2025 IB DP Mathematics - Applications and Interpretation Practice Paper with Answers

(a) Using the compound interest formula: \(A = P\left(1 + \frac{r}{100k}\right)^{kt}\), where \(P = 5000\), \(r = 4.2\), \(k = 12\), and \(t = 3\). \(A = 5000\left(1 + \frac{4.2}{1200}\right)^{36} = 5000(1.0035)^{36} \approx 5670.08\) USD. (b) To double the investment, we require \(5000(1.0035)^n = 10000\), where \(n\) is the number of months. \((1.0035)^n = 2\) \(n = \frac{\ln(2)}{\ln(1.0035)} \approx 198.39\) months. Since we require the number of complete months, we round up to 199 months.

(a) M1 for substituting into the compound interest formula. A1 for correct value: 5670.08 (accept 5670). (b) M1 for setting up the equation \(5000(1.0035)^n = 10000\) or equivalent. A1 for finding \(n \approx 198.39\). A1 for rounding up to 199 months.

(a) The initial height is at \(t = 0\): \(h(0) = -0.4(0)^2 + 4.8(0) + 1.2 = 1.2\) m. (b) The maximum height occurs at the vertex where \(t = -\frac{b}{2a} = -\frac{4.8}{2(-0.4)} = 6\) seconds. The maximum height is \(h(6) = -0.4(6)^2 + 4.8(6) + 1.2 = -14.4 + 28.8 + 1.2 = 15.6\) m. (c) The drone hits the ground when \(h(t) = 0\). Solving \(-0.4t^2 + 4.8t + 1.2 = 0\) using GDC or quadratic formula: \(t \approx 12.245\) or \(t \approx -0.245\). Since \(t \ge 0\), \(t \approx 12.2\) seconds.

(a) A1 for 1.2 m. (b) M1 for finding the time at maximum height \(t = 6\) (or using derivative \(h'(t) = 0\)). A1 for 15.6 m. (c) M1 for setting \(h(t) = 0\). A1 for 12.2 (accept 12.245).

(a) The angle \(BAC = 115^\circ - 40^\circ = 75^\circ\). (b) Using the cosine rule on triangle \(ABC\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(BAC)\). \(BC^2 = 120^2 + 180^2 - 2(120)(180)\cos(75^\circ)\). \(BC^2 = 14400 + 32400 - 43200\cos(75^\circ) \approx 35619.01\). \(BC \approx 188.73\) m, which is 189 m (to 3 s.f.). (c) To find the bearing of \(C\) from \(B\), first find the angle \(ABC\) using the sine rule: \(\frac{\sin(ABC)}{180} = \frac{\sin(75^\circ)}{188.73}\) which gives \(\sin(ABC) \approx 0.92125\). Since \(BC > AC\), angle \(ABC\) is acute, so \(ABC \approx 67.11^\circ\). The bearing of \(A\) from \(B\) is \(40^\circ + 180^\circ = 220^\circ\). Thus, the bearing of \(C\) from \(B\) is \(220^\circ - 67.11^\circ = 152.89^\circ\), which is \(153^\circ\) (to 3 s.f.).

(a) A1 for \(75^\circ\). (b) M1 for substituting correct values into cosine rule. A1 for 189 m (accept 188.73). (c) M1 for using sine rule (or cosine rule) to find angle \(ABC \approx 67.11^\circ\). A1 for subtracting \(67.11^\circ\) from \(220^\circ\) to obtain bearing of \(153^\circ\) (accept 152.89).

(a) Let \(X\) be the weight of a package. \(X \sim N(1005, 4^2)\). We want to find \(P(X < 1000)\). Using the GDC normal cumulative distribution function: \(P(X < 1000) \approx 0.10565 \approx 0.106\). (b) We need to find the value of \(w\) such that \(P(X < w) = 0.02\). Using the GDC inverse normal function with area 0.02, mean 1005, and standard deviation 4: \(w \approx 996.785\) grams. To 3 significant figures, this is 997 grams.

(a) M1 for setting up normal probability \(P(X < 1000)\) or finding \(z = -1.25\). A1 for 0.106 (accept 0.10565). (b) M1 for setting up inverse normal equation \(P(X < w) = 0.02\). M1 for finding the correct z-score \(-2.0537\) or substituting correct parameters into GDC. A1 for 997 g (accept 996.785).

