Section A: Pure Mathematics [40 marks]. 1. (i). Sketch f for the required domain
only! Note: There is only 1 intersection point at (0, 0) between f(x) and g(x). (ii).
2008 PAPER 1 Solutions 2
1.
∫
2 1
⎡ x 3 ⎤ 23 13 7 – = x dx = ⎢ ⎥ = 3 3 ⎣ 3 ⎦1 3 2
4
∫
4 a
⎡ 32 ⎤ 3 3 ⎞ 2⎛ ⎞ 2y ⎥ 2⎛ 3 y dy = ⎢ = ⎜ (4) 2 − (a ) 2 ⎟ = ⎜ 8 − (a ) 2 ⎟ ⎢ 3 ⎥ 3⎝ ⎠ 3⎝ ⎠ ⎣⎢ ⎦⎥ a
3 ⎞ 7 2⎛ 2 8 − ( a ) ⎜ ⎟= 3⎝ ⎠ 3
2
9 ⎛ 9 ⎞3 a = ⇒a =⎜ ⎟ 2 ⎝2⎠ 3 2
2.73 (to 3 sig fig) n
2. [Actually we can write
1
∑ r (2r + 1) = 6 n(n + 1)(4n + 5) so that the context is more familiar to r =1
1 you instead of Sn = n(n + 1)(4n + 5) .] 6 1 Let Pn be the statement Sn = n(n + 1)(4n + 5) , n ∈ Z+ 6 For P1, LHS = u1= 1 (2(1) + 1) = 3
∴ P1 is true. Assume Pk is true for some k ∈ Z+, i.e. Sk =
RHS =
1 (1)(1 + 1)(4(1) + 5) = 3 = LHS 6
1 k (k + 1)(4k + 5) 6
1 1 To prove that Pk + 1 is true, i.e. Sk +1= (k + 1)[(k + 1) + 1][4(k + 1) + 5] = (k + 1)(k + 2)(4k + 9) 6 6 LHS = Sk + 1 = Sk + uk+1 1 1 = k (k + 1)(4k + 5) + (k +1) (2(k +1) + 1) = k (k + 1)(4k + 5) + (k +1) (2k +3) 6 6 1 = (k + 1) [ k (4k + 5) + 6(2k + 3)] Factorize out (k+1)/6. Remember to insert 6 the 6 for the second term! 1 1 = (k + 1) ⎡⎣ 4k 2 + 5k + 12k + 18⎤⎦ = (k + 1) ⎡⎣ 4k 2 + 17k + 18⎤⎦ 6 6 1 = (k + 1)(k + 2)(4k + 9) = RHS 6 ∴ Pk is true ⇒ Pk +1 is true Since P1 is true and Pk is true ⇒ Pk +1 is true, by Mathematical Induction, Pn is true for all positive integers n.
B 3.
(i) (ii)
⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ OA = ⎜ 4 ⎟ , OB = ⎜ −1⎟ ⎜ −3 ⎟ ⎜0⎟ O ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛5⎞ ⎛6⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OP = OA + OB = ⎜ 4 ⎟ + ⎜ −1⎟ = ⎜ 3 ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎜ −3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
P
A [by parallelogram law of addition]
OA i OB = OA OB cos ∠AOB ⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 4 ⎟ i ⎜ −1 ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ 5−4+0 1 = cos ∠AOB = ⎝ ⎠ ⎝ ⎠ = 26 26 26 ⎛1⎞ ⎛5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 4 ⎟ ⎜ −1⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠
1 ) = 87.8o (angle to 1 d.p.) 26 ⎛ 1 ⎞ ⎛ 5 ⎞ ⎛ −3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Area of parallelogram = OA × OB = ⎜ 4 ⎟ × ⎜ −1⎟ = ⎜ −15 ⎟ ⎜ −3 ⎟ ⎜ 0 ⎟ ⎜ −21 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ∠AOB = cos –1 (
(iii)
=
4. (i)
(ii)
(iii) (iv)
(−3) 2 + (−15) 2 + (−21) 2 =
675 = 15 3 units2
dy 3x 3x 3 2x = 2 ⇒ ∫ dy = ∫ 2 dx ⇒ ∫ dy = ∫ 2 dx 2 x +1 x +1 x +1 dx 3 3 y = ln | x 2 + 1| + C = ln( x 2 + 1) + C Brackets or modulus of x2+1 are both acceptable. 2 2 3 When x = 0, y = 2, 2 = ln1 + C ⇒ C = 2 2 3 The required particular solution: y = ln( x 2 + 1) + 2 2 dy → 0. The gradient tends towards 0. When x → ±∞ , dx 3 y y = ln x 2 + 1 + 2 2 3 y = ln x 2 + 1 2 2 x 3 0 y = ln x 2 + 1 − 2 2 −2 The gradient at x = 0 is 0 when you sub x = 0 into dy/dx. So turning points (min) should be drawn rather than pointed cusps. Graphs should also show that the gradient at the two ends tends towards 0.
