2025 CCEA AS-Level Mathematics 2210 模拟试题及答案详解

Substituting \( y=2x-3 \) into the quadratic equation: \( 2x-3 = x^2-4x+5 \). Rearranging: \( x^2-4x+5-2x+3=0 \Rightarrow x^2-6x+8=0 \). Factorising: \( (x-2)(x-4)=0 \Rightarrow x=2 \) or \( x=4 \). When \( x=2 \): \( y=2(2)-3=1 \). When \( x=4 \): \( y=2(4)-3=5 \).

Final answer: the points of intersection are \( (2,1) \) and \( (4,5) \).

[1] correct substitution to eliminate y; [1] correctly rearranged into the form \( x^2-6x+8=0 \); [2] correct factorisation or use of the quadratic formula, giving \( x=2 \) and \( x=4 \) (1 mark each value, ECF from a correct method); [1] correct \( y=1 \) when \( x=2 \); [1] correct \( y=5 \) when \( x=4 \); [1] both coordinate pairs clearly stated as final answer.

Let \( u=x^2 \). The equation becomes a quadratic in u: \( u^2-5u+4=0 \). Factorising: \( (u-1)(u-4)=0 \Rightarrow u=1 \) or \( u=4 \). Since \( u=x^2 \): if \( x^2=1 \), \( x=\pm1 \); if \( x^2=4 \), \( x=\pm2 \).

Final answer: \( x=-2,\ -1,\ 1,\ 2 \).

[1] correct substitution \( u=x^2 \) to form a quadratic in u; [1] correctly reduced equation \( u^2-5u+4=0 \); [1] correct factorisation/solution giving \( u=1 \) and \( u=4 \); [1] correctly converts \( u=1 \) to \( x=\pm1 \); [1] correctly converts \( u=4 \) to \( x=\pm2 \); [1] all four solutions clearly stated (lose this mark if any extraneous or missing root, e.g. only positive roots given).

(a) \( y=f(x+3) \) translates the graph of \( y=f(x) \) by 3 units in the negative x-direction (left), moving the minimum to \( (2-3,-5)=(-1,-5) \). Then subtracting 4, \( y=f(x+3)-4 \), translates this down by 4 units, moving the minimum to \( (-1,-5-4)=(-1,-9) \).

(b) \( y=2f(x) \) stretches the graph vertically by scale factor 2 relative to the x-axis; the x-coordinate of the minimum is unchanged, and the y-coordinate is doubled: \( (2,\,2\times(-5))=(2,-10) \).

Final answer: (a) \( (-1,-9) \); (b) \( (2,-10) \).

(a) [1] correctly identifies horizontal translation by 3 to the left giving x-coordinate -1; [1] correctly identifies the further translation down by 4 giving y-coordinate -9 (i.e. final point \( (-1,-9) \)). (b) [1] correctly identifies x-coordinate unchanged (2); [1] correctly identifies y-coordinate doubled to -10 (i.e. final point \( (2,-10) \)).

(a) \( \dfrac{dy}{dx} = 3x^2-6x \)

(b) At \( x=1 \): \( y = (1)^3-3(1)^2+2 = 1-3+2 = 0 \), so the point is \( (1,0) \). Gradient at \( x=1 \): \( \dfrac{dy}{dx}=3(1)^2-6(1)=3-6=-3 \). Using \( y-y_1=m(x-x_1) \): \( y-0=-3(x-1) \Rightarrow y=-3x+3 \).

(c) The normal is perpendicular to the tangent, so its gradient is \( -\dfrac{1}{-3}=\dfrac{1}{3} \). Using the same point \( (1,0) \): \( y-0=\dfrac{1}{3}(x-1) \Rightarrow y=\dfrac{1}{3}x-\dfrac{1}{3} \).

Final answer: tangent \( y=-3x+3 \); normal \( y=\dfrac{1}{3}x-\dfrac{1}{3} \).

(a) [1] correct differentiation of \( x^3 \) and \( -3x^2 \) terms; [1] correct full derivative \( 3x^2-6x \) (constant term correctly differentiates to 0). (b) [1] correct y-coordinate at x=1 (y=0); [1] correct gradient at x=1 (m=-3); [1] correct tangent equation \( y=-3x+3 \) (ECF). (c) [1] correctly identifies perpendicular gradient rule; [1] correct normal gradient \( \frac13 \) (ECF from (b)); [1] correct normal equation \( y=\frac13x-\frac13 \) (ECF).