(a) Rewrite the curve equation as \(y = 3x^2 - 8x^{-1} + 2\). Differentiating with respect to \(x\): \(\frac{dy}{dx} = 6x - 8(-1)x^{-2} = 6x + \frac{8}{x^2}\). (b) At \(x = 2\), the y-coordinate of the point is: \(y = 3(2)^2 - \frac{8}{2} + 2 = 12 - 4 + 2 = 10\). The gradient of the tangent is \(m = 6(2) + \frac{8}{2^2} = 12 + 2 = 14\). Using the point-slope formula \(y - y_1 = m(x - x_1)\): \(y - 10 = 14(x - 2) \implies y - 10 = 14x - 28 \implies y = 14x - 18\).

(a) M1 for differentiating \(3x^2\) to \(6x\). A1 for differentiating \(-\frac{8}{x}\) to \(\frac{8}{x^2}\). (b) M1 for finding the y-coordinate \(y = 10\). M1 for finding the gradient \(m = 14\). A1 for the correct tangent equation \(y = 14x - 18\).

(a) The total length of the fencing is \( 2x + y = 120 \), where \( y \) is the length of the paddock parallel to the river. Rearranging for \( y \): \( y = 120 - 2x \). The area \( A \) of the rectangle is given by: \( A = x \cdot y = x(120 - 2x) = 120x - 2x^2 \). (b) Differentiating \( A(x) \) with respect to \( x \): \( \frac{dA}{dx} = 120 - 4x \). (c) To find the maximum area, set \( \frac{dA}{dx} = 0 \): \( 120 - 4x = 0 \Rightarrow 4x = 120 \Rightarrow x = 30\text{ m} \). The maximum area is: \( A(30) = 120(30) - 2(30)^2 = 3600 - 1800 = 1800\text{ m}^2 \).

(a) M1: For expressing the perimeter/fencing equation: \( 2x + y = 120 \) (or equivalent). A1: For substituting \( y = 120 - 2x \) into \( A = xy \) and simplifying to obtain \( 120x - 2x^2 \) (must show intermediate step). (b) A1: \( \frac{dA}{dx} = 120 - 4x \). (c) M1: Setting their derivative to 0: \( 120 - 4x = 0 \). A1: \( x = 30 \). A1: \( \text{Area} = 1800\text{ m}^2 \).

Let \( X \sim N(150, 18^2) \). (a) Use a GDC to find \( P(130 < X < 165) \approx 0.664188... \approx 0.664 \). (b) We are given \( P(X < k) = 0.15 \). Using the inverse normal function on a GDC: \( k \approx 131.345... \approx 131\text{ g} \). (c) First find \( P(X > 165) \approx 0.202328... \). The expected number of apples is given by: \( \text{Expected value} = n \times p = 120 \times 0.202328... \approx 24.279... \approx 24.3 \).

(a) M1: For a correct probability expression, e.g., \( P(130 < X < 165) \). A1: \( 0.664 \) (accept \( 0.664188... \)). (b) M1: For a correct probability expression involving \( k \), e.g., \( P(X < k) = 0.15 \). A1: \( 131 \) (accept \( 131.345... \)). (c) M1: For finding \( P(X > 165) \approx 0.202 \) or showing the product \( 120 \times P(X > 165) \). A1: \( 24.3 \) (accept 24 or \( 24.279... \)).

(a) The value of the car, \( V \), after 5 years is modeled by: \( V = 24000 \times (1 - 0.12)^5 = 24000 \times (0.88)^5 \approx 12665.61 \). To the nearest dollar, the value is $12666. (b) The savings account value, \( S \), after 5 years (60 months) is: \( S = 15000 \times \left(1 + \frac{0.045}{12}\right)^{60} \approx 18777.01 \). To the nearest dollar, the value is $18777. (c) Let \( n \) be the number of complete years. We solve the inequality: \( 15000 \times \left(1 + \frac{0.045}{12}\right)^{12n} > 24000 \times (0.88)^n \). For \( n = 2 \): \( C = 18585.60 \) and \( S \approx 16410.15 \) (car is worth more). For \( n = 3 \): \( C \approx 16355.33 \) and \( S \approx 17164.55 \) (savings are worth more). Thus, it takes 3 complete years.

