### SECTION A (20%)
1. Given the points P(2,−3) and Q(−2,5), find the coordinates of the mid-point A of PQ and its distance from B(−1,3). [05marks]
2. Calculate the value of M of the distribution with probability density function f(x) = M sin x, where 0 ≤ x ≤ π. [04marks]
3. Calculate, P(A) given that P(B) = 0.72, P(A∪B) = 0.98, if A and B are mutually exclusive events. [03marks]
4. Find the equation of the circle whose centre is (−2,1) and radius √5. [04marks]
5. Find the eccentricity and foci of the ellipse x²/25 + y²/4 = 1. [04marks]
---
### SECTION B: CO-ORDINATE GEOMETRY
6. (a). Show that the equation y² − 2y − 4x + 5 = 0 represents a parabola. Find the coordinates of its vertex, focus and the equations of its directrix and axis. [10marks]
(b). Find the equation of the circle which passes through the points (1,2), (1,0) and (0,2). [10marks]
7. (a). Show that the equation x² + 4y² + 2x + 8y + 1 = 0 represents an ellipse. Hence, find the value of its eccentricity, centre and the equations of its directrix. [10marks]
(b). Show that the locus of points which move so that its distance from the origin is twice its distance from the point (3,0) is a circle. [10marks]
8. (a). Find the equations of the tangents to the hyperbola x²/16 − y²/4 = 1 from the point (−2,1). [10marks]
(b). Sketch the curve r = 2(1 − sinθ) in polar form, for 0 ≤ θ ≤ 360°. [10marks]
---
### SECTION C: STATISTICS
9. (a). The number of deaths in a small town follows a poison law with an av LPerage of 2 deaths a week. Find the probability that in a certain week:
(i) at least two deaths occur; (ii) at most two deaths occur; (iii) there are exactly two deaths. [12marks]
(b). If a card is randomly selected from a deck of cards, find the probability that it is:
(i) A queen or a Diamond; (ii) A king of Diamond. [08marks]
10. (a). If 20% of the bolts produced by a machine are defective, determine, using binomial distribution, the probability that out of 5 bolts selected at random:
(i) 1; (ii) 0 and (iii) less than 2 will be defective. [12marks]
(b). The age of students in a certain class is normally distributed with mean 18 years and standard deviation of 4 years, determine the probability of a student with the age between 15 years and 25 years. [08marks]
11. The following are marks obtained by eight students out of a maximum of 10 marks for each subject in a test:
| Statistics X | 6 | 8 | 7 | 6 | 9 | 5 | 8 | 7 |
|---|---|---|---|---|---|---|---|---|
| Economics Y | 7 | 6 | 8 | 8 | 5 | 9 | 7 | 6 |
