2020 IJMB mathematics paper 2

IJMB Statistics & Coordinate Geometry — Practice Paper

SECTION A (20%)

1. Given the points $P(4,-2)$ and $Q(-2,6)$, find the coordinates of the mid-point $A$ of $PQ$ and its distance from $B(3,5)$. [05 marks]

2. Calculate the value of $M$ of the distribution with probability density function $f(x) = M\sin x$, where $0\le x\le \pi/3$. [04 marks]

3. Calculate $P(A)$ given that $P(B)=0.65$, $P(A\cup B)=0.95$, if $A$ and $B$ are mutually exclusive events. [03 marks]

4. Find the equation of the circle whose centre is $(3,-2)$ and radius $\sqrt7$. [04 marks]

5. Find the eccentricity and foci of the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{9}=1$. [04 marks]

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SECTION B: CO-ORDINATE GEOMETRY

6. (a) Show that the equation $y^2-4y-6x+10=0$ represents a parabola. Find the coordinates of its vertex, focus, and the equations of its directrix and axis. [10 marks]

(b) Find the equation of the circle which passes through the points $(2,3)$, $(2,1)$, and $(0,3)$. [10 marks]

7. (a) Show that the equation $x^2+9y^2+4x+18y+4=0$ represents an ellipse. Hence, find the value of its eccentricity, centre, and the equations of its directrices. [10 marks]

(b) Show that the locus of points which move so that its distance from the origin is three times its distance from the point $(4,0)$ is a circle. [10 marks]

8. (a) Find the equations of the tangents to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$ from the point $(-2,-1)$. [10 marks]

(b) Sketch the curve $r = 3(1+\cos\theta)$ in polar form, for $0\le\theta\le360°$. [10 marks]

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SECTION C: STATISTICS

9. (a) The number of accidents at a busy junction follows a Poisson law with an average of 3 accidents a week. Find the probability that in a certain week:

(i) at least three accidents occur; (ii) at most one accident occurs; (iii) there are exactly three accidents. [12 marks]

(b) If a card is randomly selected from a deck of cards, find the probability that it is:

(i) A Jack or a Club; (ii) The Ace of Spades. [08 marks]

10. (a) If 15% of the components produced by a machine are defective, determine, using the binomial distribution, the probability that out of 6 components selected at random:

(i) 1; (ii) 0; and (iii) less than 2 will be defective. [12 marks]

(b) The weight of packages from a factory is normally distributed with mean 20 kg and standard deviation of 5 kg. Determine the probability that a package weighs between 10 kg and 30 kg. [08 marks]

11. The following are marks obtained by eight students out of a maximum of 10 marks for each subject in a test:

Mathematics $X$	5	7	6	4	8	5	7	6
Physics $Y$	7	5	8	7	4	8	6	3

(i) Verify that for the above data $\bar{x}=\bar{y}=6$

(ii) Calculate the correlation coefficient, $r$.

(iii) Interpret the value of this coefficient.

(iv) Calculate Spearman’s rank correlation coefficient.

(v) Interpret the value of the coefficient.

(vi) Do the values in (ii) and (iv) above show any correspondence?

(vii) Plot the points $(x,y)$ on the scatter diagram.

(viii) Does the scatter diagram indicate the same type of result as in (ii) and (iv)?

(ix) Calculate the equation of the line of $x$ on $y$. [20 marks]

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COMPLETE SOLUTIONS

SECTION A

Q1. Midpoint $A$ of $PQ$ and Distance from $B(3,5)$

Given: $P(4,-2)$, $Q(-2,6)$

Midpoint $A$:

$$A = \left(\frac{4+(-2)}{2}, \frac{-2+6}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right)$$

$$\boxed{A = (1,2)}$$

Distance $AB$ where $B(3,5)$:

$$AB = \sqrt{(1-3)^2+(2-5)^2} = \sqrt{4+9}$$

$$\boxed{AB = \sqrt{13}\ \text{units}}$$

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Q2. Find $M$ for $f(x) = M\sin x$, $0\le x\le \pi/3$