(a) M1: For writing a correct expression for depreciation: \( 24000 \times (0.88)^5 \). A1: $12666 (must be rounded to the nearest dollar). (b) M1: For writing a correct expression for compound interest: \( 15000 \times \left(1 + \frac{0.045}{12}\right)^{60} \). A1: $18777 (must be rounded to the nearest dollar). (c) M1: For setting up an inequality/equation or evaluating both values for at least one year \( n \ge 2 \). A1: 3 (complete years).

(a) Using the cosine rule on triangle \( ABC \): \( AC^2 = 75^2 + 110^2 - 2 \cdot 75 \cdot 110 \cdot \cos(65^\circ) \approx 10751.82 \). Thus, \( AC = \sqrt{10751.82} \approx 103.69... \approx 104\text{ m} \). (b) The area is given by: \( \text{Area} = \frac{1}{2} \cdot 75 \cdot 110 \cdot \sin(65^\circ) \approx 3738.53... \approx 3740\text{ m}^2 \). (c) Let \( h \) be the height of the flagpole at \( A \). Using right-angled triangle trigonometry: \( \tan(8^\circ) = \frac{h}{AC} \Rightarrow h = 103.69 \cdot \tan(8^\circ) \approx 14.572... \approx 14.6\text{ m} \).

(a) M1: For correct substitution into the cosine rule: \( 75^2 + 110^2 - 2(75)(110)\cos(65^\circ) \). A1: \( 104 \) (m) (accept \( 103.69... \)). (b) M1: For correct substitution into the area formula: \( \frac{1}{2}(75)(110)\sin(65^\circ) \). A1: \( 3740 \) (\( \text{m}^2 \)) (accept \( 3738.53... \)). (c) M1: For a correct trigonometric equation linking the height, angle, and \( AC \), e.g., \( \tan(8^\circ) = \frac{h}{AC} \). A1: \( 14.6 \) (m) (accept \( 14.572... \) or follow-through using \( 104 \) leading to \( 14.6 \)).

(a) At \( t = 0 \), \( T(0) = 85 \Rightarrow a + b e^{0} = 85 \Rightarrow a + b = 85 \). (b) At \( t = 15 \), \( T(15) = 51 \Rightarrow a + b e^{-0.05 \times 15} = 51 \Rightarrow a + b e^{-0.75} = 51 \). Substituting \( a = 85 - b \): \( 85 - b + b e^{-0.75} = 51 \Rightarrow b(1 - e^{-0.75}) = 34 \Rightarrow b = \frac{34}{1 - e^{-0.75}} \approx 64.43867... \approx 64.4 \). Hence, \( a = 85 - 64.43867... \approx 20.56133... \approx 20.6 \). (c) After 30 minutes: \( T(30) = 20.56133 + 64.43867 e^{-0.05 \times 30} \approx 34.939... \approx 34.9^\circ\text{C} \). (d) As \( t \to \infty \), \( e^{-0.05t} \to 0 \), so \( T(t) \to a \). The ambient room temperature is \( 20.6^\circ\text{C} \).

(a) A1: \( a + b = 85 \). (b) M1: For substituting \( t = 15 \) and \( T = 51 \) into the model: \( a + b e^{-0.75} = 51 \). M1: For an algebraic method to solve the simultaneous equations, e.g., substituting \( a = 85 - b \). A1: For showing both values match the given ones clearly: \( b \approx 64.438... \Rightarrow b = 64.4 \) and \( a \approx 20.561... \Rightarrow a = 20.6 \). (c) A1: \( 34.9 \) (\( ^\circ\text{C} \)) (accept \( 34.939... \)). (d) A1: \( 20.6 \) (\( ^\circ\text{C} \)) (accept \( 20.56... \)).

(a) Let \(W \sim \text{N}(150, \sigma^2)\). We are given \(P(W > 165) = 0.15\), which means \(P(W \le 165) = 0.85\). Using the standard normal distribution \(Z = \frac{W - 150}{\sigma}\), we have \(P\left(Z \le \frac{165 - 150}{\sigma}\right) = 0.85\). From the inverse normal function on a GDC, the \(z\)-score is \(1.03643\). Thus, \(\frac{15}{\sigma} \approx 1.03643 \implies \sigma \approx 14.4727 \approx 14.5\text{ g}\).

(b) Using \(\sigma \approx 14.4727\), we find \(P(W < 130)\) via the normal cumulative function on a GDC with a lower bound of \(-\infty\), an upper bound of \(130\), a mean of \(150\), and a standard deviation of \(14.4727\). This yields \(P(W < 130) \approx 0.083495 \approx 0.0835\).