(i) Verify that for the above data x̄ = ȳ = 7
(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. [20marks]
---
# COMPLETE SOLUTIONS
## SECTION A
### Q1. Midpoint A of PQ and Distance from B(−1,3)
**Given:** P(2,−3), Q(−2,5)
**Midpoint A:**
A = ((2+(−2))/2, (−3+5)/2) = (0/2, 2/2)
$$\boxed{A = (0, 1)}$$
**Distance AB where B(−1,3):**
AB = √[(0−(−1))² + (1−3)²]
= √[1 + 4]
$$\boxed{AB = \sqrt{5} \text{ units}}$$
---
### Q2. Find M for f(x) = M sin x, 0 ≤ x ≤ π
For a valid probability density function:
$$\int_0^{\pi} f(x)\,dx = 1$$
$$M\int_0^{\pi} \sin x\,dx = 1$$
$$M[-\cos x]_0^{\pi} = 1$$
$$M[-\cos\pi + \cos 0] = 1$$
$$M[1 + 1] = 1$$
$$2M = 1$$
$$\boxed{M = \frac{1}{2}}$$
---
### Q3. Find P(A) given Mutually Exclusive Events
**Given:**
- P(B) = 0.72
- P(A∪B) = 0.98
- A and B are **mutually exclusive** → P(A∩B) = 0
**Formula:** P(A∪B) = P(A) + P(B) − P(A∩B)
0.98 = P(A) + 0.72 − 0
P(A) = 0.98 − 0.72
$$\boxed{P(A) = 0.26}$$
---
### Q4. Equation of Circle — Centre (−2,1), Radius √5
**Standard form:** (x−h)² + (y−k)² = r²
(x−(−2))² + (y−1)² = (√5)²
$$(x+2)^2 + (y-1)^2 = 5$$
**Expanding:**
x² + 4x + 4 + y² − 2y + 1 = 5
$$\boxed{x^2 + y^2 + 4x - 2y = 0}$$
---
### Q5. Eccentricity and Foci of x²/25 + y²/4 = 1
**Comparing with** x²/a² + y²/b² = 1:
- a² = 25 → a = 5
- b² = 4 → b = 2
- Since a > b, major axis is along x-axis
**c² = a² − b² = 25 − 4 = 21**
c = √21
**Eccentricity:**
$$e = \frac{c}{a} = \frac{\sqrt{21}}{5}$$
$$\boxed{e = \frac{\sqrt{21}}{5} \approx 0.917}$$
**Foci:** (±c, 0)
$$\boxed{\text{Foci} = (\sqrt{21},\ 0) \text{ and } (-\sqrt{21},\ 0)}$$
---
## SECTION B: CO-ORDINATE GEOMETRY
### Q6(a). Show y² − 2y − 4x + 5 = 0 is a Parabola
**Complete the square in y:**
y² − 2y = 4x − 5
(y − 1)² − 1 = 4x − 5
(y − 1)² = 4x − 4
**(y − 1)² = 4(x − 1)**
This is of the form **(Y)² = 4pX** where Y = y−1, X = x−1, 4p = 4 → **p = 1**
This is a **parabola** opening to the right. ✓
---
**Vertex:** X = 0, Y = 0 → x = 1, y = 1
$$\boxed{\text{Vertex} = (1, 1)}$$
**Focus:** X = p = 1 → x − 1 = 1 → x = 2, y = 1
$$\boxed{\text{Focus} = (2, 1)}$$
**Directrix:** X = −p → x − 1 = −1
$$\boxed{x = 0}$$
**Axis of symmetry:** Y = 0 → y − 1 = 0
$$\boxed{y = 1}$$
---
### Q6(b). Circle Through (1,2), (1,0) and (0,2)
**General equation:** x² + y² + Dx + Ey + F = 0
**Substituting (1,2):**
1 + 4 + D + 2E + F = 0
D + 2E + F = −5 ...(1)
**Substituting (1,0):**
1 + 0 + D + 0 + F = 0
D + F = −1 ...(2)
**Substituting (0,2):**
0 + 4 + 0 + 2E + F = 0
