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(a) Let X be the weight of an apple in grams, where \(X \sim N(150, 12^2)\). To find \(P(X > 165)\), we use a GDC normal cumulative distribution function with lower bound 165, upper bound 1E99, mean 150, and standard deviation 12. This yields \(P(X > 165) \approx 0.105649...\) which rounds to 0.106 (or 10.6%). (b) We want to find the weight \(w\) such that \(P(X < w) = 0.05\). Using the inverse normal function on a GDC with area 0.05, mean 150, and standard deviation 12, we find \(w \approx 130.261...\) grams. Thus, the minimum weight of an accepted apple is 130 grams (or 130.3 grams).

(a) M1 for writing a correct probability expression \(P(X > 165)\) or standardizing to obtain \(P(Z > 1.25)\). A1 for 0.106 (accept 0.1056... or 10.6%). (b) M1 for setting up the equation \(P(X < w) = 0.05\) or showing a sketch indicating the lower 5% region. M1 for an inverse normal calculation setup, such as standardizing to \(\frac{w - 150}{12} = -1.645\). A1 for 130 (accept 130.3 or 130.26).

(a) Using the compound interest formula: \(FV = PV \times \left(1 + \frac{r}{100k}\right)^{kt}\) where \(PV = 8000\), \(r = 4.5\), \(k = 12\), and \(t = 5\).
\(FV = 8000 \times \left(1 + \frac{4.5}{1200}\right)^{12 \times 5} = 8000 \times (1.00375)^{60} \approx 10014.37\) USD.

(b) Let \(t\) be the number of years.
Chloe's investment value after \(t\) years is \(8000 \times (1.00375)^{12t}\).
Jack's car value after \(t\) years is \(15000 \times (1 - 0.12)^t = 15000 \times (0.88)^t\).
We solve for \(t\) when Chloe's value exceeds Jack's car value:
\(8000 \times (1.00375)^{12t} > 15000 \times (0.88)^t\).
Using a graphic display calculator (GDC) to find the intersection of \(y_1 = 8000 \times (1.00375)^{12x}\) and \(y_2 = 15000 \times (0.88)^x\), we find the intersection at \(t \approx 3.64\) years.
For \(t = 3\): Chloe's value is \(8000 \times (1.00375)^{36} \approx 9153.98\) and Jack's car value is \(15000 \times (0.88)^3 = 10222.08\).
For \(t = 4\): Chloe's value is \(8000 \times (1.00375)^{48} \approx 9574.33\) and Jack's car value is \(15000 \times (0.88)^4 = 8995.43\).
Therefore, it will take 4 complete years for Chloe's investment to exceed the value of Jack's car.

(a) [3 marks]
M1 for substituting the correct values into the compound interest formula or representing the correct GDC financial solver inputs (e.g., \(N=60\), \(I\%=4.5\), \(PV=-8000\), \(P/Y=12\), \(C/Y=12\)).
A1 for the correct substituted expression: \(8000 \times (1.00375)^{60}\).
A1 for the correct value \(10014.37\) (accept \(10014\) or \(10000\) to 3 s.f.).

(b) [3 marks]
M1 for setting up an appropriate inequality or equation comparing the two values, e.g., \(8000(1.00375)^{12t} > 15000(0.88)^t\).
A1 for finding the critical value \(t \approx 3.64\) (or \(3.6388...\)).
A1 for the correct number of complete years: 4.

(a) The midpoint of \(AB\) is \(\left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4, 2)\). Since the line segment \(AB\) is horizontal (both y-coordinates are 2), the perpendicular bisector is a vertical line passing through \(x = 4\). Thus, the equation of the perpendicular bisector of \(AB\) is \(x = 4\).

(b) The midpoint of \(AC\) is \(\left(\frac{1+4}{2}, \frac{2+8}{2}\right) = (2.5, 5)\). The gradient of \(AC\) is \(m_{AC} = \frac{8-2}{4-1} = \frac{6}{3} = 2\). The gradient of the perpendicular bisector is the negative reciprocal: \(m_{\perp} = -\frac{1}{2} = -0.5\). Using the point-slope form: \(y - 5 = -0.5(x - 2.5)\) which simplifies to \(y = -0.5x + 6.25\) (or equivalent form).

(c) The circumcenter is the intersection of the perpendicular bisectors. Substituting \(x = 4\) into \(y = -0.5x + 6.25\) gives \(y = -0.5(4) + 6.25 = 4.25\). Thus, the coordinates are \((4, 4.25)\) or \(\left(4, \frac{17}{4}\right)\).

(a) [2 marks]
M1 for finding the midpoint of \(AB\) as \((4, 2)\) or recognizing that the line is vertical.
A1 for the correct equation \(x = 4\).

