression line of \(D\) on \(T\) is \(D = 3.9T - 8.5\). Part (c): For \(T = 28\), we substitute: \(D = 3.9(28) - 8.5 = 100.7\) drinks. Estimating to the nearest whole drink gives 101 drinks. This estimate is highly reliable because the value of \(r\) is extremely close to 1, indicating a very strong positive linear relationship, and 28°C is within the range of the given data (interpolation).

(a) (A1) for 0.998. (b) (M1) for using the GDC's linear regression feature. (A1) for \(D = 3.9T - 8.5\). (c) (A1) for 101 (or 100.7). (R1) for stating that it is highly reliable due to strong correlation and interpolation.

Part (a): Since \(V(0) = 32000\), we have \(A = 32000\). Part (b)(i): Since \(V(3) = 18500\), we write \(18500 = 32000 \times b^3\). Solving for \(b\) yields \(b^3 = 0.578125\), so \(b = \sqrt[3]{0.578125} \approx 0.833\). Part (b)(ii): The percentage annual rate of depreciation is \((1 - b) \times 100 = (1 - 0.833036) \times 100 \approx 16.7\%\). Part (c): We set up the inequality \(32000 \times (0.833036)^t < 5000\). Solving for \(t\) using a GDC or logarithms: \(t > \frac{\ln(5000/32000)}{\ln(0.833036)} \approx 10.2\) years.

(a) (A1) for 32000. (b)(i) (M1) for substituting the values into the equation. (A1) for 0.833. (b)(ii) (A1) for 16.7%. (c) (M1) for setting up the inequality or equation. (A1) for 10.2.

Part (a): In triangle \(PAB\), we can find the interior angle \(\angle PAB\). The bearing from \(A\) back to \(P\) is \(60^\circ + 180^\circ = 240^\circ\). The bearing from \(A\) to \(B\) is \(130^\circ\). Thus, the interior angle \(\angle PAB = 240^\circ - 130^\circ = 110^\circ\). Using the cosine rule to find the side \(PB\): \(PB^2 = 15^2 + 22^2 - 2(15)(22)\cos(110^\circ) \approx 934.73\). Taking the square root, \(PB \approx 30.6\) km. Part (b): Using the sine rule to find the angle \(\angle APB\): \(\frac{\sin(\angle APB)}{22} = \frac{\sin(110^\circ)}{30.573}\), which yields \(\sin(\angle APB) \approx 0.67618\), so \(\angle APB \approx 42.5^\circ\). The bearing of \(B\) from \(P\) is \(60^\circ + 42.5^\circ = 102.5^\circ \approx 103^\circ\).

(a) (M1) for finding the angle \(\angle PAB = 110^\circ\). (M1) for applying the cosine rule correctly. (A1) for 30.6 km. (b) (M1) for applying the sine rule to find \(\angle APB\). (A1) for finding \(\angle APB \approx 42.5^\circ\). (A1) for the final bearing of 103°.

(a) Let \(X\) be the weight of an apple, where \(X \sim N(150, \sigma^2)\). We are given \(P(X > 180) = 0.08\). This implies that \(P(X \le 180) = 0.92\). Standardizing this gives \(P\left(Z \le \frac{180 - 150}{\sigma}\right) = 0.92\). Using the inverse normal function on a GDC, we find \(Z_{0.92} \approx 1.40507\). Therefore, \(\frac{30}{\sigma} = 1.40507 \implies \sigma = \frac{30}{1.40507} \approx 21.351 \approx 21.4\) grams. (b) We want to find \(P(X < 135)\) where \(X \sim N(150, 21.351^2)\). Using the normal cumulative distribution function on a GDC with lower bound \(-\infty\) (or a very small number like \(-9999\)), upper bound \(135\), mean \(150\), and standard deviation \(21.351\), we find \(P(X < 135) \approx 0.24115 \approx 0.241\).

