MS3
£3.00
WELSH JOINT EDUCATION COMMITTEE
CYD-BWYLLGOR ADDYSG CYMRU
General Certificate of Education
Advanced Subsidiary/Advanced
Tystysgrif Addysg Gyffredinol
Uwch Gyfrannol/Uwch
MARKING SCHEMES
SUMMER 2006
MATHEMATICS
M1-M3 and S1-S3
INTRODUCTION
The marking schemes which follow were those used by the WJEC for the 2006 examination
in GCE MATHEMATICS. They were finalised after detailed discussion at examiners'
conferences by all the examiners involved in the assessment. The conferences were held
shortly after the papers were taken so that reference could be made to the full range of
candidates' responses, with photocopied scripts forming the basis of discussion. The aim of
the conferences was to ensure that the marking schemes were interpreted and applied in the
same way by all examiners.
It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conferences, teachers
may have different views on certain matters of detail or interpretation.
The WJEC regrets that it cannot enter into any discussion or correspondence about these
marking schemes.
MATHEMATICS M1
1.
Accelerating a = 3
Constant speed a = 0
(a)
o.e.
M1
A1
R = 6g – 6 × 3
= 40·8N
c.a.o.
A1
R = 6g
= 58·8N
Decelerating a = − 2
2.
N2L applied to object
6g – R = 6a
R= 6+6×2
= 70·8N
B1
c.a.o.
Use of v = u + at with u =0, a = 0·4, v = 1·5
16
= 40s
t=
0⋅4
A1
M1
A1
(b)
B1
B1
B1
B1
(c)
1
1
× 40 × 16 + 16T +
× 20 ×16
2
2
16T + 4800
T
=
2400
=
=
2400
120s
M1, B1
f.t.(a)
c.ao.
A1
A1
used
B1
M1
dim. correct
M1 A1
c.a.o.
A1
3.
(a)
R = 20g
F = limiting friction
= μR
= 0·3 × 20g
= 58·8N
N2L
T–F
a
(b)
= 20a
65 − 58 ⋅ 8
=
20
= 0·31 ms-2
T < limiting friction
∴F=T
= 45N
B1
1
4.
N2L applied to A and B
3.1g – T = 3·1a
M1
B1
T – 1.8g = 1·8a
A1
N2L applied to B
Adding
3.1g = 1.8g = 4.9a
a = 2·6ms-1
T = 1·8 (2.6 + 9.8)
= 22·32N
5.
(a)
1 2
at with s = 0, u = 22·05, a = (±) 9·8
2
1
0 = 22·05t − × 9·8 t2
2
t = 4.5s
Using s = ut +
c.a.o.
m1
A1
c.a.o.
A1
o.e.
M1
A1
A1
v = 22·05ms−1
(b)
(c)
B1
Using v2 = u2 + 2as with v = 0, u = 22·05, a = (±) 9·8
0 = 22·052 – 2 × 9·8s
s = 24·8 (0625)m
o.e.
c.a.o.
M1
A1
A1
Using v = u + at with u = 22·05, a = (±) 9·8, t = 3
v = 22·05 – 9·8 × 3
v = − 7·35
o.e.
f.t. (a) if used
M1
A1
Speed
= 7·35ms-1
Direction is downwards
A1
B1
6.
Moments about C
all forces dim. correct
4·4g × 1·5 + 2·2g × 3·4 = RD × 2·2
RD = 62·72N (6.4g)
Resolve ↑ RC + RD = 4·4g × 2·2g
RC = 0·2g = 1·96N
2
M1
c.a.o.
A2 B1
A1
f.t. RD
M1
A1
7.
(a)
Conservation of momentum
used
M1
0·1 × 10 + 0·6 × 2 = 0·1 VA + 0·6 VB
VA + 6VB = 22
A1
Restitution
used
M1
3
VB − VA = − (2 – 10)
4
− VA + VB = 6
(b)
(c)
A1
Adding 7VB = 28
VB = 4 ms-1
VA = −2 ms-1
dep. on both Ms
c.a.o.
c.a.o.