(b) [3 marks]
M1 for finding the midpoint of \(AC\) as \((2.5, 5)\) and the gradient of \(AC\) as \(2\).
M1 for using the negative reciprocal gradient \(-0.5\) in a line equation.
A1 for a correct final equation: \(y = -0.5x + 6.25\) (accept equivalent forms, e.g., \(y - 5 = -0.5(x - 2.5)\) or \(2x + 4y = 25\)).

(c) [1 mark]
A1 for the correct coordinates \((4, 4.25)\) or \(\left(4, \frac{17}{4}\right)\).

Let \(X\) be the weight of an apple in grams. \(X \sim N(150, 12^2)\).

(a) We want to find \(P(X > 165)\). Using a GDC (normalcdf with lower bound 165, upper bound \(\infty\), \(\mu = 150\), \(\sigma = 12\)): \(P(X > 165) \approx 0.105649... \approx 0.106\) (to 3 s.f.).

(b) We are given \(P(X < w) = 0.15\). Using the inverse normal function on a GDC: \(w = \text{invNorm}(0.15, 150, 12) \approx 137.56... \approx 138\) grams (to 3 s.f.).

(c) Let \(Y\) be the number of apples weighing more than 165 grams in a box of 8. Then \(Y \sim B(8, p)\), where \(p = 0.105649...\). We want to find \(P(Y = 2)\).
Using a GDC (binompdf with \(n = 8\), \(p = 0.105649...\), \(x = 2\)): \(P(Y = 2) = \binom{8}{2} p^2 (1-p)^6 \approx 0.16048... \approx 0.160\) (to 3 s.f.).
If using the rounded value \(p = 0.106\), we get \(P(Y = 2) \approx 0.161\).

(a) [2 marks]
M1 for writing a correct probability statement \(P(X > 165)\) or showing a correct GDC normalcdf setup.
A1 for \(0.106\) (accept \(0.105649...\)).

(b) [2 marks]
M1 for setting up the equation \(P(X < w) = 0.15\) or showing a correct inverse normal GDC setup.
A1 for \(138\) (accept \(137.6\) or \(137.56...\)).

(c) [2 marks]
M1 for recognizing the binomial model \(Y \sim B(8, p)\) and attempting to calculate \(P(Y = 2)\).
A1 for \(0.160\) (accept \(0.161\) if \(p = 0.106\) was used).

**(a)**
To find the maximum rate, we find the derivative of \(R(t)\) with respect to \(t\):
\(R'(t) = 12 - 6t\)

Set the derivative to zero to find the stationary point:
\(12 - 6t = 0 \implies t = 2\)

Since \(R''(t) = -6 < 0\), this value of \(t\) represents a maximum.
Therefore, the water is flowing at the maximum rate when \(t = 2\) hours.

**(b)**
**Method 1 (Analytical integration):**
The volume of water in the tank, \(V(t)\), is the integral of the rate of flow:
\(V(t) = \int (12t - 3t^2) \, dt = 6t^2 - t^3 + C\)

Since the tank is initially empty, \(V(0) = 0 \implies C = 0\).
Thus, the volume function is:
\(V(t) = 6t^2 - t^3\)

Substitute \(t = 4\) to find the volume after 4 hours:
\(V(4) = 6(4)^2 - (4)^3 = 96 - 64 = 32\text{ m}^3\).

**Method 2 (Definite integration / GDC):**
The total volume after 4 hours is the definite integral of the rate from \(t = 0\) to \(t = 4\):
\(V(4) = \int_{0}^{4} (12t - 3t^2) \, dt\)

Evaluating this definite integral gives:
\(V(4) = 32\text{ m}^3\).

**(a)**
\(R'(t) = 12 - 6t\) *(M1)* for an attempt to differentiate \(R(t)\).
\(12 - 6t = 0\) *(M1)* for setting their derivative equal to 0.
\(t = 2\) (hours) *(A1)*

**(b)**
**Method 1:**
\(\int (12t - 3t^2) \, dt = 6t^2 - t^3 (+ C)\) *(M1)(A1)* for integrating at least one term correctly *(M1)* and the full integrated expression correct *(A1)*.
\(V(0) = 0 \implies C = 0\) *(M1)* for applying the initial boundary condition to find \(C\).
\(32\text{ m}^3\) *(A1)*

**Method 2:**
\(\int_{0}^{4} (12t - 3t^2) \, dt\) *(M2)* for setting up the correct definite integral with correct limits.
\(32\text{ m}^3\) *(A2)* for the correct evaluation of the definite integral.

(a) Let \(V \sim N(500, 8^2)\).
We want to find \(P(V < 490)\).
Using a graphic display calculator (GDC):
\(P(V < 490) \approx 0.10565 \approx 0.106\) (to 3 s.f.)

(b) Let \(X\) be the number of bottles with less than \(490\) ml.
\(X \sim B(3, 0.10565)\).
We want to find \(P(X = 2)\).
Using the binomial probability formula or GDC:
\(P(X = 2) = \binom{3}{2} (0.10565)^2 (1 - 0.10565) \approx 0.02995 \approx 0.0300\) (to 3 s.f.)