(a) \( \dfrac{dy}{dx} = 6x^2-18x+12 \). At stationary points, \( \dfrac{dy}{dx}=0 \): \( 6x^2-18x+12=0 \Rightarrow x^2-3x+2=0 \Rightarrow (x-1)(x-2)=0 \Rightarrow x=1 \) or \( x=2 \).

(b) \( \dfrac{d^2y}{dx^2}=12x-18 \). At \( x=1 \): \( 12(1)-18=-6<0 \), so \( x=1 \) is a maximum point (with \( y=2-9+12+1=6 \)). At \( x=2 \): \( 12(2)-18=6>0 \), so \( x=2 \) is a minimum point (with \( y=16-36+24+1=5 \)).

Final answer: maximum at \( (1,6) \); minimum at \( (2,5) \).

(a) [1] correct derivative \( 6x^2-18x+12 \); [1] correctly set equal to zero and simplified to \( x^2-3x+2=0 \); [1] correct factorisation/solution; [1] \( x=1 \) and \( x=2 \) both stated. (b) [1] correct second derivative \( 12x-18 \); [1] correct evaluation and conclusion at x=1 (negative, maximum); [1] correct evaluation and conclusion at x=2 (positive, minimum).

\( 2\sin\theta-1=0 \Rightarrow \sin\theta=\dfrac{1}{2} \). The principal value is \( \theta=\sin^{-1}(0.5)=30^\circ \). Since sine is positive in the first and second quadrants, the second solution in the range \( 0^\circ\le\theta<360^\circ \) is \( \theta=180^\circ-30^\circ=150^\circ \). No further solutions lie in the given range.

Final answer: \( \theta=30^\circ \) or \( \theta=150^\circ \).

[1] correct rearrangement to \( \sin\theta=0.5 \); [1] correct principal value \( \theta=30^\circ \); [2] correct identification that sine is also positive in the second quadrant and correct second solution \( \theta=150^\circ \) (1 mark for method, 1 for correct value); [2] both solutions clearly stated within the given range with no extra/incorrect solutions listed; [1] correct working/method shown throughout (reject unsupported answers with no method).

(a) \( 1+\cos2\theta = 1+(2\cos^2\theta-1) = 2\cos^2\theta \). So \( \dfrac{\sin2\theta}{1+\cos2\theta} = \dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta \), as required.

(b) Using the identity from (a), the equation becomes \( \tan\theta=1 \). The principal value is \( \theta=45^\circ \). Since \( \tan\theta \) has period \( 180^\circ \), the next solution would be \( 45^\circ+180^\circ=225^\circ \), which is outside the given range \( 0^\circ\le\theta<180^\circ \).

Final answer: (a) shown; (b) \( \theta=45^\circ \).

(a) [1] correct substitution of the cos2θ identity into the denominator; [1] correct simplification \( 1+\cos2\theta=2\cos^2\theta \); [1] correct substitution of sin2θ identity into the numerator; [1] correct cancellation of the common factor leading to \( \tan\theta \), completing the proof. (b) [1] correctly uses part (a) to reduce the equation to \( \tan\theta=1 \); [1] correct principal value \( \theta=45^\circ \); [1] correctly identifies no other solution lies in the given range (reject an answer that incorrectly includes 225°).

(a) Using the cosine rule: \( AC^2 = AB^2+BC^2-2(AB)(BC)\cos(ABC) = 8.0^2+11.0^2-2(8.0)(11.0)\cos65^\circ = 64+121-176(0.4226) = 185-74.38 = 110.6 \). \( AC = \sqrt{110.6} = 10.5\ \text{cm} \) (3 s.f.).

(b) Using the sine rule: \( \dfrac{\sin(BAC)}{BC}=\dfrac{\sin(ABC)}{AC} \Rightarrow \sin(BAC) = \dfrac{BC\sin(ABC)}{AC} = \dfrac{11.0\sin65^\circ}{10.5} = \dfrac{9.968}{10.5} = 0.9494 \Rightarrow BAC = \sin^{-1}(0.9494) = 71.4^\circ \) (3 s.f.).