5. (i)
1 3
1 dx 1 + 9 x2 1 1 1 dx = ∫ 3 2 9 0 ⎛1⎞ 2 ⎜ ⎟ +x ⎝3⎠
∫
0
Remember make the coefficient of x2 1 by pulling out 1/9. 1
1 3 1 1⎡ 1 −1 −1 x ⎤ = ⎡⎣ tan 3x ⎤⎦ 3 = ⎢ tan 0 9 9 ⎣1/ 3 1/ 3 ⎥⎦ 0
1 ⎡π 1 ⎤ π = ⎡⎣ tan −1 ( 3) − tan −1 (0) ⎤⎦ = ⎢ − 0 ⎥ = 3⎣3 3 9 ⎦ (ii)
Exact form for tan−1 3 required by Q and should be evaluated in terms of radians.
u = ln x , v ' = xn 1 n +1 1 u ' = , v = n +1 x x
∫
e 1
x n ln x dx
e Remember to sub the limits ⎡ 1 n +1 ⎤ e 1 (ln ) x x n +1 1 ⎥⎦ – = ⎢⎣ n + 1 ( x ) x dx into the first term! 1 n + 1 ∫1 1 n +1 1 n +1 e 1 = n + 1 e (ln e) – n + 11 (ln1) – x n dx ∫ 1 n +1 e 1 n +1 1 ⎡ x n +1 ⎤ 0 e − = n +1 – ⎢ ⎥ n + 1 ⎣ n + 1⎦ 1 1 n +1 1 ⎡ en +1 − 1⎤⎦ = n +1 e – 2 ⎣ ( n + 1)
= =
6(a)
1
( n + 1)
2
⎡⎣ (n + 1)e n +1 − e n +1 + 1⎤⎦
ne n +1 + 1
( n + 1)
2
AC2 = AB2 + BC2 – 2(AB)(BC) cos θ
AC2 = 12 + 32 – 2(1)(3) cos θ AC2 = 10 – 6 cos θ
Since θ is a small angle, cos θ = 1 – AC2 ≈ 10 – 6(1 – AC ≈ (4 + 3 θ ) 2
1 2
θ
2
2
COSINE RULE. PLEASE REVISE!