(c) Let \(V'\) be the new volume distribution, where \(V' \sim N(\mu', 8^2)\).
We want \(P(V' < 490) = 0.01\).
The corresponding z-score for the 1st percentile is:
\(z \approx -2.3263\)
Using the standardisation formula \(z = \frac{x - \mu}{\sigma}\):
\(-2.3263 = \frac{490 - \mu'}{8}\)
\(-18.61 = 490 - \mu'\)
\(\mu' \approx 508.6\) ml (to 1 d.p.)

(a)
(M1) for setting up the normal cumulative distribution with correct parameters (e.g. sketching or writing \(P(V < 490)\)).
(A1) for \(0.106\) (accept \(0.1056\)). [2 marks]

(b)
(M1) for recognizing a binomial distribution \(B(3, p)\).
(M1) for correct substitution into binomial formula or GDC calculation.
(A1) for \(0.0300\) (accept \(0.0299\)). [3 marks]

(c)
(M1) for finding the z-score of \(-2.326\) (or equivalent equation using invNorm).
(M1) for setting up the equation \(\frac{490 - \mu'}{8} = -2.326\).
(A1) for \(508.6\) (must be to 1 decimal place as requested). [3 marks]

(a) Scheme A represents an arithmetic sequence with first term \(u_1 = 5000\) and common difference \(d = 600\).
The 10th term is given by:
\(u_{10} = u_1 + 9d = 5000 + 9(600) = 10400\)

(b) The sum of the first 10 terms for Scheme A is:
\(S_{10} = \frac{10}{2}(u_1 + u_{10}) = 5(5000 + 10400) = 77000\)

(c) Scheme B represents a geometric sequence with first term \(u_1 = 5000\) and common ratio \(r = 1.08\).
The sum of the first 10 terms is:
\(S_{10} = \frac{5000(1.08^{10} - 1)}{1.08 - 1} \approx 72432.81\)
So, the total donation is $72433 (or $72400 to 3 s.f.).

(d) Scheme A results in a greater total donation.
Difference:
\(77000 - 72432.81 = 4567.19\)
So, Scheme A exceeds Scheme B by $4567 (or $4570 to 3 s.f.).

(a)
(M1) for using the arithmetic sequence term formula.
(A1) for \(10400\). [2 marks]

(b)
(M1) for using the arithmetic series sum formula.
(A1) for \(77000\). [2 marks]

(c)
(M1) for identifying the geometric sequence parameters \(u_1 = 5000, r = 1.08\).
(M1) for substituting into the geometric series sum formula.
(A1) for \(72433\) (or \(72400\)). [3 marks]

(d)
(A1) for Scheme A by $4567 (accept $4570, or $4600 if using 3 s.f. rounded values from part c). [1 mark]

(a) Using Pythagoras' theorem on the right-angled triangle \(ABC\) on the base:
\(AC = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2} \approx 17.0\text{ cm}\)

(b) Since \(O\) is the center of the square base, the length of \(AO\) is half of \(AC\):
\(AO = \frac{1}{2} AC = 6\sqrt{2} \approx 8.485\text{ cm}\).
Using Pythagoras' theorem on the vertical right-angled triangle \(AOV\):
\(AV = \sqrt{AO^2 + VO^2} = \sqrt{(6\sqrt{2})^2 + 8^2} = \sqrt{72 + 64} = \sqrt{136} \approx 11.7\text{ cm}\)

(c) The angle between the edge \(AV\) and the base is the angle \(\theta = \angle VAO\):
\(\tan \theta = \frac{VO}{AO} = \frac{8}{6\sqrt{2}}\)
\(\theta = \arctan\left(\frac{4}{3\sqrt{2}}\right) \approx 43.3^\circ\) (or \(0.756\) radians)

(d) Let \(M\) be the midpoint of side \(AB\). The distance from \(O\) to \(M\) is \(6\text{ cm}\).
Using Pythagoras' theorem on the triangle \(VOM\), the slant height \(VM\) of each triangular face is:
\(VM = \sqrt{VO^2 + OM^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\).
Area of one triangular face is:
\(\text{Area}_{\text{face}} = \frac{1}{2} \times 12 \times 10 = 60\text{ cm}^2\).
Total Surface Area (TSA) is the sum of the square base and the four triangular faces:
\(\text{TSA} = 12^2 + 4(60) = 144 + 240 = 384\text{ cm}^2\).