(a) [3.15 marks]
(M1) for setting up the standardized equation: \(\frac{180 - 150}{\sigma} = z\)
(A1) for finding \(z \approx 1.40507\)
(A1) for \(\sigma \approx 21.4\) (accept \(21.3\) from early rounding of \(z\) to \(1.41\))

(b) [3 marks]
(M1) for setting up the normal cumulative probability: \(P(X < 135)\)
(A1) for substituting the mean and their standard deviation from (a)
(A1) for \(0.241\) (accept \(0.242\) if using \(\sigma = 21.4\))

(a) The initial temperature occurs when \(t = 0\). Substituting these values: \(T(0) = 22 + A e^{0} = 85 \implies 22 + A = 85 \implies A = 63\). (b) When \(t = 15\), \(T(15) = 48\). Substituting \(A = 63\): \(22 + 63 e^{-15k} = 48 \implies 63 e^{-15k} = 26 \implies e^{-15k} = \frac{26}{63}\). Taking the natural logarithm on both sides: \(-15k = \ln\left(\frac{26}{63}\right) \implies k = -\frac{1}{15}\ln\left(\frac{26}{63}\right) \approx 0.05896 \approx 0.0590\). (c) Set \(T(t) = 30\): \(22 + 63 e^{-0.05896t} = 30 \implies 63 e^{-0.05896t} = 8 \implies e^{-0.05896t} = \frac{8}{63}\). Taking the natural logarithm: \(-0.05896t = \ln\left(\frac{8}{63}\right) \implies t = -\frac{1}{0.05896}\ln\left(\frac{8}{63}\right) \approx 35.006 \approx 35.0\) minutes.

(a) [2 marks]
(M1) for substituting \(t = 0\) and equating to \(85\)
(A1) for \(A = 63\)

(b) [2.15 marks]
(M1) for substituting \(t = 15\), \(T = 48\), and their \(A\) value into the equation
(A1) for obtaining \(k \approx 0.0590\) (accept \(0.05896\))

(c) [2 marks]
(M1) for setting up the equation \(22 + 63 e^{-kt} = 30\) with their values
(A1) for \(t \approx 35.0\) (accept \(35\))

(a) Using the Cosine Rule: \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\hat{BAC})\). Substituting the known values: \(BC^2 = 110^2 + 85^2 - 2(110)(85)\cos(55^circ) = 12100 + 7225 - 18700\cos(55^\circ) \approx 19325 - 10725.88 = 8599.12\). Therefore, \(BC = \sqrt{8599.12} \approx 92.731 \approx 92.7\text{ m}\). (b) Using the sine area formula: \(\text{Area} = \frac{1}{2}(AB)(AC)\sin(\hat{BAC}) = \frac{1}{2}(110)(85)\sin(55^\circ) = 4675\sin(55^\circ) \approx 3829.53 \approx 3830\text{ m}^2\). (c) The area can also be expressed using the base \(BC\) and perpendicular height \(AD\): \(\text{Area} = \frac{1}{2} \times BC \times AD\). Substituting the known area and base: \(3829.53 = \frac{1}{2} \times 92.731 \times AD \implies AD = \frac{2 \times 3829.53}{92.731} \approx 82.595 \approx 82.6\text{ m}\).

(a) [2.15 marks]
(M1) for correct substitution into the Cosine Rule
(A1) for \(BC \approx 92.7\text{ m}\) (accept \(92.73\))

(b) [2 marks]
(M1) for correct substitution into the area formula
(A1) for \(\text{Area} \approx 3830\text{ m}^2\) (accept \(3829.5\))

(c) [2 marks]
(M1) for equating the area from part (b) to \(\frac{1}{2} \times BC \times AD\)
(A1) for \(AD \approx 82.6\text{ m}\) (accept \(82.5\) or \(82.7\) depending on rounding propagation)