M1
A1
A1
′
After B collide with wall, it is moving with speed VB towards A
′ 1
VB =
× 4 = 1 ms-1
4
′
′
Since | VB | = 1 and | VA | = 2, B will not catch up with A
f.t.
B1
I = 0·6 (4 + 1)
= 3 NS
8.
B1
f.t. VB, VB
Area
(a)
ABDE
BCD
Lamina
dist. from AB
dist. from AE
3
4
x
5
12
y
60
18
78
′
M1
A1
B1
B1 B1
B1 (areas)
Moments about AB
60 × 3 + 18 × 4
= 78x
x=
43
13
f.t. cand's values
M1 A1
c.a.o.
A1
f.t. cand's values
M1 A1
c.a.o.
A1
f.t. x
B1
Moments about AE
60 × 5 + 18 × 12 = 78 xy
86
y=
13
(b)
AX =
43
cm
13
3
μ = 0·4
9.
⊥ to plane
R =
15g cos 25º
Limiting friction
F
μR
0·4 x 15g cos 25 º
53·2909 N
=
=
=
B1
used
si
M1
M1
f.t. F
c.a.o.
M1 A1
A1
f.t. F
c.a.o.
M1 A1
A1
Max T when body on point of moving up plane
T
T
=
=
15g sin α + F
115·42 N
Min T when body on point of moving down plane
T
T
=
=
15g sin α − F
8·83 N
4
MATHEMATICS M2
1.
sin α =
P=
(a)
45 × 1000
v
M1 A1
N2L to whole system
M1
45000
− 2000g sin α − 150 – 100 = 2000 a
25
1800 – 700 – 250 = 2000a
a = 0·425 ms-2
(b)
(a)
(b)
A2
A1
N2L applied to trailer
T – 800 × 9·8 ×
2.
1
28
1
− 100 = 800 × 0·425
28
T = 720N
A2
A1
rA
=
=
(i − 10k) + t (−2i – 2j − 5k)
(1 − 2t)i + (− 2t)j + (− 10 – 5t)k
B1
rB
=
=
(7i + 9j − 6k) + t(i – 8j – 5k)
(7 + t)i + (9 – 8t)j + (− 6 – 5t)k
B1
=
=
(− 3 – 9)i + (− 4 + 7)j + (− 20 + 16)k
− 12i + 3j – 4k
M1
When t = 2
rA – rB
Distance between A and B
=
12 2 + 3 2 + 4 2
= 13
m1
A1
5
3.
s
(a)
=
∫ (12t − 3t ) dt
=
6t2 – t3 (+C)
When t = 1, s = 0
∴
0 =
C =
∴
m
=
a
=
Power
=
F.v. used
F
=
=
=
ma
3 (12 – 9)
9
=
=
=
9 × (12 × 1·5 – 3 × 1·52)
9 × 11·25
101·25W
=
PE =
=
=
Final EE
=
=
=
o.e.
6t2 – t3 – 5 m
a
Initial energy
A1
6–1+C
– 5
=
Power
4.
M1
s
(b)
(c)
2
M1
A1
dv
dt
12 − 6t
attempted
M1
A1
M1
B1
mgh
3 × 9.8 × 1.2
35·28J
f.t. F
A1
used
M1
si
A1
1 λ (1 ⋅ 2 − 0 ⋅ 8) 2
×
2
0⋅8
1 35 ⋅ 4 × 0 ⋅ 4 2
×
2
0⋅8
3·54 J
A1
35·28
M1 A1
4·6 ms-1
A1
M1
Conservation of energy
1·5v2 + 3.54 =
v
=
6
5.
(a)
Let u be initial horizontal and vertical velocities (v sin/cos 45)
o.e.
B1
t be time to hit target.