(c) Let \(X\) be the number of small apples chosen. Since the apples are selected from a large pool, \(X\) follows a binomial distribution \(X \sim \text{B}(3, 0.083495)\). We want to find \(P(X \ge 1) = 1 - P(X = 0) = 1 - (1 - 0.083495)^3 \approx 1 - 0.76993 \approx 0.230\).

(a) [3 marks]

M1 for setting up the normal equation, e.g., \(P(W < 165) = 0.85\) or the standardization equation.
A1 for the correct \(z\)-score of \(1.03643...\).
A1 for \(\sigma \approx 14.5\) (accept \(14.4\) or more precise values like \(14.4727...\)).

(b) [2.5 marks]

M1.5 for setting up normal cumulative distribution with mean \(150\) and their \(\sigma\) from (a).
A1 for \(0.0835\) (accept \(0.0830\) if using \(14.5\) directly).

(c) [3 marks]

M1 for recognizing the binomial context or the complement probability approach, \(1 - P(\text{no small apples})^3\).
M1 for evaluating \((1 - 0.0835)^3\) or \((1 - 0.083495)^3\).
A1 for \(0.230\) (accept \(0.230\) or \(0.231\) depending on rounding).

(a) The volume of a rectangular box with a square base of side length \(x\) and height \(h\) is \(V = x^2 h\). Since the volume is \(4000\text{ cm}^3\), we have \(x^2 h = 4000 \implies h = \frac{4000}{x^2}\). The total external surface area \(A\) of the open-topped box consists of one square base and four vertical sides: \(A = x^2 + 4xh\). Substituting \(h\) into the area equation gives: \(A = x^2 + 4x\left(\frac{4000}{x^2}\right) = x^2 + \frac{16000}{x}\).

(b) Rewriting the expression for area as \(A = x^2 + 16000x^{-1}\) and differentiating with respect to \(x\) gives: \(\frac{\text{d}A}{\text{d}x} = 2x - 16000x^{-2} = 2x - \frac{16000}{x^2}\).

(c) To find the minimum surface area, set the derivative to zero: \(2x - \frac{16000}{x^2} = 0 \implies 2x^3 = 16000 \implies x^3 = 8000 \implies x = 20\text{ cm}\).

(d) Substitute \(x = 20\) back into the formula for \(A\): \(A = 20^2 + \frac{16000}{20} = 400 + 800 = 1200\text{ cm}^2\).

(a) [3 marks]

M1 for writing \(x^2 h = 4000\) and expressing \(h\) as \(\frac{4000}{x^2}\).
M1 for writing the surface area formula \(A = x^2 + 4xh\).
A1 for showing a clear substitution to obtain the given expression \(A = x^2 + \frac{16000}{x}\).

(b) [2 marks]

M1 for differentiating at least one term correctly (either \(x^2 \to 2x\) or \(16000/x \to -16000/x^2\)).
A1 for \(2x - \frac{16000}{x^2}\).

(c) [2 marks]

M1 for setting their derivative equal to 0.
A1 for \(x = 20\text{ cm}\).

(d) [1.5 marks]

M0.5 for substituting their \(x\) value from (c) into the equation for \(A\).
A1 for \(1200\text{ cm}^2\) (or simply \(1200\)).

(a) The volume of a cylinder is given by \(V = \pi r^2 h\). Since the volume is \(500\text{ cm}^3\), we have:
\(500 = \pi r^2 h\)
Solving for \(h\) gives:
\(h = \frac{500}{\pi r^2}\)

(b) The total surface area of the top and bottom is \(2\pi r^2\).
The cost for the top and bottom is \(0.02 \times 2\pi r^2 = 0.04\pi r^2\) USD.
The surface area of the curved side is \(2\pi r h\).
The cost for the curved side is \(0.015 \times 2\pi r h = 0.03\pi r h\) USD.
Substituting \(h = \frac{500}{\pi r^2}\) into the curved side cost gives:
\(0.03\pi r \left(\frac{500}{\pi r^2}\right) = \frac{15}{r}\) USD.
Therefore, the total cost \(C\) is:
\(C = 0.04\pi r^2 + \frac{15}{r}\)

(c) We write \(C = 0.04\pi r^2 + 15r^{-1}\).
Differentiating term by term with respect to \(r\):
\(\frac{\mathrm{d}C}{\mathrm{d}r} = 2 \times 0.04\pi r - 15r^{-2} = 0.08\pi r - \frac{15}{r^2}\)

(d) To find the minimum cost, set \(\frac{\mathrm{d}C}{\mathrm{d}r} = 0\):
\(0.08\pi r - \frac{15}{r^2} = 0\)
\(0.08\pi r^3 = 15\)
\(r^3 = \frac{15}{0.08\pi} \approx 59.683\)
\(r = \sqrt[3]{\frac{15}{0.08\pi}} \approx 3.91\text{ cm}\) (3 s.f.)