θ2 2
3
θ
) = 4 + 3θ 2
(Shown)
1
1
AC ≈ 4 2 (1 +
3θ 2 12 1 ⎛ 3θ 2 ⎞ ) =2[1+ ⎜ ⎟ + …] 4 2⎝ 4 ⎠
3θ 2 +… 4 3 ∴ a = 2, b = 4 =2+
(b)
f(x) = tan (2x +
π
4
)
f ′ (x) = 2 sec2 (2x +
f ′′ (x) = 4 sec (2x + = 8 tan (2x +
π 4
π
4
π 4
(
or f ′ (x) = 2 1 + (f ( x)) 2
)
)[2 tan (2x + ) sec 2(2x +
π 4
π
4
)sec (2x + )
π 4
)
)] or f ''(x) = 2 ( 2f ( x)f '( x) )
f ′ (0) = 4 f ′′ (0) = 16 16 f(x) = 1 + 4x + x2 + … 2! = 1 + 4x + 8x2 + …
f(0) = 1
1 ⎛x⎞ 7. Total hours = 180 = 3(x + 2y) + 9( )(2 π ⎜ ⎟ ) 2 ⎝2⎠ 9π x 180 = 3x + 6y + 2 3π x 60 = x + 2y + 2 3π x 2y = 60 – x – 2 x 3π x y = 30 – – 4 2 Let A be the area of the flower – bed. 2 ⎛x⎞ π⎜ ⎟ x 3π x π x2 2 ]+ A = xy + ⎝ ⎠ = x[30 – – 4 2 8 2 2 2 2 2 x 3π x πx x 5π x 2 – + = 30x – – = 30x – 4 8 2 8 2 dA 5π x = 30 – x – dx 4
Replace y in terms of x so that A is in terms of one variable, that is x only.
dA =0. dx 5π x 30 – x – =0 4 5π 30 = (1 + ) x 4 30 x= ≈ 6.09 (to 3 sig fig) 5π (1 + ) 4 6.0889 3π (6.0889) – y = 30 – ≈ 12.6 (to 3 sig fig) 2 4 d2 A 5π =–1– 0 for all real values of x. = 2 2 ( cx + d ) ( 2 x + 1)
(iii) For f(x) =
∴y=
3x − 7 has a positive gradient at all points of the graph. 2x +1
(iv) y y=
3x − 7 2x +1 y=
7 3 −7
1 x=− 2
x=−
1 2 y
3 2 x
The gradient at x = 7/3 (which is originally an x‐intercept in y original graph) is // to the y‐axis. You will be penalized if you draw it pointed.
y2 =
3x − 7 2x +1 3 2
y= 7 3
x y=−
3 2
10 (i) 10, 13, 16, 19, … a = 10, d = 3 n n [ 2a + (n − 1)d ] > 2000 ⇒ [ 2(10) + (n − 1)3] > 2000 2 2 n [17 + 3n] > 2000 ⇒ 3n2 + 17n − 4000 > 0 2 −39.46
33.79
n < – 39.46 or n > 33.79 Least n = 34 months or 2 years 10 months ∴ She will first saved over $2000 on 1st Oct 2011. (ii) (a)
Her original $10 will get her 10(1.02)24 after 24 months [that is 2 years] Therefore, compound interest = 10(1.02)24 – 10 = $6.084 ≈ $6.08 (to 3 sig fig)
(b)
End of 1st month = 10(1.02) End of 2nd month = [10(1.02) + 10](1.02) = 10(1.02)2 + 10(1.02) End of 3rd month = [10(1.02)2 + 10(1.02) + 10](1.02) = 10(1.02)3 + 10(1.02)2 + 10(1.02) End of 24th month = 10(1.02)24 + 10(1.02)23 + … + 10(1.02) = 10(1.02) + 10(1.02)2 + … 10(1.02)23 + 10(1.02)24 = 10 ((1.02) + (1.02)2 + … (1.02)23 + (1.02)24) ⎡1.02(1.0224 − 1) ⎤ = 10 ⎢ ⎥ 1.02 − 1 ⎣ ⎦ ≈ $310.30 ≈ $310 (to 3 sig fig)
(c)
n
End of nth month = 10 ((1.02) + (1.02)2 + … + (1.02) ) ⎡1.02(1.02n − 1) ⎤ = 10 ⎢ ⎥ ⎣ 1.02 − 1 ⎦ = 510 (1.02n – 1) 510(1.02n – 1) > 2000 ⎛ 200 ⎞ + 1⎟ 1.02n > ⎜ ⎝ 51 ⎠ ⎛ 200 ⎞ + 1⎟ n ln 1.02 > ln ⎜ ⎝ 51 ⎠ Inequality sign does not n > 80.5 change as ln 1.02 is positive. Least n = 81 months
11.