Horizontal motion
ut = 120
M1 A1
Vertical motion
Using s = ut +
1 2
at with s = 41·6, u = u (c), a = (±) 9·8
2
41·6 = 120 +
1
⎛ 120 ⎞
× (− 9·8) ⎜
⎟
2
⎝ u ⎠
M1
2
m1 (subst.)
u = 30 ms−1
Speed of projection
=
t
(b)
A1
2u 2
= 30 2 = 42·4264 ms-1
120
=
u
= 4s
B1
convincing
Using v = u + at with u = 30 (c), a = (±) 9·8, t = 4
v
= 30 – 9·8 × 4
= − 9·2
Resultant speed
=
=
M1
f.t. u
f.t. u
(−9 ⋅ 2) 2 + 30 2 (c)
31·38 ms−1
A1
A1
A1
M1
f.t. u
⎛ 9⋅2 ⎞
Direction of motion = tan−1 ⎜⎜ 30(c) ⎟⎟
⎝
⎠
A1
M1
[No if +9.2 used]
i.e. angle above horizontal
= 17·05º
7
A1
6.
(a)
(b)
r
=
cos 3t i + sin 3t j
v
=
d
(r )
dt
=
−3 sin 3t i + 3 cos 3t j
A2
=
(− 3 sin 3t i + 3 cos 3 t j).(cos 3t i + sin 3t j)
M1
=
− 3 sin 3t cos 3t + 3 sin 3t cos 3t
B1
=
0
Consider
v.r
used
dot product
∴ v is perpendicular to r for all values of t
(c)
7.
Speed of P =
si
(−3 sin 3t ) 2 + (3 cos 3t ) 2
=
=
9(sin 2 3t + cos 2 3t )
c.a.o.
Resolve ↑
T cos θ = mg
T = 3 × 9·8 ×
= 31·85N
M1
M1
3
(a)
(b)
A1
|v|
=
M1
A1
M1
A1
1⋅ 3
1⋅ 2
c.a.o.
A1
N2L towards C
T sin θ
=
mrω2
ω2
=
31·85 ×
ω
=
2·86 rads-1
M1
A1 B1 (a= rω2)
0⋅5
1
×
1⋅ 3 3 × 0 ⋅ 5
f.t. T
8
A1
8.
(a)
(b)
At lowest point
mv 2
r
T – mg
=
T = 2 mg, v = u, r = l mg
=
u2
=
mu 2
l
gl
u
=
gl
M1 (ma) A1
m1
convincing
A1
Conservation of energy
1 2
mu
2
=
1 2
mv = mgl ( 1 – cos θ)
2
M1
A1 (KE)
A1 (all correct)
=
gl – 2gl (1 – cos θ)
=
2gl cos θ – gl
v2
=
cos θ
=
0
1
2
θ
=
v2
(c)
(d)
At max θ,
πc
3
= 60 o
used
M1
c.a.o.
A1
At general θ
T – mg cos θ =
mv 2
l
M1(ma)
⎛ mv 2
M1 ⎜⎜
⎝ r
m
m
2gl cos θ + mg cos θ –
gl
l
l
= 3 mg cos θ – mg
2
=
3
= 48·2º, 0·84º
Subst. for v2 and r = l T =
T = mg
cos
θ
9
m1
A1
⎞
⎟
⎟
⎠
MATHEMATICS M3
1.
(a)
dv
dx
⎛ B ⎞
−2
=⎜
⎟ − B ( x + A)
⎝ x + A⎠
a=v
[
=
(b)
used
]
M1
A1
− B2
( x + A) 3
A1
t = 0, v = 12, x = 0
∴ B = 12 A
M1
t = 0, a = − 16, x = 0
− B2 = − 16 A3
A1
144A2 = 16 A3
A= 9
B = 108
(c)
v=
as required
convincing
dx 108
=
dt x + 9
A1
M1
∫ (x + 9)dx = 108∫ dt
2
x
+ 9 x = 108t + C
2
A1
t = 0, x = 0 ⇒ C = 0
f.t. minor error
∴ 216t = x2 + 18x
1
t=
x( x + 18)
216
2.