For a valid probability density function:

$$\int_0^{\pi/3} f(x)\,dx = 1$$

$$M\int_0^{\pi/3}\sin x\,dx = 1$$

$$M[-\cos x]_0^{\pi/3} = 1$$

$$M\left[-\cos\tfrac{\pi}{3}+\cos0\right] = 1$$

$$M\left[-\tfrac12+1\right] = 1$$

$$M\cdot\frac12 = 1$$

$$\boxed{M = 2}$$

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Q3. Find $P(A)$ given Mutually Exclusive Events

Given:

• $P(B) = 0.65$
• $P(A\cup B) = 0.95$
• $A$ and $B$ are mutually exclusive $\Rightarrow P(A\cap B)=0$

Formula: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$

$$0.95 = P(A) + 0.65 - 0$$

$$\boxed{P(A) = 0.30}$$

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Q4. Equation of Circle — Centre $(3,-2)$, Radius $\sqrt7$

Standard form: $(x-h)^2+(y-k)^2=r^2$

$$(x-3)^2+(y+2)^2 = (\sqrt7)^2 = 7$$

Expanding:

$$x^2-6x+9+y^2+4y+4 = 7$$

$$\boxed{x^2+y^2-6x+4y+6=0}$$

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Q5. Eccentricity and Foci of $\dfrac{x^2}{36}+\dfrac{y^2}{9}=1$

Comparing with $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$:

• $a^2=36 \Rightarrow a=6$
• $b^2=9 \Rightarrow b=3$
• Since $a>b$, major axis is along the $x$-axis

$$c^2 = a^2-b^2 = 36-9 = 27 \Rightarrow c=3\sqrt3$$

Eccentricity:

$$e = \frac{c}{a} = \frac{3\sqrt3}{6} = \frac{\sqrt3}{2}$$

$$\boxed{e = \frac{\sqrt3}{2} \approx 0.866}$$

Foci: $(\pm c, 0)$

$$\boxed{\text{Foci} = (3\sqrt3,0)\ \text{and}\ (-3\sqrt3,0)}$$

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SECTION B: CO-ORDINATE GEOMETRY

Q6(a). Show $y^2-4y-6x+10=0$ is a Parabola

Complete the square in $y$:

$$y^2-4y = 6x-10$$

$$(y-2)^2-4 = 6x-10$$

$$(y-2)^2 = 6x-6$$

$$(y-2)^2 = 6(x-1)$$

This is of the form $(Y)^2=4pX$ where $Y=y-2$, $X=x-1$, $4p=6 \Rightarrow p=\dfrac32$

This is a parabola opening to the right. ✓

Vertex: $X=0,Y=0 \Rightarrow x=1,y=2$

$$\boxed{\text{Vertex} = (1,2)}$$

Focus: $X=p=\dfrac32 \Rightarrow x-1=\dfrac32 \Rightarrow x=\dfrac52,y=2$

$$\boxed{\text{Focus} = \left(\frac52,\ 2\right)}$$

Directrix: $X=-p \Rightarrow x-1=-\dfrac32$

$$\boxed{x = -\frac12}$$

Axis of symmetry: $Y=0 \Rightarrow y-2=0$

$$\boxed{y=2}$$

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Q6(b). Circle Through $(2,3)$, $(2,1)$, and $(0,3)$

General equation: $x^2+y^2+Dx+Ey+F=0$

Substituting $(2,3)$:

$$4+9+2D+3E+F=0 \Rightarrow 2D+3E+F=-13\quad(1)$$

Substituting $(2,1)$:

$$4+1+2D+E+F=0 \Rightarrow 2D+E+F=-5\quad(2)$$

Substituting $(0,3)$:

$$0+9+0+3E+F=0 \Rightarrow 3E+F=-9\quad(3)$$

From $(1)-(2)$:

$$2E=-8 \Rightarrow E=-4$$

From $(3)$: $F = -9-3(-4) = -9+12 = 3$

From $(2)$: $2D=-5-E-F=-5+4-3=-4 \Rightarrow D=-2$

Equation:

$$\boxed{x^2+y^2-2x-4y+3=0}$$

Verification:

• $(2,3)$: $4+9-4-12+3=0$ ✓
• $(2,1)$: $4+1-4-4+3=0$ ✓
• $(0,3)$: $0+9-0-12+3=0$ ✓

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Q7(a). Show $x^2+9y^2+4x+18y+4=0$ is an Ellipse