p1 : 2x – 5y + 3z = 3 p2 : 3x + 2y – 5z = – 5 p3 : 5x – 20.9 y + 17z = 16.6 4 4 7 Using GC, x = − , y = − , z = 11 11 11
4 7⎞ ⎛ 4 Coordinates of the intersection point = ⎜ − , − , ⎟ ⎝ 11 11 11 ⎠ (i)
p1 : 2x – 5y + 3z = 3 p2 : 3x + 2y – 5z = – 5 Using GC, z = α , α ∈ R y= –1+α x= –1+α ⎛ x ⎞ ⎛ −1 + α ⎞ ⎛ −1 ⎞ ⎛ 1⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ r = ⎜ y ⎟ = ⎜ −1 + α ⎟ = ⎜ −1 ⎟ + α ⎜ 1 ⎟ ⎜ 1⎟ ⎜z⎟ ⎜ α ⎟ ⎜ 0 ⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎛ −1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ Vector equation of the line l: r = ⎜ −1⎟ + α ⎜1⎟ , α ∈ R ⎜1⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠
(ii)
Eqn should be written as r = … NOT line = … or line = r = …
⎛5⎞ ⎜ ⎟ p3 : 5x + λ y + 17z = μ or r i ⎜ λ ⎟ = μ ⎜17 ⎟ ⎝ ⎠ Since l lies in p3, 1. the direction vector of l is perpendicular to the normal of p3: ⎛1⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ . ⎜ λ ⎟ = 0 ⇒ λ = – 22 ⎜1⎟ ⎜17 ⎟ ⎝ ⎠ ⎝ ⎠
2. a point on the line (−1, −1, 0) should also lie on the plane p3: ⎛ −1 ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −1⎟ . ⎜ −22 ⎟ = μ ⇒ μ = 17 ⎜ 0 ⎟ ⎜ 17 ⎟ ⎝ ⎠ ⎝ ⎠ Therefore, μ = 17 and λ = – 22.
Must know the concepts well for (ii) and (iii).
(iii)
Since there are no point in common, l is just // to p3 but does not lie on p3. 1. the direction vector of l is perpendicular to the normal of p3: ⎛1⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ . ⎜ λ ⎟ = 0 ⇒ λ = – 22 ⎜ ⎟ ⎜ ⎟ ⎝1⎠ ⎝17 ⎠ 2. a point on the line (−1, −1, 0) should NOT lie on the plane p3: ⎛ −1 ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ −1⎟ . ⎜ −22 ⎟ ≠ μ ⇒ μ ≠ 17 ⎜ 0 ⎟ ⎜ 17 ⎟ ⎝ ⎠ ⎝ ⎠ Therefore, μ ≠ 17 and λ = – 22.
(iv)
⎛ 1 ⎞ ⎛ −1 ⎞ ⎜ ⎟ ⎜ ⎟ Another direction vector = ⎜ −1⎟ − ⎜ −1⎟ = ⎜3⎟ ⎜0⎟ ⎝ ⎠ ⎝ ⎠ ⎛1⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ n = ⎜ 1 ⎟ × ⎜ 0 ⎟ = ⎜ −1 ⎟ ⎜ 1 ⎟ ⎜ 3 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 3 ⎞ ⎛ −1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r . ⎜ −1 ⎟ = ⎜ −1 ⎟ . ⎜ −1 ⎟ = – 2 ⎜ −2 ⎟ ⎜ 0 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3x – y – 2z = – 2
2008 Paper 2 Solutions Section A: Pure Mathematics [40 marks] 1 (i)
Sketch f for the required domain only!