(c) Angles in a triangle sum to \( 180^\circ \): \( BCA = 180^\circ - 65^\circ - 71.4^\circ = 43.6^\circ \).

Final answer: \( AC=10.5\ \text{cm} \), angle BAC \( =71.4^\circ \), angle BCA \( =43.6^\circ \).

(a) [1] correct substitution into the cosine rule; [1] correct value under the square root (accept 110–111); [1] \( AC=10.5\ \text{cm} \) to 3 s.f. (b) [1] correct substitution into the sine rule (ECF from (a)); [1] correct value of sin(BAC); [1] \( BAC=71.4^\circ \) to 3 s.f. (accept 71.3–71.5, ECF); reject the obtuse alternative \( 180-71.4=108.6^\circ \) without justification, since BC is not the longest side opposite the largest given angle context here (must check triangle validity, but 71.4 is the valid solution as angle sum constraint (65+71.4+43.6=180) confirms). (c) [1] correct use of angle sum of triangle, \( BCA=43.6^\circ \) (ECF).

(a) Using the law \( \log_a m+\log_a n=\log_a(mn) \): \( \log_2\left[(x+3)(x-1)\right]=5 \). Converting from log form: \( (x+3)(x-1)=2^5=32 \Rightarrow x^2+2x-3=32 \Rightarrow x^2+2x-35=0 \Rightarrow (x+7)(x-5)=0 \Rightarrow x=-7 \) or \( x=5 \). For the original logarithms to be defined, we need \( x+3>0 \) and \( x-1>0 \), i.e. \( x>1 \). Since \( x=-7 \) does not satisfy \( x>1 \), it is rejected. So \( x=5 \).

(b) Let \( \log_a b = k \), so \( a^k=b \). Then \( \dfrac{1}{b}=\dfrac{1}{a^k}=a^{-k} \). Taking \( \log_a \) of both sides: \( \log_a\left(\dfrac{1}{b}\right) = \log_a(a^{-k}) = -k = -\log_a b \), as required. (Alternatively, using the quotient law with \( \log_a 1=0 \): \( \log_a\left(\dfrac{1}{b}\right)=\log_a 1-\log_a b = 0-\log_a b = -\log_a b \).)

(c) Taking logarithms (base 10, or ln) of both sides: \( \ln(7^x)=\ln50 \Rightarrow x\ln7=\ln50 \Rightarrow x=\dfrac{\ln50}{\ln7} = \dfrac{3.912}{1.946} = 2.01 \) (3 s.f.).

Final answer: (a) \( x=5 \); (c) \( x=2.01 \).

(a) [1] correct use of the addition law to combine the logs; [1] correct conversion from log form to \( (x+3)(x-1)=32 \); [1] correctly expanded/rearranged to \( x^2+2x-35=0 \) and solved (factorisation or formula) to give \( x=-7 \) and \( x=5 \); [1] correctly states the domain requirement (x>1, or equivalently that logs of negative numbers are undefined); [1] correctly rejects \( x=-7 \) and states final answer \( x=5 \). (b) [1] correct method set-up (e.g. \( \log_ab=k \Rightarrow a^k=b \), or use of quotient law with \( \log_a1=0 \)); [1] correct algebraic manipulation shown; [1] correctly reaches \( -\log_ab \), completing the proof with valid reasoning throughout. (c) [1] correct method (taking logs of both sides); [1] correct rearrangement \( x=\ln50/\ln7 \) (or equivalent base); [1] correct evaluation of the two logarithms; [1] \( x=2.01 \) to 3 s.f. (accept 2.00–2.02).

(a) Using the binomial expansion \( (1+2x)^5 = \sum_{r=0}^{5}\binom{5}{r}(2x)^r \): \( \binom{5}{0}=1 \); \( \binom{5}{1}(2x)^1=5(2x)=10x \); \( \binom{5}{2}(2x)^2=10(4x^2)=40x^2 \); \( \binom{5}{3}(2x)^3=10(8x^3)=80x^3 \). So \( (1+2x)^5 = 1+10x+40x^2+80x^3+\ldots \)

(b) The coefficient of \( x^3 \) in \( (2-x)(1+10x+40x^2+80x^3+\ldots) \) comes from two products: \( 2\times80x^3 \) (giving \( 160x^3 \)) and \( (-x)\times40x^2 \) (giving \( -40x^3 \)). Adding these: \( 160-40=120 \).