A1
A1
Auxiliary equation m2 + 2m + 10 = 0
− 2 ± 4 − 40
m =
2
= −1 ± 3i
B1
B1
∴ C.F. - is x = e-t (A sin 3t + B cos 3t)
B1
For P.I. try x = at + b
dx
=a
dt
M1
∴ 2a + 10 (at + b) = 5t – 14
10a = 5
1
a=
2
comp. coeff.
10
A1
M1
2a + 10b = − 14
3
b = −
2
both c.a.o.
∴ General solution is x = e-t (A sin 3 t + B cos 3t) +
1
3
t −
2
2
1 dx
1
When t = 0, x = 4 ,
= 3
2 dt
2
1
3
4 = B −
2
2
B=6
dx
1
= −e-t (A sin 3t + B cos 3t) + e−t (3 A cos 3t – 3 B sin 3t) +
dt
2
1
1
3 = −B+3A+
2
2
A =3
1
3
∴x = 3e−t (sin 3t + 2 cos 3t) + t −
2
2
A1
B1
used
M1
f.t. a.b.
A1
B1
c.a.o.
A1
3.
(a)
(b)
− 0·01v2 – 0·4g = 0·4a
dv
0·4 v
= − 3·92 – 0·01 v2
dx
dv
× 100
40 v
= − (392 + v2)
dx
N2L
vdv
⌠
40⎮
= − dx
2
⌡ (392 + v )
20 ln (392 + v2) = − x + C
∫
A1
convincing
A1
sep. var.
M1
A1 A1
t = 0, v = 17, x = 0
∴ 20 ln (392 + 172) = C
C = 20 ln (681)
x = 20ln (681) – 20 ln (392 + v2)
⎛ 681 ⎞
= 20 ln ⎜
⎟
2
⎝ 392 + v ⎠
At greatest height, v = 0
⎛ 681 ⎞
∴ x = 20 ln ⎜
⎟
⎝ 392 ⎠
= 11.05 m
(c)
M1
dv
a=v
dx
m1
f.t. minor error
A1
m1
c.a.o.
A1
Speed of ball when it returns to O is less than 17 ms−1
B1
because energy used (lost) in overcoming air resistance.
B1
11
4.
(a)
2π
Period =
ω
ω=
(b)
(c)
=4
M1
π
A1
2
Using vMAX
=
aω with vMAX = 3 π, ω =
3π
=
a×
a
=
6m
=
ω2 (a2 – x2) with ω =
Using v2
v2
=
v
=
=
Let x =
π
π
M1
2
2
c.a.o.
π
2
(c), a = 6 (c), x = 4·8
π2
(36 − 4·82)
4
1·8π
5·65 ms−1
A1
M1
f.t. a, ω
A1
f.t. a, ω
A1
f.t. a, ω
B1
distance from O, y = 0, p is at O
⎛π ⎞
6 sin ⎜ t ⎟
⎝2 ⎠
⎛π ⎞
4·8 = 6 sin ⎜ t ⎟
⎝2 ⎠
⎛π ⎞ 4 ⋅8
= 0⋅8
sin ⎜ t ⎟ =
6
⎝2 ⎠
2
t=
sin −1 (0·8)
x =
M1
π
f.t. a, ω
= 0·59s
(d)
Max acceleration when x = a
M1
ω2a
π2
M1
| Max acceleration | =
=
=
=
(e)
A1
×6
4
3π 2
2
14·8 ms−1
=
12
oscillations
4
3 × (4a)
=
72m
Distance travelled =
f.t. a, ω
A1
M1
f.t. a
12
A1
5.