Note: There is only 1 intersection point at (0, 0) between f(x) and g(x). f(x) = e x sin x (ii) x 2 x3 x3 + + …)(x – + …) = (1 + x + 2 6 6
=x–
Standard series used here is faster!
x3 x3 x3 + x2 + + … = x + x2 + + … 6 2 3
Your graph of g is a cubic curve and you should exhibit understanding that they are close near x = 0 since g is an approx to f.
x3 = g(x) 3
(iii)
See above diagram for y = x + x 2 +
(iv)
| f(x) – g(x) | < 0.5 is the same as | f(x) – g(x) | − 0.5 < 0. Sketch y = |ex sin x – ( x + x2 +
x3 )| − 0.5. 3
y
y = 0.5 −1.96
x 1.56
From the graph, – 1.96 < x < 1.56 for | f(x) – g(x) | − 0.5 < 0.
2
2 (i) y = x (1 – x)
1 2
1
y = ± x 2 (1 – x)
1 4 1 2
1
1 4
1
1 2
1 4
Area of R = 2 ∫ x (1 − x) dx = 2 ∫ x (1 − x) dx = 0.9988 = 0.999 (to 3 sig fig) 0
(ii)
0
Let u = 1 – x ⇒ x = 1 – u When x = 0, u = 1 – 0 = 1 When x = 1, u = 1 – 1 = 0 du = – 1 ⇒ dx = −du dx 1
Volume = π ∫ y 2 dx
Limits change correspondingly. Upper limit is x = 1 which changes to u = 0.
0
1
1
= π ∫ x(1 − x) 2 dx 0
1
0
= π ∫ (1 − u )u 2 (−du ) 1
0
0
1 2
= −π ∫ u − u 1
3 2
5 ⎤ ⎡ 3 ⎢u2 u2 ⎥ du = − π ⎢ − ⎥ 3 5 ⎢ ⎥ 2 ⎦1 ⎣2
⎡2 2⎤ 4 π units3 = π ⎢ − ⎥= ⎣ 3 5 ⎦ 15 1
(iii) y2 = x (1 – x) 2 2y
1 1 1 − dy 1 x ⎛1⎞ = (1 – x) 2 + x ⎜ ⎟ (1 – x) 2 (– 1) = (1 – x) 2 − dx 2 1− x ⎝2⎠
1 1 − dy 1 2 = [(1 – x) − x (1 – x) 2 ] / 2y dx 2
When
1 dy 1 x = 0, (1 – x) 2 − =0 dx 2 1− x 1 x 1 1− x = ⇒ (1 – x) = x 2 1− x 2 2 x= 3
Exact form means you must use the analytical method which is by differentiation! You cannot use GC to find 0.66666 which is 2/3 in exact form. You can use GC to check answers though. ☺
3 (a) p =
w w*
=
( )
p 5 = ei 2θ
r =1 r 5
arg( p ) = arg
w = arg w − arg w* = θ − (−θ ) = 2θ w*
= ei10θ = cos (10 θ ) + i sin(10 θ )
For p5 to be real, sin(10 θ ) = 0, i.e. 10 θ = π , 2 π , 3 π , 4 π , … π π 3π 2π π θ= , , , [Since 0 < θ < ] 10 5 10 5 2 For p5 to be positive ⇒ cos (10 θ ) > 0 . Therefore θ =
θ =
π 5
,
π 10
(rejected),
π 3π ,
5 10
(rejected),
2π . 5
2π 5
OR Shorter Method: For p5 to be real and positive, arg (p) = 10 θ must be multiples of 2π . 10 θ = 2 π , 4 π , … π 2π π Im θ = , [Since 0 < θ < ] 2 5 5 (b)
(8,6)
z ≤6
(4,3)
Re
x2 + y 2 = 36
(i) (ii)
z = z − 8 − 6i
See diagram. OA = 10, OB = 5, OQ = 6 = radius Let ∠POB = ∠QOB = θ = cos −1
OB 5 = cos −1 6 OQ
6 3 = tan −1 8 4 3 5 Least value of arg z = tan −1 − cos −1 = 0.0058 (3 dp) 4 6 3 5 Greatest value of arg z = tan −1 + cos −1 = 1.229 (3 dp) 4 6 Arg (8+6i) = tan −1
Do follow the accuracy specified by Q which is 3 decimal places.