(a)
(M1) for attempting to use Pythagoras' theorem on the base.
(A1) for \(17.0\text{ cm}\) (or \(12\sqrt{2}\)). [2 marks]

(b)
(M1) for finding \(AO\) and attempting Pythagoras' theorem with height \(8\).
(A1) for \(11.7\text{ cm}\) (or \(\sqrt{136}\)). [2 marks]

(c)
(M1) for attempting to use a trigonometric ratio in triangle \(AOV\).
(A1) for \(43.3^\circ\) (or \(0.756\text{ rad}\)). [2 marks]

(d)
(M1) for finding the slant height \(10\text{ cm}\) (or calculating the area of one face).
(A1) for \(384\text{ cm}^2\). [2 marks]

(a) Differentiating the profit function term-by-term:
\(P'(x) = -6x^2 + 30x - 24\)

(b) Setting the derivative equal to zero to find the critical values:
\(-6x^2 + 30x - 24 = 0\)
\(-6(x^2 - 5x + 4) = 0\)
\(-6(x - 1)(x - 4) = 0\)
This yields critical points at \(x = 1\) and \(x = 4\).
Testing with the second derivative \(P''(x) = -12x + 30\):
At \(x = 4\), \(P''(4) = -18 < 0\), confirming a local maximum.
Thus, profit is maximized when \(x = 4\).

(c) Substituting \(x = 4\) back into the original function:
\(P(4) = -2(4)^3 + 15(4)^2 - 24(4) - 5 = -128 + 240 - 96 - 5 = 11\)
Since \(P\) is in thousands of dollars, the maximum profit is $11000.

(d) Setting \(P(x) \ge 0\) and using GDC to find the roots of the cubic equation \(-2x^3 + 15x^2 - 24x - 5 = 0\) in the interval \([0.5, 6]\):
The lower boundary root is \(x \approx 2.68614\).
To not make a loss, we require \(x \ge 2.68614\).
Since \(x\) represents the number of hundreds of items, the required number of items is:
\(100 \times 2.68614 = 268.614\) items.
Rounding up to the nearest integer to ensure a non-negative profit, we get 269 items.

(a)
(M1) for an attempt to differentiate.
(A1) for \(-6x^2 + 30x - 24\). [2 marks]

(b)
(M1) for setting their derivative equal to 0.
(A1) for \(x = 4\) (with justification or indication that this is the maximum). [2 marks]

(c)
(M1) for substituting their \(x\) value from part (b) into the original equation.
(A1) for $11000 (accept 11). [2 marks]

(d)
(M1) for setting \(P(x) = 0\) and identifying the critical root \(x \approx 2.686\).
(A1) for 269 (must be an integer, 268 is incorrect since it leads to a loss). [2 marks]

(a) \(H_0\): Choice of preferred music genre is independent of age group (or "there is no association between preferred music genre and age group").

(b) \(\text{Degrees of freedom} = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(3 - 1) = 2\).

(c) \(\text{Expected frequency} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{90 \times 55}{200} = 24.75\).

(d) Using a GDC to run a \(\chi^2\) two-way test:
(i) \(\chi^2 \approx 26.6\) (accept \(26.586...\))
(ii) \(p \approx 1.69 \times 10^{-6}\) (or \(0.00000169\))

(e) Since the \(p\)-value \(\approx 1.69 \times 10^{-6} < 0.05\) (or the calculated \(\chi^2\) value of \(26.6\) is greater than the critical value of \(5.991\)), we reject the null hypothesis \(H_0\). There is strong evidence that preferred music genre is not independent of age group.

(a) A1 for stating the null hypothesis correctly (must contain "independent" or "no association").

(b) A1 for 2.

(c) M1 for showing the correct formula setup: \(\frac{90 \times 55}{200}\).
A1 for obtaining the correct given value of 24.75.

(d) (i) A1 for \(\chi^2 \approx 26.6\).
(ii) A2 for \(p \approx 1.69 \times 10^{-6}\) (or \(0.00000169\)). Award A1 for a value that rounds to \(1.7 \times 10^{-6}\) or for an extremely small value close to 0.

(e) R1 for a valid comparison (e.g., comparing \(p\)-value to \(0.05\) or comparing \(\chi^2\) to critical value \(5.991\)).
A1 for the correct conclusion of rejecting \(H_0\) consistent with their comparison.

(a) Using the compound interest formula:
\(FV = PV \left(1 + \frac{r}{k}\right)^{kt}\)
\(FV = 15\,000 \left(1 + \frac{0.045}{12}\right)^{12 \times 5}\)
\(FV = 15\,000 (1.00375)^{60}\)
\(FV \approx \$18\,776.94\) (to the nearest cent).

(b) For the next 4 years (48 months), the initial balance is the final balance from part (a).
Using a financial calculator (TVM solver) with the following parameters:
\(N = 48\)
\(I\% = 4.5\)
\(PV = -18\,776.9366\) (or \(-18\,776.94\))
\(PMT = -200\)
\(P/Y = 12\)
\(C/Y = 12\)

Alternatively, using the annuity formula:
\(FV = 18\,776.94(1.00375)^{48} + 200 \times \frac{(1.00375)^{48} - 1}{0.00375}\)
\(FV = 22\,472.52 + 10\,496.81 = \$32\,969.33\) (to the nearest cent).