(a) Using the compound interest formula: \(FV = PV \left(1 + \frac{r}{100k}\right)^{kn}\), where \(PV = 8000\), \(r = 4.2\), \(k = 12\), and \(n = 5\). \(FV = 8000 \left(1 + \frac{4.2}{1200}\right)^{12 \times 5} = 8000 (1.0035)^{60}\). Using a GDC: \(FV \approx 9865.795\). To the nearest cent, the amount is \(\$9\,865.80\). (b) Interest earned is \(FV - PV = 9865.80 - 8000 = \$1\,865.80\). (c) To double the investment, the future value must be at least \(\$16\,000\). Let \(m\) be the number of months: \(8000(1.0035)^m = 16000 \implies 1.0035^m = 2\). Taking logarithms: \(m = \frac{\ln(2)}{\ln(1.0035)} \approx 198.38\) months. Since we are looking for the number of complete months, we round up to \(199\) (at \(198\) months, the value is \(\$15\,978.47\), which is not yet doubled; at \(199\) months, it is \(\$16\,034.40\)).

(a) [2 marks]
(M1) for correct substitution into the compound interest formula
(A1) for \(\$9\,865.80\)

(b) [2 marks]
(M1) for subtracting \(8000\) from their answer in (a)
(A1) for \(\$1\,865.80\) (allow follow-through)

(c) [2.15 marks]
(M1) for setting up the inequality or equation \(8000(1.0035)^m = 16000\)
(A1) for \(198.38\)
(A1) for rounding up to \(199\) complete months

(a) Applying the power rule to differentiate: \(P'(x) = \frac{d}{dx}(-x^3 + 12x^2 - 21x - 10) = -3x^2 + 24x - 21\). (b) To find the critical points, set \(P'(x) = 0\): \(-3x^2 + 24x - 21 = 0 \implies -3(x^2 - 8x + 7) = 0 \implies -3(x-1)(x-7) = 0\). This gives \(x = 1\) or \(x = 7\). Since the domain is \(2 \le x \le 10\), we consider \(x = 7\). To confirm it is a local maximum, we can find the second derivative: \(P''(x) = -6x + 24\). Evaluated at \(x=7\): \(P''(7) = -6(7) + 24 = -18 < 0\), confirming a local maximum. Thus, \(x = 7\). (c) Substitute \(x = 7\) into \(P(x)\): \(P(7) = -(7)^3 + 12(7)^2 - 21(7) - 10 = -343 + 588 - 147 - 10 = 88\). Since \(P\) is in thousands of dollars, the maximum weekly profit is \(88 \times 1000 = \$88\,000\).

(a) [2 marks]
(M1) for applying the power rule to at least two terms
(A1) for \(-3x^2 + 24x - 21\)

(b) [2.15 marks]
(M1) for setting their derivative equal to \(0\)
(A1) for finding \(x = 7\) (rejecting \(x = 1\) due to domain restrictions)

(c) [2 marks]
(M1) for substituting their \(x\) value from (b) into \(P(x)\)
(A1) for \(\$88\,000\) (must include correct unit/scale; accept \(88\) thousand dollars)

(a) (i) Using the depreciation formula: \(V = 85000 \times (1 - 0.12)^5 = 85000 \times 0.88^5 = 44857.22\). To the nearest dollar, the value is $44,857.

(ii) We solve \(85000 \times 0.88^t < 30000 \implies 0.88^t < \frac{30000}{85000} \approx 0.35294\). Taking logarithms on both sides: \(t \ln(0.88) < \ln(0.35294) \implies t > 8.15\) years. Since we require the number of completed years, it is 9 years.

(b) (i) Using a financial calculator (TVM solver) with: \(N = 84\), \(I\% = 6.5\), \(PV = 85000\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). This gives \(PMT = -1260.11\). So the monthly payment is $1,260.11.

(ii) Total amount paid = \(1260.11 \times 84 = 105849.24\) USD.

(iii) Total interest paid = \(105849.24 - 85000 = 20849.24\) USD.