(a)
Impulse = change in momentum
Apply to A
I =
=
−I =
=
−8 =
40 cos α =
Apply to B
∴
cos α =
α =
(b)
40 sin α =
u =
=
2v
2×4
M1
A1
− 40 cos α + 3 v
− 40 cos α + 3 × 4
− 40 cos α + 12
20
1
2
60º
M1
A1
M1
A1
3u
40 ×
3 1
×
2 3
20 3
3
A1
2
Speed of b
=
=
θ =
=
⎛ 20 3 ⎞
⎜
⎟ + 42
⎜ 3 ⎟
⎝
⎠
−1
12·22 ms
M1
A1
⎛ 20 3 ⎞
⎟
tan −1 ⎜
⎜ 3× 4 ⎟
⎝
⎠
70·89º
M1
A1
13
6.
(a)
Moments about B
dim correct, attempted equation M1
37·5g × 4 cos 60º + 75g × x cos 60º = S × 8 sin 60º
B1 A2
Resolve ↑
R
=
=
37·5 g + 75 g
112·5 g
M1
Resolved → S
F
∴ S
=
=
=
F
μR
μ. 112·5g
M1
M1
Substitute S into moment equation and μ = 0·25
x (75 g cos 60º)
=
x
=
0·25 × 112·5 g × 8 sin 60º − 37·5 g × 4 cos 60º
112 ⋅ 5 3 − 75
37 ⋅ 5
A1
3·196
A1
=
(b)
c.a.o.
Substitute x = 8 and s = μ.112·5g (c) into moment equation
M2
μ.112·5g × 8 sin 60º = 37·5g × 4 cos 60º + 75g × 8 cos 60º
A1
μ. =
300 + 75
450 3
c.a.o.
= 0·481
(c)
m1
Person modelled as particle/
Ladder modelled as rod
A1
B1
14
MATHEMATICS S1
1.
(a)
(b)
1
3
2 1 1
P(B eats red sweet) = × =
3 2 3
P(A eats red sweet) =
B1
M1A1
[Method must be shown, award M1 for either
(c)
2 1 2 1
× or × ]
3 2 3 3
2 1 1
2
× × or 1 −
3 2 1
3
1
= (no working required)
3
P(C eats red sweet) =
M1
A1
[FT on answers to (a) and (b) - Special case – award 3/5 for writing down the correct
answers with no working]
2.
(a)
(b)
P(A ∩B) = P(A) + P(B) – P(A∪B)
= 0.05
M1A1
P(A).P(B) = 0 ·12 or P(A⎢B) = 1/12 or P(B⎢A) = 1/4
A and B are not independent
B1
B1
P(exactly one of A,B) = P( A ∩ B ′) + P( A′ ∩ B)
M1
[Award M0 if independence assumed having stated not independent in (a).
If independence stated in (a), FT partially bearing in mind that the problem is now easier]
= 0·2 – 0·05 + 0·6 – 0·05
= 0·7
A1A1
A1
OR
3.
P(exactly one of A,B) = P(A∪B) – P(A∩B)
= 075 – 0·5
= 07
M1A1
A1
A1
(a)
Var(X) = 4
E(Y) = 2 × 4 + 8 = 16
Var(Y) = 4 × 4 = 16
B1
M1A1
M1A1
(b)
Because Y can only take the values 8,10,12 etc
B1
15
4.
(a)
(b)
(i)
P(X > 10) = 0·6528 (or 1 – 0·3472)
M1A1
(ii)
P(X = 15) = 0·8444 – 0·7720 or 0·2280 – 0·1556
= 0·0724 (cao)
6 ⋅ 35
P(Y = 5) = e −6.3 .
= 0 ⋅ 152
5!
⎛
6 ⋅ 32 ⎞
⎟
P(Y < 3) = e - 6.3 ⎜⎜1 + 6 ⋅ 3 +
2 ⎟⎠
⎝
= 0·0498
(cao)
B1B1
B1
(i)
(ii)
5.