(c) Total contributions made by Sarah:
\(\text{Initial Principal} + \text{Total Monthly Deposits} = 15\,000 + 48 \times 200 = \$24\,600\).

Total interest earned:
\(\text{Interest} = \text{Final Value} - \text{Total Contributions}\)
\(\text{Interest} = 32\,969.33 - 24\,600 = \$8\,369.33\).

(a) M1 for a correct substitution into the compound interest formula.
A1 for seeing \(15\,000(1.00375)^{60}\) or equivalent.
A1 for \(\$18\,776.94\) (must be rounded to the nearest cent; do not award if not rounded correctly).

(b) M1 for setting up the TVM parameters or annuity formula correctly.
A1 for the future value of the lump sum part: \(22\,472.52\) (or \(22\,472.51\)).
A1 for the future value of the annuity part: \(10\,496.81\).
A1 for the total final value: \(\$32\,969.33\) (accept \(\$32\,969.32\) from rounded intermediate steps).

(c) M1 for calculating total contributions: \(15\,000 + (48 \times 200) = 24\,600\).
A1 for \(\$8\,369.33\) (accept \(\$8\,369.32\) if consistent with part (b)).

(a) \(H_0\): Preferred weekend activity and age group are independent.

(b) Expected frequency for (Over 40, Reading) = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{95 \times 80}{300} = \frac{7600}{300} = 25.333...\) which is \(25.3\) (correct to 3 significant figures).

(c) Degrees of freedom = \((r - 1)(c - 1) = (3 - 1)(3 - 1) = 2 \times 2 = 4\).

(d) Using a GDC:
(i) \(\chi^2 \approx 36.8\) (or \(36.779...\))
(ii) \(p\text{-value} \approx 1.99 \times 10^{-7}\) (accept \(2.0 \times 10^{-7}\) or \(0.000000199\))

(e) Since the \(p\)-value \((1.99 \times 10^{-7}) < 0.05\), we reject the null hypothesis \(H_0\). There is sufficient evidence to conclude that preferred weekend activity is not independent of age group.

(f) (i) \(\text{P}(\text{both Sports}) = \frac{100}{300} \times \frac{99}{299} = \frac{1}{3} \times \frac{99}{299} = \frac{33}{299} \approx 0.110\) (or \(0.110368...\))
(ii) We are selecting from the 100 residents who prefer Sports. There are 45 in 'Under 18' and 20 in 'Over 40'.
The probability is given by: \(\frac{\binom{45}{1} \times \binom{20}{1}}{\binom{100}{2}} = \frac{45 \times 20}{4950} = \frac{900}{4950} = \frac{2}{11} \approx 0.182\).
Alternatively: \(\frac{45}{100} \times \frac{20}{99} + \frac{20}{100} \times \frac{45}{99} = \frac{1800}{9900} = \frac{2}{11} \approx 0.182\).

(a) Award A1 for stating independence clearly. Do not accept 'related' or 'associated' without qualification.
H_0: Preferred weekend activity and age group are independent. [1 mark]

(b) Award M1 for substituting the correct values into the expected value formula: \(\frac{95 \times 80}{300}\).
Award A1 for showing the unrounded value \(25.333...\) leading to \(25.3\). [2 marks]

(c) Award A1 for 4. [1 mark]

(d) (i) Award A2 for \(36.8\) (or \(36.779...\)).
(ii) Award A1 for \(1.99 \times 10^{-7}\) (or \(2.0 \times 10^{-7}\)). [3 marks]

(e) Award R1 for comparing their p-value to 0.05 (e.g. \(p < 0.05\) or \(\chi^2 > 9.488\)).
Award A1 for a consistent conclusion in context (reject \(H_0\), activity and age are not independent). [2 marks]

(f) (i) Award M1 for set-up \(\frac{100}{300} \times \frac{99}{299}\). Award A1 for \(0.110\) (accept \(0.110368...\)). [1 mark]
(ii) Award M1 for setting up the conditional probability calculation: \(\frac{45 \times 20}{\binom{100}{2}}\) or \(\frac{45}{100} \times \frac{20}{99} \times 2\).
Award A1 for \(0.182\) (accept \(\frac{2}{11}\) or \(0.181818...\)). [2 marks]

(a) (i)
\(H_0\): Gender and weekly visit frequency are independent (or "there is no association between gender and weekly visit frequency").
\(H_1\): Gender and weekly visit frequency are not independent (or "there is an association between gender and weekly visit frequency").

(ii)
Total number of females = \(35 + 65 + 45 = 145\)
Total number of high-frequency visits = \(30 + 45 = 75\)
Grand total = \(275\)

$$\text{Expected value} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{145 \times 75}{275} \approx 39.545 \approx 39.5$$

(iii)
$$\text{df} = (r - 1)(c - 1) = (2 - 1)(3 - 1) = 2$$

(iv)
Using a graphic display calculator (GDC) to run a \(\chi^2\) two-way test on the matrix:
\(\chi^2 \approx 4.28\)
\(p\)-value \approx 0.118 (or 0.1178...)