Complete the square:

$$(x^2+4x) + 9(y^2+2y) + 4 = 0$$

$$(x+2)^2-4 + 9\left[(y+1)^2-1\right] + 4 = 0$$

$$(x+2)^2-4+9(y+1)^2-9+4=0$$

$$(x+2)^2+9(y+1)^2=9$$

Dividing by 9:

$$\frac{(x+2)^2}{9} + \frac{(y+1)^2}{1} = 1$$

This is of the form $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ with $a^2=9$, $b^2=1$ $\Rightarrow$ ellipse ✓

Centre: $(-2,-1)$

$$\boxed{\text{Centre} = (-2,-1)}$$

$a=3$, $b=1$

$$c^2=a^2-b^2=9-1=8 \Rightarrow c=2\sqrt2$$

Eccentricity:

$$\boxed{e = \frac{2\sqrt2}{3} \approx 0.943}$$

Directrices (parallel to $y$-axis):

$$x+2 = \pm\frac{a}{e} = \pm\frac{3}{2\sqrt2/3} = \pm\frac{9}{2\sqrt2}$$

$$\boxed{x = -2 \pm \frac{9}{2\sqrt2} = -2 \pm \frac{9\sqrt2}{4}}$$

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Q7(b). Locus — Distance from Origin is Three Times Distance from $(4,0)$

Let $P(x,y)$ be any point on the locus.

Distance from origin: $r_1 = \sqrt{x^2+y^2}$

Distance from $(4,0)$: $r_2 = \sqrt{(x-4)^2+y^2}$

Condition: $r_1 = 3r_2$

$$\sqrt{x^2+y^2} = 3\sqrt{(x-4)^2+y^2}$$

Squaring:

$$x^2+y^2 = 9\left[(x-4)^2+y^2\right]$$

$$x^2+y^2 = 9x^2-72x+144+9y^2$$

$$0 = 8x^2-72x+144+8y^2$$

$$x^2+y^2-9x+18=0$$

Completing the square:

$$(x-4.5)^2-20.25+y^2+18=0$$

$$(x-4.5)^2+y^2=2.25$$

This is a circle with centre $(4.5,0)$ and radius $1.5$. ✓ $\blacksquare$

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Q8(a). Tangents to Hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$ from $(-2,-1)$

Equation of tangent with slope $m$ to $\dfrac{x^2}{16}-\dfrac{y^2}{4}=1$:

$$y = mx \pm\sqrt{16m^2-4}$$

Passing through $(-2,-1)$:

$$-1 = -2m \pm\sqrt{16m^2-4}$$

$$2m-1 = \pm\sqrt{16m^2-4}$$

Squaring:

$$(2m-1)^2 = 16m^2-4$$

$$4m^2-4m+1 = 16m^2-4$$

$$0 = 12m^2+4m-5$$

Quadratic formula:

$$m = \frac{-4\pm\sqrt{16+240}}{24} = \frac{-4\pm\sqrt{256}}{24} = \frac{-4\pm16}{24}$$

$$m_1 = \frac{12}{24} = \frac12,\qquad m_2 = \frac{-20}{24} = -\frac56$$

Tangent 1 ($m=1/2$):

$$y+1 = \frac12(x+2)$$

$$2y+2 = x+2$$

$$\boxed{x-2y=0}$$

Tangent 2 ($m=-5/6$):

$$y+1 = -\frac56(x+2)$$

$$6y+6 = -5x-10$$

$$\boxed{5x+6y+16=0}$$

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Q8(b). Sketch $r=3(1+\cos\theta)$, $0\le\theta\le360°$

Key values table:

$\theta$	$\cos\theta$	$r=3(1+\cos\theta)$
$0°$	$1$	$6$
$60°$	$0.5$	$4.5$
$90°$	$0$	$3$
$120°$	$-0.5$	$1.5$
$180°$	$-1$	$0$
$240°$	$-0.5$	$1.5$
$270°$	$0$	$3$
$300°$	$0.5$	$4.5$
$360°$	$1$	$6$

This is a cardioid with its cusp at $\theta=180°$ ($r=0$, pointing left) and maximum $r=6$ at $\theta=0°$ (pointing right).