Final answer: (a) \( 1+10x+40x^2+80x^3 \); (b) coefficient of \( x^3 \) is \( 120 \).

(a) [1] correct binomial coefficients identified (1,5,10,10,...) e.g. via Pascal's triangle or \( \binom{5}{r} \); [1] correct powers of 2x used; [1] correct x term (10x); [1] correct x^2 term (40x^2); [1] correct x^3 term (80x^3). (b) [1] correctly identifies both relevant products (\( 2\times80x^3 \) and \( -x\times40x^2 \)) needed to form the x^3 term of the product (ECF from (a)); [1] correct final coefficient, 120.

(a) Volume of a cuboid \( = x^2h \). Since the volume must equal \( 500\ \text{cm}^3 \): \( x^2h=500 \Rightarrow h=\dfrac{500}{x^2} \), as required.

(b) The base has area \( x^2 \). Each of the four (vertical, rectangular) sides has area \( xh \), so the total area of the four sides is \( 4xh \). No top is included. So \( S = x^2+4xh = x^2+4x\left(\dfrac{500}{x^2}\right) = x^2+\dfrac{2000}{x} \), as required.

(c) \( \dfrac{dS}{dx} = 2x-\dfrac{2000}{x^2} \). Setting \( \dfrac{dS}{dx}=0 \): \( 2x=\dfrac{2000}{x^2} \Rightarrow 2x^3=2000 \Rightarrow x^3=1000 \Rightarrow x=\sqrt[3]{1000}=10 \).

(d) \( \dfrac{d^2S}{dx^2} = 2+\dfrac{4000}{x^3} \). At \( x=10 \): \( \dfrac{d^2S}{dx^2}=2+\dfrac{4000}{1000}=2+4=6 \). Since this is positive, the stationary point is a minimum.

(e) \( S = (10)^2+\dfrac{2000}{10} = 100+200 = 300\ \text{cm}^2 \). \( h=\dfrac{500}{(10)^2}=\dfrac{500}{100}=5\ \text{cm} \).

Final answer: minimum surface area \( S=300\ \text{cm}^2 \) when \( x=10\ \text{cm} \) and \( h=5\ \text{cm} \).

(a) [1] correct volume formula for a cuboid with square base, \( V=x^2h \); [1] correct rearrangement to \( h=500/x^2 \), completing the 'show that'. (b) [1] correctly identifies base area \( x^2 \); [1] correctly identifies four sides each of area xh, total 4xh; [1] correct substitution of h from (a); [1] correct simplification to \( S=x^2+2000/x \), completing the 'show that'. (c) [1] correct differentiation of \( x^2 \) term; [1] correct differentiation of \( 2000/x \) term (\( -2000/x^2 \)); [1] correct derivative \( dS/dx=2x-2000/x^2 \); [1] correctly sets derivative to zero; [1] correct rearrangement to \( x^3=1000 \); [1] \( x=10 \). (d) [1] correct second derivative \( 2+4000/x^3 \); [1] correct substitution of x=10; [1] correctly concludes minimum since \( d^2S/dx^2>0 \) (ECF from (c)). (e) [1] correct substitution into S to find minimum surface area; [1] \( S=300\ \text{cm}^2 \); [1] correct value of h=5 cm from part (a) (ECF).

(a) Setting \( y=0 \): \( 6x-x^2=0 \Rightarrow x(6-x)=0 \Rightarrow x=0 \) or \( x=6 \), so the curve crosses the x-axis at these two values, as required.

(b) Area \( = \displaystyle\int_0^6 (6x-x^2)\,dx = \left[3x^2-\dfrac{x^3}{3}\right]_0^6 = \left(3(36)-\dfrac{216}{3}\right)-(0) = 108-72 = 36 \) square units.

(c) Setting the curve and line equal: \( 2x = 6x-x^2 \Rightarrow x^2-4x=0 \Rightarrow x(x-4)=0 \Rightarrow x=0 \) or \( x=4 \).