(v)
Since the \(p\)-value (\(0.118\)) is greater than the significance level (\(0.05\)), we fail to reject the null hypothesis \(H_0\).
There is insufficient evidence to suggest that a member's weekly visit frequency is associated with their gender.

(b) (i)
Let \(X \sim N(55, 12^2)\).
Using GDC normCDF(\(60, \infty, 55, 12\)):
$$P(X > 60) \approx 0.339 \quad (0.33853...)$$

(ii)
This follows a binomial distribution \(Y \sim B(5, p)\), where \(p = 0.33853...\).
We want to find \(P(Y = 3)\):
$$P(Y = 3) = \binom{5}{3} (0.33853)^3 (1 - 0.33853)^2 \approx 0.170 \quad (0.1697...)$$

(iii)
We want to find \(k\) such that \(P(X > k) = 0.15\), which is equivalent to \(P(X \le k) = 0.85\).
Using GDC invNorm(\(0.85, 55, 12\)):
$$k \approx 67.437\text{ minutes}$$
To the nearest minute, the minimum time is \(67\) minutes.

(iv)
We require the conditional probability \(P(X < 65 \mid X > 45)\).
$$P(X < 65 \mid X > 45) = \frac{P(45 < X < 65)}{P(X > 45)}$$
Using GDC:
$$P(45 < X < 65) = \text{normCDF}(45, 65, 55, 12) \approx 0.59534$$
$$P(X > 45) = \text{normCDF}(45, \infty, 55, 12) \approx 0.79767$$
$$\text{Conditional Probability} = \frac{0.59534}{0.79767} \approx 0.746 \quad (0.74635...)$$

(a) (i)
A1: for stating both correct null and alternative hypotheses. Must include independence/association in context.

(ii)
M1: for a correct fraction/setup utilizing totals.
A1: for the correct expected value of 39.5 (or 39.545...).

(iii)
A1: for the correct degrees of freedom (2).

(iv)
A1: for the correct p-value of 0.118 (or 0.1178...).

(v)
R1: for comparing their p-value to 0.05.
A1: for a consistent conclusion in context (do not reject H0, no significant association).

(b) (i)
M1: for standardizing or setting up the normal CDF on GDC.
A1: for 0.339.

(ii)
M1: for identifying a binomial distribution setup with n = 5 and their p-value from (b)(i).
M1: for substituting values into binomial formula or GDC binomial PDF.
A1: for 0.170.

(iii)
M1: for a correct probability statement like P(X < k) = 0.85 or P(X > k) = 0.15.
A1: for 67.437...
A1: for rounding to the nearest minute (67 minutes).

(iv)
M1: for setting up the conditional probability formula.
M1: for finding both numerator and denominator probabilities (0.595 and 0.798).
A1: for 0.746.

(a)
The volume of a rectangular prism with a square base of side \(x\) and height \(h\) is given by:
$$V = x^2 h$$
Given that the volume is \(12\text{ m}^3\):
$$12 = x^2 h \implies h = \frac{12}{x^2}$$

(b)
Let's calculate the cost of each component:
- Base area = \(x^2\). Cost of base = \(5 \times x^2 = 5x^2\).
- Outer side walls: There are four walls, each of area \(x h\).
Cost of outer walls = \(3 \times 4xh = 12xh\).
- Internal partition: One vertical partition of dimensions \(x\) by \(h\).
Cost of partition = \(2 \times xh = 2xh\).

Total Cost, \(C\), is the sum of these costs:
$$C = 5x^2 + 12xh + 2xh = 5x^2 + 14xh$$

Substitute \(h = \frac{12}{x^2}\) into the total cost equation:
$$C(x) = 5x^2 + 14x \left(\frac{12}{x^2}\right) = 5x^2 + \frac{168}{x}$$

(c)
$$C'(x) = 10x - 168x^{-2} = 10x - \frac{168}{x^2}$$

(d)
To find the minimum, set \(C'(x) = 0\):
$$10x - \frac{168}{x^2} = 0 \implies 10x^3 = 168 \implies x^3 = 16.8$$
$$x = \sqrt[3]{16.8} \approx 2.5612 \approx 2.56\text{ m}$$

To justify that this is a minimum, we find the second derivative \(C''(x)\):
$$C''(x) = 10 + \frac{336}{x^3}$$
Substitute \(x = 2.5612\):
$$C''(2.5612) = 10 + \frac{336}{16.8} = 10 + 20 = 30$$
Since \(C''(x) > 0\) for \(x > 0\), the value \(x \approx 2.56\) represents a local minimum.

(e)
Substitute \(x = 2.5612\) into the cost function:
$$C(2.5612) = 5(2.5612)^2 + \frac{168}{2.5612} \approx 32.80 + 65.59 = 98.39$$
To the nearest dollar, the minimum cost is \(\$98\).