              θ=90° (r=3)
                  *
            *            *
       *                       *
  θ=180°(r=0)                θ=0°(r=6)
       *                       *
            *            *
                  *
              θ=270° (r=3)

(Plot in polar coordinates showing a rightward-pointing cardioid)

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SECTION C: STATISTICS

Q9(a). Poisson Distribution — Accidents

Given: $\lambda = 3$ accidents/week, $X\sim\text{Poisson}(3)$

Poisson formula: $P(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$

$$e^{-3} = 0.0498$$

$$P(X=0) = 0.0498$$

$$P(X=1) = 0.0498\times3 = 0.1494$$

$$P(X=2) = 0.0498\times4.5 = 0.2241$$

$$P(X=3) = \frac{e^{-3}\cdot3^3}{3!} = 0.0498\times4.5 = 0.2241$$

(i) $P(X\ge3)$ — at least three accidents:

$$P(X\ge3) = 1-P(X=0)-P(X=1)-P(X=2) = 1-0.0498-0.1494-0.2241$$

$$\boxed{P(X\ge3) = 0.5767}$$

(ii) $P(X\le1)$ — at most one accident:

$$P(X\le1) = P(X=0)+P(X=1) = 0.0498+0.1494$$

$$\boxed{P(X\le1) = 0.1992}$$

(iii) $P(X=3)$ — exactly three accidents:

$$\boxed{P(X=3) = 0.2241}$$

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Q9(b). Card Probability

Deck: 52 cards, 4 suits, 13 cards each

(i) $P(\text{Jack or Club})$:

Using the addition rule:

$$P(J\cup C) = P(J)+P(C)-P(J\cap C) = \frac{4}{52}+\frac{13}{52}-\frac{1}{52} = \frac{16}{52}$$

$$\boxed{P(\text{Jack or Club}) = \frac{4}{13} \approx 0.308}$$

(ii) $P(\text{Ace of Spades})$:

Only 1 ace of spades in the deck:

$$\boxed{P(\text{Ace of Spades}) = \frac{1}{52} \approx 0.019}$$

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Q10(a). Binomial Distribution — Defective Components

Given: $n=6$, $p=0.15$ (defective), $q=0.85$

$X\sim B(6,0.15)$

Formula: $P(X=r) = \binom{n}{r}p^r q^{n-r}$

(i) $P(X=1)$ — exactly 1 defective:

$$P(X=1) = \binom{6}{1}(0.15)^1(0.85)^5 = 6\times0.15\times0.4437$$

$$\boxed{P(X=1) = 0.3993}$$

(ii) $P(X=0)$ — none defective:

$$P(X=0) = \binom{6}{0}(0.15)^0(0.85)^6 = (0.85)^6 = 0.4437\times0.85$$

$$\boxed{P(X=0) = 0.3771}$$

(iii) $P(X<2)$ — less than 2 defective:

$$P(X<2) = P(X=0)+P(X=1) = 0.3771+0.3993$$

$$\boxed{P(X<2) = 0.7764}$$

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Q10(b). Normal Distribution — Package Weights

Given: $\mu=20$, $\sigma=5$

Find $P(10\le X\le30)$

Standardizing:

$$Z_1 = \frac{10-20}{5} = -2$$

$$Z_2 = \frac{30-20}{5} = +2$$

$P(-2\le Z\le2)$:

$$= P(Z\le2) - P(Z\le-2) = 0.9772-0.0228$$

$$\boxed{P(10\le X\le30) = 0.9544 \approx 95.44\%}$$

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Q11. Correlation and Regression

Data:

Student	$X$	$Y$	$X^2$	$Y^2$	$XY$
1	5	7	25	49	35
2	7	5	49	25	35
3	6	8	36	64	48
4	4	7	16	49	28
5	8	4	64	16	32
6	5	8	25	64	40
7	7	6	49	36	42
8	6	3	36	9	18
$\Sigma$	48	48	300	312	278

$n=8$

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(i) Verify $\bar{x}=\bar{y}=6$

$$\bar{x} = \frac{\Sigma X}{n} = \frac{48}{8} = 6\ \checkmark$$

$$\bar{y} = \frac{\Sigma Y}{n} = \frac{48}{8} = 6\ \checkmark$$

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(ii) Correlation Coefficient $r$

Formula:

$$r = \frac{n\Sigma XY - \Sigma X\Sigma Y}{\sqrt{[n\Sigma X^2-(\Sigma X)^2][n\Sigma Y^2-(\Sigma Y)^2]}}$$