(e) (i) Substitute \(r \approx 3.9079\) into the cost equation:
\(C = 0.04\pi (3.9079)^2 + \frac{15}{3.9079} \approx 1.9193 + 3.8384 = 5.76\text{ USD}\) (3 s.f.)
(ii) Substitute \(r \approx 3.9079\) into the height equation:
\(h = \frac{500}{\pi (3.9079)^2} \approx 10.4\text{ cm}\) (3 s.f.)

(a) M1 for using the cylinder volume formula \(V = \pi r^2 h\) with \(V = 500\). A1 for showing the required algebraic rearrangement to obtain \(h = \frac{500}{\pi r^2}\). [2 marks]

(b) A1 for cost of top/bottom \(0.04\pi r^2\). A1 for cost of curved side \(0.03\pi r h\). M1 for substituting the expression for \(h\) to obtain the final formula. [3 marks]

(c) M1 for applying the power rule to \(r^2\) (obtaining \(0.08\pi r\)). M1 for applying the power rule to \(r^{-1}\) (obtaining \(-15r^{-2}\)). A1 for the correct fully simplified expression \(0.08\pi r - \frac{15}{r^2}\). [3 marks]

(d) M1 for setting their derivative equal to 0. M1 for algebraic manipulation to isolate \(r^3\) (e.g., \(0.08\pi r^3 = 15\)). A1 for \(r \approx 3.91\text{ cm}\) (accept 3.908). [3 marks]

(e) (i) A1 for minimum cost \(5.76\text{ USD}\). (ii) A1 for height \(10.4\text{ cm}\). [2 marks]

(a) \(H_0\): Preferred exercise type is independent of age group.
\(H_1\): Preferred exercise type is not independent of age group.

(b) Expected frequency for (Under 35, Yoga) = \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{100 \times 60}{200} = 30\).

(c) (i) Degrees of freedom \(df = (r-1)(c-1) = (2-1)(3-1) = 2\).
(ii) \(\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(45-40)^2}{40} + \frac{(35-30)^2}{30} + \frac{(20-30)^2}{30} + \frac{(35-40)^2}{40} + \frac{(25-30)^2}{30} + \frac{(40-30)^2}{30} = 0.625 + 0.8333 + 3.3333 + 0.625 + 0.8333 + 3.3333 = 9.5833 \approx 9.58\).
(iii) The \(p\)-value \(= P(\chi^2_2 \ge 9.5833) \approx 0.00830\).

(d) Since the \(p\)-value \(\approx 0.00830 < 0.05\) (or the calculated \(\chi^2\) value \(9.58 > 5.991\)), we reject the null hypothesis \(H_0\). There is sufficient evidence to suggest that preferred exercise type is not independent of age group.

(e) (i) Let \(T \sim N(55, 12^2)\).
\(P(T > 65) = P\left(Z > \frac{65-55}{12}\right) = P(Z > 0.8333) \approx 0.202\) (or 0.2023).
(ii) We want to find \(P(T < 80 \mid T > 65) = \frac{P(65 < T < 80)}{P(T > 65)}\).
First, \(P(65 < T < 80) = P(T < 80) - P(T \le 65) \approx 0.9814 - 0.7977 = 0.1837\).
Then, \(P(T < 80 \mid T > 65) = \frac{0.183717}{0.202328} \approx 0.908\) (or 0.9080).