(f)
Any reasonable practical limitation, such as:
- Material waste produced when cutting the wood sheets is not accounted for.
- The model assumes the panels have zero thickness, which is physically impossible.
- It does not account for labor, joining components (screws, glue), or finishing costs.

(a)
M1: for writing down the correct volume formula \(V = x^2 h\).
A1: for correctly isolating \(h\) to obtain \(h = \frac{12}{x^2}\).

(b)
M1: for writing an expression for the cost of the base (\(5x^2\)) and outer walls (\(12xh\)).
M1: for writing a correct term for the partition cost (\(2xh\)).
M1: for substituting \(h = \frac{12}{x^2}\) into a valid cost equation.
A1: for showing the correct final given algebraic form.

(c)
M1: for attempting to differentiate \(5x^2\) or \(168/x\).
A1: for the correct derivative \(10x - \frac{168}{x^2}\).

(d)
M1: for setting \(C'(x) = 0\).
A1: for solving for \(x\) to find \(x \approx 2.56\).
M1: for finding the second derivative expression \(C''(x) = 10 + \frac{336}{x^3}\).
R1: for showing \(C''(2.56) > 0\) and concluding it is a minimum.

(e)
M1: for substituting their value of \(x\) back into the cost function.
A1: for \(\$98\) (must be rounded to the nearest dollar).

(f)
R2: for any valid practical limitation stated clearly in context (1 mark for a vague attempt, 2 marks for a fully realized practical point).

(a)
This is a standard compound interest problem:
$$A = P \left(1 + \frac{r}{100k}\right)^{kn}$$
where \(P = 1000\), \(r = 4.5\), \(k = 12\) (compounded monthly), and \(n = 5\) years.
$$A = 1000 \left(1 + \frac{0.045}{12}\right)^{60} = 1000 (1.00375)^{60} \approx 1251.7957$$
To the nearest cent, the value is \(\$1251.80\).

(b) (i)
This forms a geometric sequence with first term \(u_1 = 500\) and common ratio \(r = 1.05\).
We want to find \(u_{10}\):
$$u_{10} = u_1 \times r^9 = 500 \times (1.05)^9 \approx 775.664$$
Elena's deposit at the start of Year 10 is \(\$775.66\).

(ii)
We want to find the sum of the first 10 terms of this geometric sequence:
$$S_{10} = \frac{u_1(r^{10} - 1)}{r - 1} = \frac{500(1.05^{10} - 1)}{1.05 - 1} = \frac{500(1.6288946 - 1)}{0.05} \approx 6288.946$$
The total amount deposited up to the start of Year 10 is \(\$6288.95\).

(c)
Using the TVM solver on a GDC:
- \(N = 8 \times 4 = 32\) (number of quarters)
- \(I\% = 3.6\) (annual interest rate)
- \(PV = \text{to find}\)
- \(PMT = 800\) (withdrawals/receipts are positive, or vice versa depending on sign convention)
- \(FV = 0\)
- \(P/Y = 4\)
- \(C/Y = 4\)

Alternatively, using the formula for the present value of an ordinary annuity:
$$PV = PMT \times \frac{1 - (1 + i)^{-n}}{i}$$
where \(i = \frac{0.036}{4} = 0.009\) and \(n = 32\).
$$PV = 800 \times \frac{1 - (1.009)^{-32}}{0.009} \approx 800 \times \frac{1 - 0.750731}{0.009} \approx 800 \times 27.69655 \approx 22157.24$$
Using a precise TVM solver:
\(PV \approx -22157.34\)
Thus, the required lump sum deposit is \(P = \$22157.34\).

(d)
Total payments received by Elena:
$$\text{Total Received} = 32 \times 800 = \$25600.00$$

Total interest earned is the difference between the total money received and the initial deposit \(P\):
$$\text{Interest} = \text{Total Received} - P = 25600 - 22157.34 = \$3442.66$$

(a)
M1: for setting up the compound interest formula with correct values.
A1: for finding \(1251.80\) (accept \(1251.79\) or \(1251.80\) with a correct rounding method).

(b) (i)
M1: for writing a correct geometric term expression \(500 \times (1.05)^9\).
A1: for \(\$775.66\).

(ii)
M1: for attempting to use the sum of a geometric sequence formula.
A1: for substituting correct values into the formula.
A1: for \(\$6288.95\).

(c)
M3: for inputting correct values into TVM Solver (Award M1 for correct N, M1 for correct I, M1 for PMT/FV setup) OR for using the correct analytical formula setup.
A1: for \(\$22157.34\) (accept answers close due to rounding, e.g., \(22157.24\) if formula was used with intermediate rounding).

(d)
M1: for calculating total payments received (\(32 \times 800 = 25600\)).
M1: for subtracting \(P\) from the total payments received.
A1: for \(\$3442.66\) (follow through from their part (c)).