Final answer: (b) area \( =36 \) square units; (c) \( x=0 \) and \( x=4 \).

(a) [1] correctly sets y=0 and factorises \( x(6-x)=0 \); [1] correctly states both roots x=0 and x=6. (b) [1] correct integral expression set up with correct limits; [1] correct integration of \( 6x \) term (\( 3x^2 \)); [1] correct integration of \( -x^2 \) term (\( -x^3/3 \)); [1] correct substitution of upper limit x=6; [1] correct substitution of lower limit x=0 (=0); [1] final area \( =36 \) square units with correct units stated. (c) [1] correctly sets the two expressions equal and rearranges to \( x^2-4x=0 \); [1] correctly factorises/solves to give \( x=0 \) and \( x=4 \) (both required for the mark).

(a) \( \mathbf{v} = \dfrac{d\mathbf{r}}{dt} = (6t-2)\mathbf{i}+(3t^2-4)\mathbf{j} \)

(b) \( \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 6\mathbf{i}+6t\mathbf{j} \)

(c) At \( t=2 \): \( \mathbf{v} = (6(2)-2)\mathbf{i}+(3(2)^2-4)\mathbf{j} = 10\mathbf{i}+8\mathbf{j} \) (m s^-1). Speed \( = \sqrt{10^2+8^2} = \sqrt{164} = 12.8\ \text{m s}^{-1} \) (3 s.f.).

(d) At \( t=2 \): \( \mathbf{a} = 6\mathbf{i}+6(2)\mathbf{j} = 6\mathbf{i}+12\mathbf{j} \) (m s^-2). Magnitude \( = \sqrt{6^2+12^2} = \sqrt{180} = 13.4\ \text{m s}^{-2} \) (3 s.f.).

Final answer: \( \mathbf{v}(2)=10\mathbf{i}+8\mathbf{j} \), speed \( =12.8\ \text{m s}^{-1} \); \( \mathbf{a}(2)=6\mathbf{i}+12\mathbf{j} \), magnitude \( =13.4\ \text{m s}^{-2} \).

(a) [1] correct differentiation of i-component; [1] correct differentiation of j-component. (b) [1] correct differentiation of i-component of v; [1] correct differentiation of j-component of v. (c) [1] correct substitution of t=2 into v; [1] \( \mathbf{v}=10\mathbf{i}+8\mathbf{j} \); [1] correct speed \( =12.8\ \text{m s}^{-1} \) (ECF). (d) [1] correct substitution of t=2 into a; [1] \( \mathbf{a}=6\mathbf{i}+12\mathbf{j} \); [1] correct magnitude \( =13.4\ \text{m s}^{-2} \) (ECF).

(a) Section 1: \( v_1=\dfrac{100-0}{20-0}=5.0\ \text{m s}^{-1} \). Section 2: \( v_2=\dfrac{100-100}{50-20}=0\ \text{m s}^{-1} \). Section 3: \( v_3=\dfrac{0-100}{70-50}=-5.0\ \text{m s}^{-1} \) (i.e. 5.0 m s^-1 back towards home).

(b) The cyclist is stationary (at rest), remaining 100 m from home, for this 30 s period.

(c) Total distance travelled \( = 100+0+100=200\ \text{m} \); average speed \( =\dfrac{200}{70}=2.86\ \text{m s}^{-1} \) (3 s.f.). Total displacement \( = 0 \) m (cyclist returns to the starting point); average velocity \( =\dfrac{0}{70}=0\ \text{m s}^{-1} \). These differ because average speed depends on the total distance travelled (a scalar, always positive, counting the outward and return legs separately), whereas average velocity depends only on the net displacement (a vector, here zero since the cyclist finishes at the same place they started).

Final answer: (a) 5.0, 0, -5.0 m s^-1; (c) average speed 2.86 m s^-1, average velocity 0 m s^-1.

(a) [1] each for correct velocity in each of the three sections (5.0, 0, -5.0 m s^-1, with correct sign/direction for section 3); max [3]. (b) [1] correctly describes the cyclist as stationary/at rest (not moving) during this interval. (c) [1] correct average speed calculated as total distance/total time (2.86 m s^-1); [1] correct average velocity (0 m s^-1) with valid explanation of the distance vs. displacement distinction.