(a) A1 for a correct null hypothesis stating independence. A1 for a correct alternative hypothesis stating non-independence. [2 marks]

(b) M1 for using the formula \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\). A1 for showing that \(\frac{100 \times 60}{200} = 30\). [2 marks]

(c) (i) A1 for \(df = 2\). (ii) A1 for \(\chi^2 \approx 9.58\) (accept 9.58 or 9.5833). (iii) A1 for \(p\)-value \(\approx 0.00830\) (accept 0.0083 or 0.00830). [3 marks]

(d) R1 for comparing their \(p\)-value with 0.05 (or critical value). A1 for a consistent and correct conclusion (reject \(H_0\) and state that exercise type and age are associated). [2 marks]

(e) (i) A1 for \(P(T > 65) \approx 0.202\). (ii) M1 for writing the correct conditional probability expression. A1 for finding \(P(65 < T < 80) \approx 0.184\). A1 for the final ratio \(\approx 0.908\). [3 marks]

a) At \(t=0\), \(V(0) = -(0)^3 + 18(0)^2 + 120(0) + 500 = 500\) litres. b) Differentiating \(V(t)\) with respect to \(t\) yields: \(\frac{dV}{dt} = -3t^2 + 36t + 120\). c) Substitute \(t=5\) into the derivative: \(\frac{dV}{dt}\Big|_{t=5} = -3(5)^2 + 36(5) + 120 = -75 + 180 + 120 = 225\) litres per hour. d) To find the maximum volume, set \(\frac{dV}{dt} = 0\): \(-3t^2 + 36t + 120 = 0\). Dividing by \(-3\) gives \(t^2 - 12t - 40 = 0\). Solving for \(t\) using the quadratic formula: \(t = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(-40)}}{2} = \frac{12 \pm \sqrt{304}}{2}\). Since \(t \ge 0\), we select the positive root: \(t = 6 + \sqrt{76} \approx 14.718\) hours. Thus, \(t \approx 14.7\) hours. e) Substitute \(t \approx 14.718\) back into \(V(t)\): \(V(14.718) = -(14.718)^3 + 18(14.718)^2 + 120(14.718) + 500 \approx 2979.22\) litres. Thus, the maximum volume is 2980 litres (to 3 significant figures). f) The rate of change is \(V'(t) = -3t^2 + 36t + 120\). To find its maximum, differentiate again to get \(V''(t) = -6t + 36\). Setting \(V''(t) = 0\) yields \(-6t + 36 = 0 \implies t = 6\) hours. g) At \(t = 24\), \(V(24) = -(24)^3 + 18(24)^2 + 120(24) + 500 = -13824 + 10368 + 2880 + 500 = -76\) litres. This model is not valid for \(t = 24\) because the volume of water cannot be negative.

a) A1 for 500. [1 mark] b) M1 for attempting to differentiate, A1 for correct derivative \(-3t^2 + 36t + 120\). [2 marks] c) M1 for substituting \(t = 5\) into their derivative, A1 for 225. [2 marks] d) M1 for setting their derivative to 0, M1 for attempting to solve the quadratic equation, A1 for \(t \approx 14.7\) (or exact \(6 + \sqrt{76}\)). [3 marks] e) M1 for substituting their value of \(t\) from (d) into \(V(t)\), A1 for \(2980\) (or \(2979\)). [2 marks] f) M1 for setting the second derivative to 0 (or finding vertex of the quadratic rate function), A1 for finding \(V''(t) = -6t + 36\), A1 for setting to 0, A1 for \(t = 6\). [4 marks] g) M1 for substituting \(t=24\) into \(V(t)\), A1 for obtaining \(-76\), R1 for stating the volume is negative, R1 for concluding the model is not valid. [4 marks]