2E + F = −4 ...(3)
**From (1) − (2):**
2E = −4 → **E = −2**
**From (3):** F = −4 − 2(−2) = −4 + 4 = **0**
**From (2):** D = −1 − F = −1 − 0 = **−1**
**Equation:**
$$\boxed{x^2 + y^2 - x - 2y = 0}$$
**Verification:**
- (1,2): 1+4−1−4 = 0 ✓
- (1,0): 1+0−1−0 = 0 ✓
- (0,2): 0+4−0−4 = 0 ✓
---
### Q7(a). Show x² + 4y² + 2x + 8y + 1 = 0 is an Ellipse
**Complete the square:**
(x² + 2x) + 4(y² + 2y) + 1 = 0
(x+1)² − 1 + 4[(y+1)² − 1] + 1 = 0
(x+1)² − 1 + 4(y+1)² − 4 + 1 = 0
(x+1)² + 4(y+1)² = 4
Dividing by 4:
$$\frac{(x+1)^2}{4} + \frac{(y+1)^2}{1} = 1$$
This is of the form **(X²/a²) + (Y²/b²) = 1** with a² = 4, b² = 1 → **ellipse** ✓
---
**Centre:** (−1, −1)
$$\boxed{\text{Centre} = (-1,\ -1)}$$
**a = 2, b = 1**
c² = a² − b² = 4 − 1 = 3 → c = √3
**Eccentricity:**
$$\boxed{e = \frac{\sqrt{3}}{2} \approx 0.866}$$
**Directrices** (along x-axis, major axis):
x + 1 = ±a/e = ±2/(√3/2) = ±4/√3
$$\boxed{x = -1 \pm \frac{4}{\sqrt{3}} = -1 \pm \frac{4\sqrt{3}}{3}}$$
---
### Q7(b). Locus — Distance from Origin is Twice Distance from (3,0)
**Let P(x,y) be any point on the locus.**
Distance from origin: r₁ = √(x² + y²)
Distance from (3,0): r₂ = √[(x−3)² + y²]
**Condition:** r₁ = 2r₂
√(x² + y²) = 2√[(x−3)² + y²]
**Squaring:**
x² + y² = 4[(x−3)² + y²]
x² + y² = 4[x² − 6x + 9 + y²]
x² + y² = 4x² − 24x + 36 + 4y²
0 = 3x² − 24x + 36 + 3y²
3x² + 3y² − 24x + 36 = 0
x² + y² − 8x + 12 = 0
**Completing the square:**
(x−4)² − 16 + y² + 12 = 0
**(x−4)² + y² = 4**
This is a **circle** with centre (4,0) and radius 2. ✓ □
---
### Q8(a). Tangents to Hyperbola x²/16 − y²/4 = 1 from (−2,1)
**Equation of tangent** with slope m to x²/16 − y²/4 = 1:
y = mx ± √(16m² − 4)
**Passing through (−2,1):**
1 = m(−2) ± √(16m² − 4)
1 + 2m = ±√(16m² − 4)
**Squaring:**
(1 + 2m)² = 16m² − 4
1 + 4m + 4m² = 16m² − 4
0 = 12m² − 4m − 5
**Quadratic formula:**
m = [4 ± √(16 + 240)] / 24
= [4 ± √256] / 24
= [4 ± 16] / 24
**m₁ = 20/24 = 5/6**
**m₂ = −12/24 = −1/2**
**Tangent 1 (m = 5/6):**
y − 1 = (5/6)(x + 2)
6y − 6 = 5x + 10
$$\boxed{5x - 6y + 16 = 0}$$
**Tangent 2 (m = −1/2):**
y − 1 = −(1/2)(x + 2)
2y − 2 = −x − 2
$$\boxed{x + 2y = 0}$$
---
### Q8(b). Sketch r = 2(1 − sinθ), 0 ≤ θ ≤ 360°
**Key values table:**
| θ | sinθ | r = 2(1−sinθ) |
|---|---|---|
| 0° | 0 | 2 |
| 30° | 0.5 | 1 |
| 90° | 1 | 0 |
| 150° | 0.5 | 1 |
| 180° | 0 | 2 |
| 210° | −0.5 | 3 |
| 270° | −1 | 4 |
| 330° | −0.5 | 3 |
| 360° | 0 | 2 |
This is a **cardioid** pointing **downward** (minimum at θ=90°, maximum at θ=270°).
```
θ=90° (r=0)