(a) Forces on the lift: tension T acting upward, weight \( mg \) acting downward. Taking upward as positive, Newton's second law gives \( T-mg=ma \Rightarrow T=m(g+a) \). \( T = 800(9.8+1.5) = 800(11.3) = 9040\ \text{N} \).

(b) The lift is still moving upward but decelerating, so its acceleration is directed downward (\( a=-2.0\ \text{m s}^{-2} \) with upward positive). Using the same equation: \( T=m(g+a) = 800(9.8-2.0) = 800(7.8) = 6240\ \text{N} \).

Final answer: (a) \( T=9040\ \text{N} \); (b) \( T=6240\ \text{N} \).

(a) [1] correct free body diagram/identification of the two forces (tension up, weight down); [1] correct application of Newton's second law, \( T-mg=ma \); [1] correct substitution; [1] \( T=9040\ \text{N} \). (b) [1] correctly identifies the acceleration is directed downward (deceleration while moving up); [1] correct equation \( T-mg=ma \) with \( a=-2.0\ \text{m s}^{-2} \); [1] correct substitution; [1] \( T=6240\ \text{N} \).

(a) Weight \( mg=5.0\times9.8=49\ \text{N} \). Component of weight parallel to (down) the plane: \( mg\sin20^\circ=49\times0.342=16.8\ \text{N} \). Normal reaction (perpendicular to plane): \( R=mg\cos20^\circ=49\times0.940=46.0\ \text{N} \). Maximum available friction: \( F_{max}=\mu R=0.50\times46.0=23.0\ \text{N} \). Since the component of weight down the plane (16.8 N) is less than the maximum available friction (23.0 N), friction can supply exactly the force needed to prevent sliding, so the block remains in equilibrium.

(b) At the point of sliding (limiting equilibrium), friction acts at its maximum value and the component of weight down the slope exactly equals the maximum friction: \( mg\sin\theta_c=\mu \, mg\cos\theta_c \Rightarrow \tan\theta_c=\mu=0.50 \Rightarrow \theta_c=\tan^{-1}(0.50)=26.6^\circ \) (3 s.f.).

(c) At \( 35^\circ \) (steeper than the critical angle of \( 26.6^\circ \)), the block would slide down without support, so friction now acts down the plane, opposing the impending motion up the plane. For limiting equilibrium moving up the plane: \( P = mg\sin\theta+\mu \, mg\cos\theta = 5.0(9.8)\sin35^\circ+0.50(5.0)(9.8)\cos35^\circ = 49\times0.5736+0.50\times49\times0.8192 = 28.1+20.1 = 48.2\ \text{N} \) (3 s.f.).

Final answer: (a) shown, block remains in equilibrium; (b) \( \theta_c=26.6^\circ \); (c) \( P=48.2\ \text{N} \).

(a) [1] correct component of weight down the plane, \( mg\sin20^\circ=16.8\ \text{N} \); [1] correct normal reaction, \( R=mg\cos20^\circ=46.0\ \text{N} \); [1] correct maximum friction, \( F_{max}=\mu R=23.0\ \text{N} \); [1] correct comparison and conclusion that the block remains in equilibrium since 16.8 N < 23.0 N. (b) [1] correctly sets up the limiting equilibrium condition \( mg\sin\theta_c=\mu mg\cos\theta_c \); [1] correct rearrangement to \( \tan\theta_c=\mu \); [1] \( \theta_c=26.6^\circ \). (c) [1] correctly identifies friction now acts down the plane (opposing impending motion up the slope); [1] correct equation \( P=mg\sin\theta+\mu mg\cos\theta \); [1] correct substitution; [1] \( P=48.2\ \text{N} \) (accept 48.0–48.4).

(a) Reason 1: heart rate here is continuous numerical data with many distinct values (up to 53 different possible integer values across the range), so a pie chart with one slice per distinct value would have far too many, extremely thin slices to be readable or meaningful. Reason 2: a pie chart is designed to show how a total is split between categories of a categorical (or discrete, few-category) variable, showing proportion of a whole — it is not designed to show the shape/distribution (e.g. spread, skewness, central tendency) of a continuous numerical variable, which is usually of more interest here.