(a)
(b)
(a)
M1A1
A1
P(DY) = 0 ⋅ 4 × 0 ⋅ 04 + 0 ⋅ 35 × 0 ⋅ 05 + 0 ⋅ 25 × 0 ⋅ 06
= 0·0485
0 ⋅ 4 × 0 ⋅ 04
(i)
P(A⏐DY) =
0 ⋅ 0485
= 0·330
B1
0 ⋅ 35 × 0 ⋅ 05
0 ⋅ 0485
= 0·361
P(C⏐DY) = 1 – 0·691 = 0·309
So most likely to have come from Farm B.
B1
B1
B1
(ii)
6.
M1A1
M1A1
A1
B1B1
P(B⏐DY) =
(si)
(i)
B(50,0.2)
B1
(ii)
Mean = 50 × 0·2 = 10
Var = 50 × 0·2 × 0·8
SD = 2·83 (2√2)
M1A1
M1
A1
(iii)
P(8 ≤ X ≤ 12) = 0·8139 – 0·1904 or 0·8096 – 0·1861
= 0·6235 (cao)
B1B1
B1
[For candidates summing probs award M1A1 for an expression involving
binomial probs and A1 for the answer]
7.
(b)
X is approx Po(10).
P(X < 10) = 0·4579 (or 1 – 0·5421)
M1A1
M1A1
(a)
Sum of probabilities = 15k
B1
1
15k = 1 so k =
15
B1
(b)
E(X) = 1 ×
=
1
2
5
+ 2 × + ... + 5 ×
15
15
15
M1
11
3
A1
1
2
5
+ 4 × + ... + 25 ×
15
15
15
(= 15)
E(X2) = 1 ×
⎛ 11 ⎞
Var(X) = 15 − ⎜ ⎟
⎝3⎠
= 1.56
M1A1
2
M1
A1
16
(c)
8.
(a)
The possibilities are (1,5), (2,4),(3,3).
2
1 5
2 4 ⎛3⎞
P(Y = 6) = 2 × × + 2 × × + ⎜ ⎟
15 15
15 15 ⎝ 15 ⎠
7
=
45
(i)
P(0·25 ≤ X ≤ 0·5) = F(0·5) – F(0·25)
1
1
= 0 ⋅ 5 2 + 0 ⋅ 5 − (0 ⋅ 25 2 + 0 ⋅ 25)
2
2
(
=
(ii)
)
7
(0·219)
32
B1
M1
A1
A1
The median satisfies
1 2
1
m +m =
2
2
2
m + m −1 = 0
−1+ 5
m=
2
= 0·618
(
B1B1B1
)
M1
A1
M1
A1
[Special case for candidates integrating F(x):For obtaining 2m 3 + 3m 2 − 6 = 0
M1A1]
(b)
(i)
f ( x) = F ′( x)
1
= (2 x + 1)
2
M1
A1
1
(ii)
E(X)=
1
x(2 x + 1)dx
20
∫
M1A1
[FT on candidate’s f(x) from (i). If candidates use F(x) instead of f(x),
not having obtained an answer in (i), award M1]
1
1 ⎡ 2x3 x 2 ⎤
= ⎢
+ ⎥
2⎣ 3
2 ⎦0
7
(0·583)
=
12
A1
A1
17
MATHEMATICS S2
62 ⋅ 6
(= 6·26)
10
0 ⋅1
SE of x =
(= 0·0316)
10
95% conf limits are
6·26 ± 1.96 × 0·0316
M1A1
[M1 correct form, A1 1·96]
giving [6·20,6·32]
Yes because 6·3 is within the interval.
A1
B1
x=
1.
Variance =
2.
Mean =
B1
B1
(b − a ) 2
=3
12
(b − a ) 2 = 36
b–a=6
a+b
= 10
2
b + a = 20
M1A1
A1
AG
M1
A1
Solving, a = 7, b = 13.
3.
(a)
(i)
(ii)
(b)
4.