*
* *
* *
* θ=0°(r=2) *
* *
* *
*
θ=270° (r=4) ← maximum
```
*(Plot in polar coordinates showing downward-pointing cardioid)*
---
## SECTION C: STATISTICS
### Q9(a). Poisson Distribution — Deaths
**Given:** λ = 2 deaths/week, X ~ Poisson(2)
**Poisson formula:** P(X=k) = e^(−λ)·λᵏ/k!
e^(−2) = 0.1353
---
**P(X=0)** = e^(−2)·2⁰/0! = 0.1353
**P(X=1)** = e^(−2)·2¹/1! = 0.2707
**P(X=2)** = e^(−2)·2²/2! = 0.2707
---
**(i) P(X ≥ 2) — at least two deaths:**
P(X≥2) = 1 − P(X=0) − P(X=1)
= 1 − 0.1353 − 0.2707
$$\boxed{P(X \geq 2) = 0.5940}$$
---
**(ii) P(X ≤ 2) — at most two deaths:**
P(X≤2) = P(X=0) + P(X=1) + P(X=2)
= 0.1353 + 0.2707 + 0.2707
$$\boxed{P(X \leq 2) = 0.6767}$$
---
**(iii) P(X = 2) — exactly two deaths:**
$$\boxed{P(X=2) = \frac{e^{-2} \cdot 4}{2} = 0.2707}$$
---
### Q9(b). Card Probability
**Deck: 52 cards, 4 suits, 13 cards each**
**(i) P(Queen OR Diamond):**
Using addition rule:
P(Q∪D) = P(Q) + P(D) − P(Q∩D)
= 4/52 + 13/52 − 1/52
= 16/52
$$\boxed{P(\text{Queen or Diamond}) = \frac{4}{13} \approx 0.308}$$
**(ii) P(King of Diamond):**
Only 1 king of diamonds in the deck:
$$\boxed{P(\text{King of Diamond}) = \frac{1}{52} \approx 0.019}$$
---
### Q10(a). Binomial Distribution — Defective Bolts
**Given:** n=5, p=0.20 (defective), q=0.80
X ~ B(5, 0.2)
**Formula:** P(X=r) = ⁿCᵣ · pʳ · q^(n−r)
---
**(i) P(X=1) — exactly 1 defective:**
P(X=1) = ⁵C₁ · (0.2)¹ · (0.8)⁴
= 5 × 0.2 × 0.4096
$$\boxed{P(X=1) = 0.4096}$$
---
**(ii) P(X=0) — none defective:**
P(X=0) = ⁵C₀ · (0.2)⁰ · (0.8)⁵
= 1 × 1 × 0.32768
$$\boxed{P(X=0) = 0.3277}$$
---
**(iii) P(X < 2) — less than 2 defective:**
P(X<2) = P(X=0) + P(X=1)
= 0.3277 + 0.4096
$$\boxed{P(X < 2) = 0.7373}$$
---
### Q10(b). Normal Distribution — Student Ages
**Given:** μ = 18, σ = 4
Find P(15 ≤ X ≤ 25)
**Standardizing:**
Z₁ = (15−18)/4 = −3/4 = **−0.75**
Z₂ = (25−18)/4 = 7/4 = **+1.75**
**P(−0.75 ≤ Z ≤ 1.75):**
= P(Z ≤ 1.75) − P(Z ≤ −0.75)
= 0.9599 − 0.2266
$$\boxed{P(15 \leq X \leq 25) = 0.7333 \approx 73.33\%}$$
---
### Q11. Correlation and Regression
**Data:**
| Student | X | Y | X² | Y² | XY |
|---|---|---|---|---|---|
| 1 | 6 | 7 | 36 | 49 | 42 |
| 2 | 8 | 6 | 64 | 36 | 48 |
| 3 | 7 | 8 | 49 | 64 | 56 |
| 4 | 6 | 8 | 36 | 64 | 48 |
| 5 | 9 | 5 | 81 | 25 | 45 |
| 6 | 5 | 9 | 25 | 81 | 45 |
| 7 | 8 | 7 | 64 | 49 | 56 |
| 8 | 7 | 6 | 49 | 36 | 42 |
| **Σ** | **56** | **56** | **404** | **404** | **382** |
n = 8
---
### (i) Verify x̄ = ȳ = 7
x̄ = ΣX/n = 56/8 = **7** ✓
ȳ = ΣY/n = 56/8 = **7** ✓
---
### (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:**
nΣXY − ΣXΣY = 8(382) − (56)(56) = 3056 − 3136 = **−80**
**Denominator:**
nΣX² − (ΣX)² = 8(404) − 56² = 3232 − 3136 = **96**
nΣY² − (ΣY)² = 8(404) − 56² = 3232 − 3136 = **96**
√(96 × 96) = 96
$$\boxed{r = \frac{-80}{96} = -0.833}$$
---
### (iii) Interpretation of r
r = −0.833 indicates a **strong negative correlation** between Statistics and Economics marks. Students who score high in Statistics tend to score low in Economics and vice versa.