(c) The lump sum is the present value of the remaining 48 payments. Using TVM solver with: \(N = 48\), \(I\% = 6.5\), \(PMT = -1260.11\), \(FV = 0\), \(P/Y = 12\), \(C/Y = 12\). This gives \(PV = 53141.61\). To the nearest dollar, the lump sum is $53,142. (If using unrounded payment of 1260.1068, \(PV = 53141.48 \approx 53141\)).

(a) (i) (M1) for substituting into depreciation formula. (A1) for 44857.22. (A1) for nearest dollar response: $44,857. [3 marks]
(ii) (M1) for setting up inequality or equation. (A1) for 8.15. (A1) for 9 years. [3 marks]
(b) (i) (M1) for setting up TVM parameters. (A2) for $1,260.11. [3 marks]
(ii) (M1) for multiplying payment by 84. (A1) for $105,849.24. [2 marks]
(iii) (M1) for total paid minus loan principal. (A1) for $20,849.24. [2 marks]
(c) (M1) for identifying 48 remaining payments. (M1) for setting up TVM to find PV. (A1) for $53,142 (accept $53,141). [3 marks]

(a) Initial population is at \(t = 0\): \(P(0) = \frac{1200}{1 + 14 e^0} = \frac{1200}{15} = 80\).

(b) After 10 years: \(P(10) = \frac{1200}{1 + 14 e^{-1.5}} \approx \frac{1200}{1 + 14 \times 0.22313} \approx 290.99 \approx 291\).

(c) We find the derivative \(P'(5)\) using a GDC: \(P'(5) \approx 20.5\) birds per year.

(d) We solve \(P(t) = 800 \implies \frac{1200}{1 + 14 e^{-0.15 t}} = 800 \implies 1 + 14 e^{-0.15 t} = 1.5 \implies 14 e^{-0.15 t} = 0.5 \implies e^{-0.15 t} = \frac{1}{28} \implies t = \frac{\ln(28)}{0.15} \approx 22.2\) years.

(e) As \(t \to \infty\), \(e^{-0.15 t} \to 0\). Therefore, \(P(t) \to \frac{1200}{1 + 0} = 1200\). The maximum population is 1200.

(f) The linear model is within 10% of \(P(t)\) if \(P(t) \le L(t) \le 1.1 P(t)\) (since \(L(t) \ge P(t)\) for all \(t \ge 0\)). Since \(L(0) = P(0) = 80\), the lower bound is \(t = 0\). Solving \(L(t) = 1.1 P(t) \implies 45t + 80 = 1.1 \times \frac{1200}{1 + 14 e^{-0.15 t}}\) using the solver on the GDC gives \(t \approx 0.248\). Thus, the range of values is \(0 \le t \le 0.248\) (or up to 0.25 years).

(a) (M1) for evaluating \(P(0)\). (A1) for 80. [2 marks]
(b) (M1) for substituting \(t = 10\). (A1) for 291. [2 marks]
(c) (M1) for setup of numerical derivative. (A2) for 20.5 (or 20.6 with other methods). [3 marks]
(d) (M1) for setting equation equal to 800. (M1) for applying logs or GDC solver. (A1) for 22.2. [3 marks]
(e) (R1) for stating limit as \(t \to \infty\). (A1) for 1200. [2 marks]
(f) (M1) for identifying inequality \(L(t) \le 1.1 P(t)\). (M1) for setting up the equation \(45t + 80 = 1.1 P(t)\). (A1) for finding intersection at 0.248. (A1) for final range \(0 \le t \le 0.248\). [4 marks]

(a) Midpoint of \(AB\): \(M_{AB} = \left(\frac{8+1}{2}, \frac{10+9}{2}\right) = (4.5, 9.5)\).
Gradient of \(AB\): \(m_{AB} = \frac{9-10}{1-8} = \frac{1}{7}\).
Perpendicular gradient: \(m_{\perp} = -7\).
Equation of perpendicular bisector:
\(y - 9.5 = -7(x - 4.5) \implies y = -7x + 31.5 + 9.5 \implies 7x + y = 41\).