M1A1
34 − 30
28 − 30
= 2 ; z2 =
= −1
2
2
Prob = 0·97725 – 0·15866 or 0·8413 – 0·02275
= 0·819 (cao)
z1 =
Reqd weight = 25 + 2·326 × 1·8
= 29·2
[M1 for 25 ± zσ]
M1A1
B1B1
B1
M1A1
A1
X - Y is N(5,7·24)
We require P(X - Y > 0)
5
= ( ±)1.86
z=
7 ⋅ 24
Prob = 0.969 (cao)
B1B1
A1
M1A1
(a)
Prob of 1 crash on a computer
= 0 ⋅ 8 × e −0.8
= 0·3595
Prob of 1 crash on each of 5 computers = 0·35955
= 0·006
M1
A1
M1
A1
(b)
Distribution of Total is Po(4).
45
P(Total = 5) = e − 4 .
5!
= 0·156 (cao)
B1
M1A1
A1
18
5.
(a)
H 0 : μ = 2 ⋅ 4 versus H 1 : μ > 2 ⋅ 4 (Accept μ = 12]
In 5 days, number of passengers Y is Poi(12) under H0.
[M1A0 for normal approx]
B1
si
p-value = P( Y ≥ 18 ) = 0·0630
We cannot conclude that the mean has increased.
(b)
B1
M1 A1
B1
Under H0 the number of passengers in 100 days is Po(240) ≈ N(240,240)
B1B1
z=
279 ⋅ 5 − 240
M1A1
240
6.
(a)
= 2·55
Either p-value = 0·00539 or CV = 2·326
[No cc gives z = 2·58, p = 0·00494, wrong cc gives z = 2·61, p = 0·00453]
We conclude at the 1% level that the mean has increased.
A1
A1
(i)
B1
M1
A1
(ii)
(b)
7.
(a)
(b)
X is B(50,p)
(si)
Sig level = P(X ≤ 14⏐p = 0·4)
= 0·0540 (cao)
We require
P(X ≥ 15⏐p = 0·3) = 0·5532
B1
M1A1
Under H0, X is now B(500,0·4) ≈ N(200,120)
185 ⋅ 5 − 200
z=
120
= −1·32
p-value = 0·0934
A1
A1
[No cc gives z = −1·37, p = 0·0853, wrong cc gives z = −1·41, p = 0·0793]
Insufficient evidence to support the agent’s belief. (oe).
B1
H 0 : μ A = μ B versus H 1 : μ A ≠ μ B
B1
501
= 83 ⋅ 5
6
489
xB =
= 81 ⋅ 5
6
The appropriate test statistic is
x−y
TS =
1 1
+
σ
m n
83 ⋅ 5 − 81 ⋅ 5
=
1 1
1.5
+
6 6
= 2.31 (cao)
Prob from tables = 0·01044
p-value = 0·021
(i)
Accept H 0 (or the fuel consumptions are the same) at 1% SL
xA =
(ii)
Accept H 1 (or the fuel consumptions are not the same) at 5% SL
19
B1B1
M1A1
B1
B1
M1
1A1
A1
A1
B1
B1
B1
8.
(a)
(b)
The mean of a large (random) sample from any distribution is
(approximately) normally distributed.
E( X ) = 3.5, Var ( X ) =
z=
3 − 3⋅5
35
600
(si)
= −2 ⋅ 07
B1
B1B1
M1A1
35 / 600
Prob = 0·981
A1
20
MATHEMATICS S3
1.
(a)
The possible combinations are given in the following table.
Combination
112
113
114
123
124
134
234
[Accept a table with 123, 124 and 134 repeated]
M1A1
The sampling distribution of the mean is
Mean
Prob
4/3
0·1
5/3
0·1
2
0·3
7/3
0·2
8/3
0·2
3
0·1
(correct means)
(correct probs)
B1
B2
(correct medians)
(correct probs)
B1
B2
[Award B1 for 4 correct probs]
The sampling distribution of the median is
Median
Prob
1
0.3
2
0.4
3
0.3
[For each table award B1 for correct probs with replacement]
2.