---
### (iv) Spearman's Rank Correlation Coefficient
**Ranking (1 = highest):**
| Student | X | Rank X (Rx) | Y | Rank Y (Ry) | d = Rx−Ry | d² |
|---|---|---|---|---|---|---|
| 1 | 6 | 6.5 | 7 | 4.5 | 2 | 4 |
| 2 | 8 | 2.5 | 6 | 7 | −4.5 | 20.25 |
| 3 | 7 | 4.5 | 8 | 1.5 | 3 | 9 |
| 4 | 6 | 6.5 | 8 | 1.5 | 5 | 25 |
| 5 | 9 | 1 | 5 | 8 | −7 | 49 |
| 6 | 5 | 8 | 9 | 1→ | 7 | 49 |
| 7 | 8 | 2.5 | 7 | 4.5 | −2 | 4 |
| 8 | 7 | 4.5 | 6 | 7 | −2.5 | 6.25 |
| | | | | **Σd²** | | **166.5** |
*(Tied ranks are averaged)*
**Formula:**
$$r_s = 1 - \frac{6\Sigma d^2}{n(n^2-1)}$$
$$r_s = 1 - \frac{6 \times 166.5}{8(64-1)}$$
$$r_s = 1 - \frac{999}{504}$$
$$r_s = 1 - 1.982 = -0.982$$
$$\boxed{r_s \approx -0.982}$$
---
### (v) Interpretation of rₛ
rₛ = −0.982 indicates a **very strong negative rank correlation** — there is almost perfect inverse ranking between Statistics and Economics performance.
---
### (vi) Correspondence Between r and rₛ
**Yes**, both r = −0.833 and rₛ = −0.982 show **strong negative correlation**, indicating correspondence in their conclusions. Both confirm that students performing well in Statistics tend to perform poorly in Economics. The Spearman coefficient is slightly stronger due to its rank-based nature.
---
### (vii) Scatter Diagram
Plot points (X, Y):
```
Y
9 | *
8 | * *
7 | * *
6 | * *
5 | *
|___________________
5 6 7 8 9 X
```
Points: (6,7),(8,6),(7,8),(6,8),(9,5),(5,9),(8,7),(7,6)
The points show a **downward trend** from left to right.
---
### (viii) Scatter Diagram vs Calculated Coefficients
**Yes**, the scatter diagram confirms the same result as (ii) and (iv). The downward slope of the points from top-left to bottom-right is consistent with the **negative correlation** found in both r = −0.833 and rₛ = −0.982.
---
### (ix) Equation of Line of x on y
**Formula:** x − x̄ = r(σₓ/σᵧ)(y − ȳ)
**Computing σₓ and σᵧ:**
$$\sigma_x^2 = \frac{\Sigma X^2}{n} - \bar{x}^2 = \frac{404}{8} - 49 = 50.5 - 49 = 1.5$$
$$\sigma_y^2 = \frac{\Sigma Y^2}{n} - \bar{y}^2 = \frac{404}{8} - 49 = 1.5$$
σₓ = σᵧ = √1.5
**Regression coefficient of x on y:**
$$b_{xy} = r \cdot \frac{\sigma_x}{\sigma_y} = -0.833 \times \frac{\sqrt{1.5}}{\sqrt{1.5}} = -0.833$$
**Line of x on y:**
x − 7 = −0.833(y − 7)
x − 7 = −0.833y + 5.833
$$\boxed{x = -0.833y + 12.833}$$
or equivalently: **x + 0.833y = 12.833**