y y=x
4 (i) y = f −1 (x)
4
y = f(x)
1 1
(ii)
x 4
Let y = f(x) y = (x – 4)2 + 1, x > 4 y – 1 = (x – 4)2, x > 4 x – 4 = ± y −1 x = 4 − y − 1 (rejected ∵ x > 4) or x = 4 + y − 1
f – 1 ( x) = 4 + x − 1 , x > 1
D f −1 = R f = (1, ∞ )
(iii)
Refer to diagram
(iv)
y = f(x) is reflected about the line y = x to obtain y = f –1 (x) . To find the intersection between f (x) = f –1 (x), it is equivalent to find the intersection between f(x) = x. Note x > 4 (domain of f). (x – 4)2 + 1 = x x2 – 8x + 17 = x x2 – 9x + 17 = 0 −(−9) ± (−9) 2 − 4(1)(17) 9 ± 13 = x= 2(1) 2
x=
9 − 13 (rejected ∵ x > 4) 2
Exact value of x =
9 + 13 2
Exact value required. Numerical solution gains no credit.
Section B: Statistics [60 marks]
950 = 19. Randomly select a student from 1 to 19 and 50 every 19th student thereafter until a sample of 50 students is obtained.
5 Number the students from 1 to 950.
Important to mention the random start point and numbering of students.
Reason Stratified sampling (eg different level as stratas) gives a sample that is more representative of the student population in the school. Students from different levels may have different responses in terms of sports facilities. Important to mention the context by giving example on what the strata might be. Key thing is that it is more representative sample. Do not accept “allows the opinions of diff stratas to be considered separately”
6 Let X be the random variable “the mass of calcium in a 1 litre bottle” ∑ x = 1026 = 68.4 x= n 15 Unbiased estimate of population variance s 2 2 2 ⎡ x) ⎤ 1 ⎡ 1026.0 ) ⎤ ( ( 1 ⎢ ∑ 2 = ∑ x − n ⎥⎥ = 14 ⎢⎢77265.90 − 15 ⎥⎥ = 506.25 n −1 ⎢ ⎣ ⎦ ⎣ ⎦ H0 : μ = 78
H1 : μ ≠ 78 T=
X −μ 2
s n
=
68.4 − 78 = −1.652 506.25 15
Get some method mark from this step if your p‐value happens to be wrong due to carelessness.
p – value = 0.121 Since p – value = 0.121 > 0.05, we do not reject Ho and conclude that there is insufficient evidence to conclude that the mean mass of calcium in a bottle has changed at 5% significance level.
A
0.7
7
0.2 A
B
0.3
0.6
0.8 0.7
0.2
0.4
A
Realize that A wins two sets or B wins two sets, the third set is not played.
B A
A
0.3
B
B
0.8 B st
nd
1 set
3rd set
2 set
(i) P(A wins the second set) = 0.6(0.7) + 0.4(0.2) = 0.5 (ii) P(A wins the match) = 0.6(0.7) + 0.6(0.3)(0.2) + 0.4(0.2)(0.7) = 0.512 (iii) P(B won the first set | A wins the match) 0.4(0.2)(0.7) 7 P( B won the first set AND A wins the match) = = = 0.109 or P( A wins the match) 0.512 64
8 (i)
(ii)
From GC, r = 0.970. (to 3 sig fig) Since r is close to 1, this indicates that there is a strong positive linear correlation between x and t, thus a linear model is appropriate. Answer to the Q whether it is appropriate or not. Do not just say strong t correlation! 10 8 6
Scatter diagram should be drawn carefully, indicating the scale and axes. x is the independent variable so it must be plotted on the horizontal axis.