(b) To find \(V_1\), solve the system of equations:
\(7x + y = 41\)
\(x + 3y = 23\)
From the first equation, \(y = 41 - 7x\). Substituting into the second:
\(x + 3(41 - 7x) = 23 \implies x + 123 - 21x = 23 \implies -20x = -100 \implies x = 5\).
Then, \(y = 41 - 35 = 6\).
So the coordinates of \(V_1\) are \((5, 6)\).

(c) (i) Using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\):
- \(HA = \sqrt{(8-6)^2 + (10-5)^2} = \sqrt{29} \approx 5.39\text{ km}\)
- \(HB = \sqrt{(1-6)^2 + (9-5)^2} = \sqrt{41} \approx 6.40\text{ km}\)
- \(HC = \sqrt{(9-6)^2 + (2-5)^2} = \sqrt{18} \approx 4.24\text{ km}\)
- \(HD = \sqrt{(5-6)^2 + (1-5)^2} = \sqrt{17} \approx 4.12\text{ km}\).

(ii) Tower \(D\) services this house because it is the closest (distance of 4.12 km).

(d) The coordinates of \(V_1\) are \((5, 6)\) and \(D\) are \((5, 1)\). The line segment \(V_1 D\) is a vertical segment of length \(6 - 1 = 5\). If we treat this as the base of the triangle \(V_1 V_2 D\), the height is the horizontal distance from the vertex \(V_2(5.78, 6.39)\) to the line \(x = 5\):
\(h = 5.78 - 5 = 0.78\).
Therefore, the Area of \(V_1 V_2 D\) is:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 0.78 = 1.95\).
(Using precise coordinates of \(V_2 = (\frac{52}{9}, \frac{57.5}{9})\), \(h = \frac{7}{9}\) and \(\text{Area} = \frac{35}{18} \approx 1.94\)). Both 1.94 and 1.95 are accepted.

(a) (M1) for finding midpoint of \(AB\). (M1) for finding gradient of \(AB\). (M1) for taking negative reciprocal of gradient. (A1) for showing the steps leading to the given equation. [4 marks]
(b) (M1) for attempting to solve the system of simultaneous equations. (A1) for \(x = 5\), (A1) for \(y = 6\). [3 marks]
(c) (i) (M1) for using distance formula. (A2) for calculating any three distances correctly, (A1) for all four distances correct. [4 marks]
(ii) (A1) for stating Tower \(D\). [1 mark]
(d) (M1) for identifying the length of \(V_1 D\) as 5. (M1) for identifying the height as \(5.78 - 5 = 0.78\) (or using other trigonometric methods). (M1) for substituting into area of triangle formula. (A1) for 1.94 (or 1.95). [4 marks]

(a) (i) \(P(\text{Swimming}) = \frac{100}{300} = \frac{1}{3} \approx 0.333\).

(ii) \(P(\text{Over 50} \cap \text{Yoga}) = \frac{30}{300} = 0.1\).

(iii) \(P(\text{Under 30} \mid \text{Running}) = \frac{55}{110} = 0.5\).

(b) (i) \(H_0\): Preferred type of exercise and age group are independent.

(ii) \(\text{Degrees of freedom} = (r-1)(c-1) = (3-1)(3-1) = 4\).

(iii) \(\text{Expected Frequency} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{80 \times 110}{300} \approx 29.3\).

(iv) Using the GDC \(\chi^2\) two-way test:
\(\chi^2\text{ statistic} \approx 27.5\),
\(p\text{-value} \approx 1.61 \times 10^{-5}\).

(v) Since the \(p\text{-value} \approx 1.61 \times 10^{-5} < 0.05\), we reject the null hypothesis. There is sufficient evidence at the 5% significance level to conclude that preferred type of exercise and age group are not independent (they are associated).