(b) A more appropriate method would be to group the data into class intervals (e.g. 50–59, 60–69, ..., 100–109 bpm) and display it as a histogram, since a histogram is specifically designed to show the distribution/shape (spread, skew, modal class) of grouped continuous data, with bar area representing frequency (or frequency density for unequal class widths).

Final answer: histogram of grouped data is more suitable than a pie chart for this continuous distribution.

(a) [1] each for any two valid, distinct reasons (e.g. too many categories/slices for a pie chart to be readable; pie chart unsuitable for showing distribution/shape of continuous data; pie chart designed for categorical/proportion-of-whole data not continuous numerical data); max [2]; [1] additional mark for clear, correctly-reasoned explanation (not just naming an issue) in at least one of the two points. (b) [1] valid alternative graph named (histogram of grouped/class-interval data, or equivalent e.g. stem-and-leaf/box plot for a fuller justification); [1] correct justification linking the chosen graph to showing the distribution of continuous data appropriately.

(a) \( \text{mean} = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{4(5)+8(15)+10(25)+6(35)+2(45)}{30} = \dfrac{20+120+250+210+90}{30} = \dfrac{690}{30} = 23.0\ \text{minutes} \)

(b) \( \text{variance} = \dfrac{\Sigma fx^2}{\Sigma f}-\bar{x}^2 \). \( \Sigma fx^2 = 4(5^2)+8(15^2)+10(25^2)+6(35^2)+2(45^2) = 100+1800+6250+7350+4050 = 19\,550 \). \( \text{variance} = \dfrac{19\,550}{30}-23.0^2 = 651.67-529 = 122.67 \). Standard deviation \( = \sqrt{122.67} = 11.1\ \text{minutes} \) (3 s.f.).

(c) The sample only includes patients attending during a specific 2-hour morning window on a single day (a Monday); waiting times could vary considerably at other times of day (e.g. busier afternoon clinics) or on other days of the week (e.g. Mondays might be unusually busy after the weekend), so this sample may not be representative of waiting times across the whole week/all clinic hours, introducing bias into any estimate based on it.

Final answer: (a) mean \( \approx23.0 \) minutes; (b) standard deviation \( \approx11.1 \) minutes.

(a) [1] correct method (Σfx/Σf) with correct Σfx=690 shown; [1] mean = 23.0 minutes. (b) [1] correct method for Σfx^2 (=19550) or an equivalent correct variance method; [1] correct variance (122.67, accept 122–123); [1] correct standard deviation, 11.1 minutes (accept 11.0–11.2, ECF). (c) [1] identifies the sample is restricted to a limited time window/single day, not representative of all clinic operating times; [1] correct reasoning as to why this could bias the estimate (e.g. waiting times may differ systematically at other times/days).

(a) Let F = studies French, S = studies Spanish. The Venn diagram has two overlapping circles within a rectangle representing all 30 students: the overlap region (both F and S) contains 8 students; the 'French only' region (F but not S) contains \( 18-8=10 \) students; the 'Spanish only' region (S but not F) contains \( 15-8=7 \) students; the region outside both circles (neither) contains \( 30-(10+8+7)=5 \) students.

(b) French only \( =18-8=10 \) students. Neither \( =30-(F\cup S) \), where \( F\cup S=18+15-8=25 \), so neither \( =30-25=5 \) students.

(c) \( P(F)=\dfrac{18}{30}=0.600 \), \( P(S)=\dfrac{15}{30}=0.500 \), so \( P(F)\times P(S)=0.600\times0.500=0.300 \). The actual probability of both: \( P(F\cap S)=\dfrac{8}{30}=0.267 \) (3 s.f.). Since \( P(F\cap S)\neq P(F)\times P(S) \) (0.267 ≠ 0.300), the two events are NOT statistically independent.

Final answer: French only = 10 students; neither = 5 students; the events are not independent (0.267 ≠ 0.300).

(a) [1] correct overlap value (8) placed in the intersection; [1] correct remaining two individual regions and correctly structured diagram description (French only, Spanish only clearly distinguished from the overlap). (b) [1] French only = 10 (correct method 18-8); [1] neither = 5 (correct method, e.g. 30-(10+8+7) or 30-(18+15-8)). (c) [1] correct P(F) and P(S) calculated, and their product (0.300) found; [1] correct \( P(F\cap S)=8/30=0.267 \) found; [1] correct comparison and conclusion that the events are NOT independent, since the two values are unequal.