P (4.8, 7.6)
4 2 2
4
6
8
10
x
(iii)
With P (4.8, 7.6) removed, the remaining points lies on a ln-type curve.
You may also draw in the ln curve onto the same scatter diagram to illustrate as shown below. t 10 8 6
P (4.8, 7.6)
4 2 2
4
6
8
10
x
(iv)
Using GC, t = 1.42 + 4.40 ln x. a = 1.42 and b = 4.40
(iv)
When x = 4.8, t = 1.42 + 4.40 ln (4.8) = 8.32 (to 3 sig fig)
(vi)
x = 8 is out of the data range, the value of t obtained by extrapolation may not be accurate even though the model is a good fit.
9 Let X be the random variable “the number of grand pianos sold in a week” X ~ Po (1.8) Be careful with P(X ≥ 4) = 1 – P(X ≤ 3) = 1 –0.89129 = 0.10871 = 0.109 (to 3 sig fig) P(X ≥ 4) = 1 – P(X ≤ 3) Let Y be the random variable “the number of upright pianos sold in a week” not 1 – P(X ≤ 4) Y ~ Po (2.6) X +Y ~ Po (1.8 + 2.6 = 4.4) P(X +Y = 4) = 0.191736 = 0.192 (to 3 sig fig) Let W be the random variable “the number of grand pianos sold in 50 weeks” W ~ Po (50 × 1.8) W ~ Po (90) Remember to do cc Since λ = 90 > 10, W ~ N(90, 90) approximately for this approx from c .c P(W < 80) ⎯⎯ → P(W < 79.5) = 0.13419 = 0.134 (to 3 sig fig)
poisson to normal.
The rate at which grand pianos are sold over the period of a year may not be constant. This is because the demand for piano may be higher during certain months eg year end when bonuses are given or sales. Your answers should relate to the rate within the period not constant and give a reason in the context of the Q why it is not constant in order to score a full 2 marks.
10 (i) Required number of ways = 3C2 × 4C3 × 5C3 = 120 (ii) Required number of ways = 9C8 = 9 (iii) Required number of ways = 5C4 × 7C4 + 5C5 × 7C3 = 210 (iiii) Required number of ways = 12C8 – (K and M only) – (L and M only which is part (ii)) = 12C8 – 8C8 – 9 = 485 Note: Complementary method is faster and also less complicated. Delegates from K and L = 7 so it is not possible to choose 8 from solely K and L. It is easier if you notice that we can combine K and M together in a single group and choose 8.
X~ N(50, 82) X1 + X2 ~ N(2 × 50,2 × 82) X1 + X2 ~ N(100, 128) P(X1 + X2 > 120) = 0.0385 (to 3 sig fig)
11
X1 – X2 ~ N(50 – 50, 82 + 82) X1 – X2 ~ N(0, 128) (i)
P(X1 > X2 + 15) = P(X1 – X2 > 15) = 0.0924 (to 3 sig fig)
74
Let Y ~ N( μ , σ 2 )
μ=
146
74 + 146 = 110 2
0.0668
E(Y) = μ = 110 (by symmetry)
P(Y < 74) = 0.0668 74 − 110 74 − 110 P(Z < ) = 0.0668 ⇒ = −1.5000
σ
σ
σ = 24 Var (Y) = 242 = 576
A common mistake is write Var (aX + b) = aVar (X) + b.
E(Y) = aE(X )+ b = 50a + b Var(Y) = Var (aX + b) = Var (aX) + Var(b) = a2 Var (X) + 0 = 64 a2 50a + b = 110 ------------------- (1) 64 a2 = 576 ------------------- (2) Solving (2), a = 3 (since a > 0) Solving (1) by substituting a = 3, E(Y) = 50a + b 110 = 50(3)+ b b = – 40