(c) Let \(X\) be the fitness score. \(X \sim N(68, 8.5^2)\).
Using the GDC (normalCDF):
\(P(60 \le X \le 75) \approx 0.622\).

(a) (i) (A1) for \(\frac{1}{3}\) (or 0.333). [1 mark]
(ii) (M1) for 30/300. (A1) for 0.1. [2 marks]
(iii) (M1) for dividing by 110. (A1) for 0.5. [2 marks]
(b) (i) (A1) for stating that exercise preference and age group are independent. [1 mark]
(ii) (A1) for 4. [1 mark]
(iii) (M1) for fraction formula. (A1) for 29.3. [2 marks]
(iv) (A2) for \(\chi^2 \approx 27.5\). (A1) for \(p\text{-value} \approx 1.61 \times 10^{-5}\). [3 marks]
(v) (R1) for comparing \(p\)-value to 0.05. (A1) for concluding that they are associated. [2 marks]
(c) (M1) for setting up normal distribution parameters. (A1) for 0.622. [2 marks]

(a) \(V = \pi r^2 h\).

(b) Total surface area is the sum of top, bottom, and curved side:
\(A = 2\pi r^2 + 2\pi r h\).
From the volume equation:
\(\pi r^2 h = 500 \implies h = \frac{500}{\pi r^2}\).
Substitute \(h\) into surface area equation:
\(A = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = 2\pi r^2 + \frac{1000}{r}\).

(c) \(\frac{\text{d}A}{\text{d}r} = 4\pi r - 1000 r^{-2} = 4\pi r - \frac{1000}{r^2}\).

(d) To find the minimum surface area, set \(\frac{\text{d}A}{\text{d}r} = 0\):
\(4\pi r - \frac{1000}{r^2} = 0 \implies 4\pi r^3 = 1000 \implies r^3 = \frac{250}{\pi} \approx 79.577\).
\(r = \sqrt[3]{\frac{250}{\pi}} \approx 4.30\text{ cm}\).

(e) Minimum surface area:
\(A(4.301) = 2\pi (4.301)^2 + \frac{1000}{4.301} \approx 116.19 + 232.50 = 348.69 \approx 349\text{ cm}^2\).

(f) The cost function \(C(r)\) is given by:
\(C = 0.04 \times (2\pi r^2) + 0.02 \times (2\pi r h)\)
\(C = 0.08\pi r^2 + 0.02 \times \left(\frac{1000}{r}\right) = 0.08\pi r^2 + \frac{20}{r}\).
To find the minimum cost, find \(\frac{\text{d}C}{\text{d}r}\):
\(\frac{\text{d}C}{\text{d}r} = 0.16\pi r - \frac{20}{r^2}\).
Setting \(\frac{\text{d}C}{\text{d}r} = 0 \implies 0.16\pi r^3 = 20 \implies r^3 = \frac{20}{0.16\pi} = \frac{125}{\pi}\).
\(r = \sqrt[3]{\frac{125}{\pi}} \approx 3.41\text{ cm}\).

(a) (A1) for \(\pi r^2 h\). [1 mark]
(b) (A1) for \(2\pi r^2 + 2\pi r h\). (M1) for expressing \(h\) in terms of \(r\). (A1) for substituting and showing the given formula. [3 marks]
(c) (A1) for \(4\pi r\). (M1) for differentiating the second term. (A1) for \(-1000 r^{-2}\). [3 marks]
(d) (M1) for setting derivative to 0. (M1) for solving for \(r^3\). (A1) for \(r \approx 4.30\). [3 marks]
(e) (M1) for substituting their \(r\) into \(A\). (A1) for 349. [2 marks]
(f) (M1) for setting up the cost equation. (M1) for finding the derivative of cost function. (M1) for setting derivative to 0. (A1) for \(r \approx 3.41\). [4 marks]