### Part A

(a) Let \(W\) be the weight of the tea bags, where \(W \sim N(3.2, 0.15^2)\).
Using a graphic display calculator (GDC):
\(P(W < 3.0) \approx 0.091211...\)
To 3 significant figures, \(P(W < 3.0) = 0.0912\).

(b) First, find the probability that a tea bag is overfilled:
\(P(W > 3.5) \approx 0.022750...\)
For a box of 50 tea bags, the expected number is:
\(E(X) = 50 \times 0.022750... \approx 1.1375...\)
To 3 significant figures, the expected number is \(1.14\).

(c) We are given that the new mean \(\mu_1\) satisfies:
\(P(W < 3.0) = 0.01\)
Standardizing this expression:
\(P\left(Z < \frac{3.0 - \mu_1}{0.15}\right) = 0.01\)
Using the inverse normal function on a GDC:
\(\frac{3.0 - \mu_1}{0.15} = -2.3263...\)
\(3.0 - \mu_1 = -2.3263... \times 0.15\)
\(3.0 - \mu_1 = -0.3489...\)
\(\mu_1 = 3.3489...\)
To 3 significant figures, \(\mu_1 = 3.35\text{ grams}\).

---

### Part B

(d) \(H_0\): Customer preference for tea flavor is independent of their age group (or there is no association between tea flavor preference and age group).

(e) \(\text{Degrees of freedom} = (\text{rows} - 1)(\text{columns} - 1) = (3 - 1)(3 - 1) = 2 \times 2 = 4\).

(f) \(\text{Expected frequency} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)
For Over 50 and Mint:
\(\text{Expected Frequency} = \frac{50 \times 70}{200} = 17.5\).

(g) Entering the observed frequencies matrix into a GDC and performing a \(\chi^2\) two-way test yields:
\(\chi^2 \text{ statistic} \approx 14.696... \approx 14.7\)
\(p\text{-value} \approx 0.0053738... \approx 0.00537\)

(h) Since the \(p\)-value (\(0.00537\)) is less than the significance level \(\alpha = 0.05\) (or because the critical value for \(df=4\) at \(5\%\) is \(9.488\) and \(14.7 > 9.488\)), we reject the null hypothesis \(H_0\).
There is significant evidence to suggest that tea flavor preference is dependent on age group.

---

### Part C

(i) Entering the values of \(T\) and \(B\) into lists on a GDC:
\(r \approx -0.99383...\)
To 3 significant figures, \(r = -0.994\).

(j) The equation of the regression line of \(B\) on \(T\) is of the form \(B = mT + c\).
From the GDC:
\(m \approx -1.6737...\)
\(c \approx 65.982...\)
So, the regression line equation is:
\(B = -1.67T + 66.0\) (to 3 s.f.) or \(B = -1.6737T + 65.982\).

(k) Substituting \(T = 24\) into the regression equation:
Using the unrounded model:
\(B = -1.6737(24) + 65.982 = 25.813...\)
Using the rounded model:
\(B = -1.67(24) + 66.0 = 25.92\)
Thus, the estimated number of boxes sold is \(25.8\) (or \(25.9\)), which rounds to \(26\) boxes.

### Part A
(a)
*(M1)* for setting up the normal distribution parameters or an equivalent probability statement.
*(A1)* for \(0.0912\) (or \(9.12\%\)).

(b)
*(M1)* for finding \(P(W > 3.5) = 0.0228\) and multiplying by 50.
*(A1)* for \(1.14\) (or \(1.1375...\)).

(c)
*(M1)* for establishing the standardized equation \(\frac{3.0 - \mu_1}{0.15} = -2.3263...\) or using invNorm on GDC.
*(A1)* for \(3.35\text{ g}\) (or \(3.3489...\)).

### Part B
(d)
*(A1)* for stating \(H_0\) in terms of independence between age group and flavor preference.

(e)
*(A1)* for \(4\).

(f)
*(M1)* for showing a valid calculation method: \(\frac{50 \times 70}{200}\).
*(A1)* for \(17.5\).

(g)
*(A1)* for \(\chi^2 = 14.7\) (accept \(14.696...\)).
*(A1)* for \(p\text{-value} = 0.00537\) (accept \(0.0053738...\)).

(h)
*(R1)* for comparing the \(p\)-value with \(0.05\) (i.e., \(0.00537 < 0.05\)) or comparing the \(\chi^2\) statistic with the critical value.
*(A1)* for concluding that the null hypothesis is rejected.

### Part C
(i)
*(M1)* for attempting to enter data in GDC list.
*(A1)* for \(r = -0.994\) (accept \(-0.99383...\)).

(j)
*(A1)* for \(m = -1.67\) (or \(-1.6737...\)).
*(A1)* for \(c = 66.0\) (or \(65.982...\)).
*Note: Award maximum (A1)(A0) if not written as an equation.*

(k)
*(M1)* for substituting \(T=24\) into their regression equation.
*(A1)* for \(25.8\) or \(25.9\) (accept \(26\) as a rounded integer representation).