(a) \( S_{xx}=\Sigma x^2-\dfrac{(\Sigma x)^2}{n}=91-\dfrac{21^2}{6}=91-73.5=17.5 \). \( S_{yy}=\Sigma y^2-\dfrac{(\Sigma y)^2}{n}=4702-\dfrac{158^2}{6}=4702-4160.67=541.33 \). \( S_{xy}=\Sigma xy-\dfrac{(\Sigma x)(\Sigma y)}{n}=650-\dfrac{(21)(158)}{6}=650-553=97.0 \).

(b) \( r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{97.0}{\sqrt{17.5\times541.33}}=\dfrac{97.0}{\sqrt{9473.3}}=\dfrac{97.0}{97.33}=0.997 \) (3 s.f.). Since r is very close to +1, this indicates a strong, positive linear correlation between advertising spend and sales — as advertising spend increases, sales tend to increase in an almost perfectly linear way.

(c) \( b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{97.0}{17.5}=5.54 \) (3 s.f.). \( \bar{x}=\dfrac{21}{6}=3.5 \), \( \bar{y}=\dfrac{158}{6}=26.33 \); \( a=\bar{y}-b\bar{x}=26.33-5.54(3.5)=26.33-19.40=6.93 \). Regression line: \( y=6.93+5.54x \). The gradient indicates that, on average, each additional £100 spent on advertising in a month is associated with an increase of approximately 554 items (5.54 × 100) in monthly sales.

Final answer: \( r=0.997 \) (strong positive linear correlation); regression line \( y=6.93+5.54x \).

(a) [1] correct \( S_{xx}=17.5 \); [1] correct \( S_{yy}=541.33 \); [1] correct \( S_{xy}=97.0 \). (b) [1] correct method, \( r=S_{xy}/\sqrt{S_{xx}S_{yy}} \); [1] correct value \( r=0.997 \) (accept 0.99–1.00, ECF from (a)); [1] correct interpretation: strong, positive linear correlation between advertising spend and sales. (c) [1] correct gradient \( b=5.54 \) (ECF from (a)); [1] correct intercept \( a=6.93 \) and correctly stated line \( y=6.93+5.54x \), with a correct interpretation of the gradient in context (sales increase, on average, by 5.54×100 items per extra £100 of advertising spend).

(a) \( X\sim B(15,\,0.08) \)

(b) \( P(X=2) = \binom{15}{2}(0.08)^2(0.92)^{13} = 105\times0.0064\times0.3383 = 0.227 \) (3 s.f.)

(c) \( P(X\le1) = P(X=0)+P(X=1) \). \( P(X=0)=\binom{15}{0}(0.08)^0(0.92)^{15}=(0.92)^{15}=0.286 \). \( P(X=1)=\binom{15}{1}(0.08)^1(0.92)^{14}=15\times0.08\times(0.92)^{14}=15\times0.08\times0.3108=0.373 \). \( P(X\le1)=0.286+0.373=0.660 \) (3 s.f.)

(d) Each component must have the same, constant probability of being defective (0.08), independent of which component it is or of the outcome for any other component in the sample (i.e. the probability of being defective does not vary between components, and whether one component is defective does not affect whether another is).

Final answer: (b) \( P(X=2)=0.227 \); (c) \( P(X\le1)=0.660 \).

(a) [1] correctly states \( X\sim B(15,0.08) \), both parameters correct. (b) [1] correct binomial coefficient \( \binom{15}{2}=105 \); [1] correct substitution into the binomial pmf; [1] \( P(X=2)=0.227 \) (accept 0.225–0.229). (c) [1] correct method, summing P(X=0) and P(X=1); [1] both individual probabilities correctly calculated (0.286 and 0.373, accept small rounding variation); [1] \( P(X\le1)=0.660 \) (accept 0.656–0.664, ECF). (d) [1] valid assumption stated (constant probability of defect across components; independence between components) — either point, correctly expressed, is sufficient for the mark.