Numerator:

$$n\Sigma XY - \Sigma X\Sigma Y = 8(278)-(48)(48) = 2224-2304 = -80$$

Denominator:

$$n\Sigma X^2-(\Sigma X)^2 = 8(300)-48^2 = 2400-2304 = 96$$

$$n\Sigma Y^2-(\Sigma Y)^2 = 8(312)-48^2 = 2496-2304 = 192$$

$$\sqrt{96\times192} = \sqrt{18432} = 96\sqrt2$$

$$\boxed{r = \frac{-80}{96\sqrt2} = -\frac{5}{6\sqrt2} = -\frac{5\sqrt2}{12} \approx -0.589}$$

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(iii) Interpretation of $r$

$r\approx-0.589$ indicates a moderate negative correlation between Mathematics and Physics marks. Students who score higher in Mathematics tend, on average, to score somewhat lower in Physics — though the relationship is not as strong as a near-perfect correlation.

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(iv) Spearman’s Rank Correlation Coefficient

Ranking (1 = highest):

Student	$X$	Rank $X$ ($R_x$)	$Y$	Rank $Y$ ($R_y$)	$d=R_x-R_y$	$d^2$
1	5	6.5	7	3.5	3	9
2	7	2.5	5	6	−3.5	12.25
3	6	4.5	8	1.5	3	9
4	4	8	7	3.5	4.5	20.25
5	8	1	4	7	−6	36
6	5	6.5	8	1.5	5	25
7	7	2.5	6	5	−2.5	6.25
8	6	4.5	3	8	−3.5	12.25
					$\Sigma d^2$	130

(Tied ranks are averaged)

Formula:

$$r_s = 1 - \frac{6\Sigma d^2}{n(n^2-1)}$$

$$r_s = 1 - \frac{6\times130}{8(64-1)} = 1-\frac{780}{504} = 1-\frac{65}{42}$$

$$\boxed{r_s = -\frac{23}{42} \approx -0.548}$$

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(v) Interpretation of $r_s$

$r_s\approx-0.548$ indicates a moderate negative rank correlation, consistent with the Pearson result — students ranking high in Mathematics tend to rank somewhat lower in Physics, though the relationship has noticeable exceptions.

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(vi) Correspondence Between $r$ and $r_s$

Yes, both $r\approx-0.589$ and $r_s\approx-0.548$ show a moderate negative correlation, indicating broad correspondence between the two measures. The values are close in magnitude, confirming the same general trend.

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(vii) Scatter Diagram

Plot points $(X,Y)$:

Y
8 |    *        *
7 |  *    *
6 |              *
5 |       *
4 |          *
3 |              *
  |___________________
     4  5  6  7  8   X

Points: $(5,7),(7,5),(6,8),(4,7),(8,4),(5,8),(7,6),(6,3)$

The points show a mild downward trend, with some scatter, from left to right.

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(viii) Scatter Diagram vs Calculated Coefficients

Yes, broadly. The scatter diagram shows a general downward drift consistent with the negative correlation found in both $r\approx-0.589$ and $r_s\approx-0.548$, although the relationship is visibly looser (more scatter) than a strong correlation would show.

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(ix) Equation of Line of $x$ on $y$

Formula: $x-\bar{x} = r\left(\dfrac{\sigma_x}{\sigma_y}\right)(y-\bar{y})$

Computing $\sigma_x^2$ and $\sigma_y^2$:

$$\sigma_x^2 = \frac{\Sigma X^2}{n}-\bar{x}^2 = \frac{300}{8}-36 = 37.5-36 = 1.5$$

$$\sigma_y^2 = \frac{\Sigma Y^2}{n}-\bar{y}^2 = \frac{312}{8}-36 = 39-36 = 3$$

Regression coefficient of $x$ on $y$:

$$b_{xy} = r\cdot\frac{\sigma_x}{\sigma_y} = r\cdot\sqrt{\frac{1.5}{3}} = r\cdot\sqrt{0.5} = -\frac{5\sqrt2}{12}\cdot\frac{1}{\sqrt2} = -\frac{5}{12}$$

Line of $x$ on $y$:

$$x-6 = -\frac{5}{12}(y-6)$$

$$x-6 = -\frac{5}{12}y + 2.5$$

$$\boxed{x = 8.5 - \frac{5}{12}y}$$

or equivalently: $12x + 5y = 102$