a) Let \(X \sim N(1500, 120^2)\). (i) \(P(X > 1650) = P\left(Z > \frac{1650-1500}{120}\right) = P(Z > 1.25) \approx 0.10565 \approx 0.106\). (ii) \(P(1350 < X < 1550) = P\left(\frac{1350-1500}{120} < Z < \frac{1550-1500}{120}\right) = P(-1.25 < Z < 0.4167) \approx 0.66154 - 0.10565 \approx 0.55589 \approx 0.556\). b) We want to find \(d\) such that \(P(X < d) = 0.05\). Using inverse normal distribution on GDC: \(Z = -1.64485\). Thus, \(d = 1500 + (-1.64485 \times 120) = 1500 - 197.382 \approx 1302.6 \approx 1300\) hours. c) Let \(Y\) be the number of defective bulbs in a sample of 8. Then \(Y \sim B(8, 0.05)\). We need \(P(Y \le 1) = P(Y = 0) + P(Y = 1)\). \(P(Y = 0) = (0.95)^8 \approx 0.66342\). \(P(Y = 1) = \binom{8}{1} (0.05)^1 (0.95)^7 \approx 0.27934\). \(P(Y \le 1) \approx 0.66342 + 0.27934 \approx 0.94276 \approx 0.943\). d) (i) \(H_0\): Bulb type and customer satisfaction level are independent. (ii) Row total for Satisfied = \(80 + 45 + 55 = 180\). Column total for Halogen = \(45 + 40 + 35 = 120\). Grand total = 340. Expected frequency = \(\frac{180 \times 120}{340} \approx 63.529 \approx 63.5\). (iii) Degrees of freedom = \((\text{rows} - 1)(\text{columns} - 1) = (3-1)(3-1) = 4\). (iv) Performing the Chi-squared test on GDC: \(\chi^2 \approx 26.507 \approx 26.5\) and \(p\)-value \\approx 2.50 \times 10^{-5}\). (v) Since \(p \approx 0.000025 < 0.05\), we reject the null hypothesis.

a) (i) M1 for standardizing or setting up normal CDF, A1 for 0.106. (ii) M1 for standardizing or setting up normal CDF interval, A1 for 0.556. [4 marks] b) M1 for setting up \(P(X < d) = 0.05\), M1 for attempting inverse normal, A1 for 1300 (or 1303). [3 marks] c) M1 for identifying binomial distribution with parameters \(n=8\) and \(p=0.05\), M1 for using binomial CDF for \(P(Y \le 1)\), A1 for 0.943. [3 marks] d) (i) A1 for stating the null hypothesis correctly. (ii) M1 for calculating expected value formula, A1 for 63.5. (iii) A1 for 4. (iv) A1 for \(\chi^2 \approx 26.5\), A1 for \(p \approx 2.50 \times 10^{-5}\). (v) R1 for comparing p-value with 0.05, A1 for stating rejection of the null hypothesis. [8 marks]

a) The cost of the car in 6 years' time is given by: \(25000 \times (1 - 0.04)^6 = 25000 \times 0.96^6 \approx 19568.94\) EUR. b) Using the compound interest formula: \(FV = PV \times \left(1 + \frac{r}{1200}\right)^{12n} = 12000 \times \left(1 + \frac{6}{1200}\right)^{72} = 12000 \times (1.005)^{72} \approx 17184.53\) EUR. c) Comparing the two values: \(17184.53 < 19568.94\). Therefore, Option A will not provide enough money to buy the car. d) (i) Option B is an ordinary annuity. The future value is: \(FV = P \times \frac{(1+i)^N - 1}{i}\), where \(FV = 19568.94\), \(i = \frac{4.8}{1200} = 0.004\), and \(N = 72\). \(19568.94 = P \times \frac{(1.004)^{72} - 1}{0.004}\). Since \(\frac{1.004^{72}-1}{0.004} \approx 83.140484\), we find: \(P = \frac{19568.94}{83.140484} \approx 235.37\) EUR. (ii) Total money deposited = \(72 \times 235.37197 = 16946.78\) EUR. Total interest earned = \(19568.94 - 16946.78 = 2622.16\) EUR. e) The car's price after an additional 4 years is: \(19568.94 \times 0.96^4 \approx 16620.81\) EUR (or 16600 EUR to 3 significant figures).

a) M1 for setting up the geometric decay expression, A1 for \(25000 \times 0.96^6\), A1 for 19568.94 EUR (accept 19600 EUR). [3 marks] b) M1 for identifying compound interest variables, A1 for substituting into formula, A1 for 17184.53 EUR (accept 17185 EUR). [3 marks] c) M1 for comparing their values from (a) and (b), A1 for concluding that Option A is insufficient. [2 marks] d) (i) M1 for using TVM solver or annuity formula, A1 for correct inputs (e.g., \(N = 72\), \(I\% = 4.8\), \(FV = 19568.94\)), A1 for calculating \(P\), A1 for 235.37 EUR (accept 235.31 to 235.45 depending on rounding of car value). (ii) M1 for calculating total contributions, M1 for subtracting total contributions from FV, A1 for 2622.16 EUR (accept surrounding answers from rounding). [7 marks] e) M1 for identifying decay over 4 more years, M1 for multiplying their car value by \(0.96^4\), A1 for 16620.81 EUR (or 16600 EUR). [3 marks]