498
= 0 ⋅ 415
1200
(a)
pˆ =
(b)
SE ≈
(c)
(d)
B1
0 ⋅ 415 × 0 ⋅ 585
1200
= 0·0142...
M1
A1
Approx 90% confidence limits are
0·415 ± 1·645 × 0·0142
giving [0·392, 0·438].
M1A1
A1
We are using a normal approximation to a binomial situation.
The standard error and/or p are estimated and are not exact.
B1
B1
21
3.
x A = 1 ⋅ 034; x B = 1 ⋅ 016
B1B1
s A2 = 3 ⋅ 474747... × 10 −4
M1A1
s B2
= 1 ⋅ 449664... × 10
−4
A1
[Accept division by n giving 3·44.. and 1·44..]
3 ⋅ 474747... × 10 −4 1 ⋅ 449664... × 10 −4
+
(= 0·002107)
100
150
95% conf lims are 1·034 – 1·016 ± 1·96 × 0·002107
giving [0·014,0·022]
(cao)
M1A1
A1
UE of μ = 8 ⋅ 54
B1
∑x
B1
SE =
4.
(a)
2
= 734 ⋅ 2 ,
734 ⋅ 2 85 ⋅ 4 2
−
9
9 × 10
= 0·542(666..)
UE of σ 2 =
(b)
5.
(a)
(b)
(c)
B1
M1
A1
H 0 : μ = 9 versus H1 : μ ≠ 9
8 ⋅ 54 − 9
Test statistic =
0 ⋅ 542666.. / 10
= − 1·97
DF = 9
Crit value = 2·26
B1
M1A1
A1
B1
B1
Insufficient evidence to reject the farmer’s claim at the 5% level.
[Award the final B1 only if t used]
B1
Σx = 75, Σy = 89·2, Σxy = 1270·5, Σx 2 = 1375
[B1 1 error]
B2
6 × 1270 ⋅ 5 − 75 × 89 ⋅ 2
6 × 1375 − 75 2
= 0·355
89 ⋅ 2 − 75 × 0 ⋅ 355
a=
6
= 10·4
[Note S xx = 437 ⋅ 5, S xy = 155 ⋅ 5]
b=
(i)
M1A1
A1
M1A1
A1
Est resist = 10·4 + 0·355 × 20 = 17·5
1 (20 − 12 ⋅ 5)
+
6 1375 − 75 2 / 6
= 0·217(343…)
M1 A1
2
SE = 0.4
(ii)
M1
A1
95% confidence limits are
17·5 ± 1·96 × 0·217
giving [17·1,18·0]
[Accept 17·9 as the upper limit]
M1
A1
22
(d)
Test stat =
=
6.
b−β
M1
σ / ( Σx 2 − ( Σx ) 2 / n )
0 ⋅ 355... − 0 ⋅ 4
A1A1
0.4 / (1375 − 75 2 / 6
= − 2·33 (Accept 2·34, 2·35)
A1
EITHER p-value = 2 × 0·0099 = 0·0198 OR Crit value = 2·576
The value 0·4 is consistent with his prediction at the 1% level.
A1
B1
(a)
(b)
(c)
E(U) = a(μ + 2 × 2μ)
1
= μ if a =
5
E(V) = b(2 ×μ + 2μ) = μ
1
if b =
4
M1A1
A1
M1
A1
(
)
M1A1
(
)
M1A1
1 2
13
σ + 4 × 3σ 2 = σ 2
25
25
1
7
Var(V) =
4 × σ 2 + 3σ 2 = σ 2
16
16
V is the better estimator (since it has the smaller variance).
Var(U) =
(i)
(ii)
(iii)
μ + k × 2μ
=μ
1 + 2k
[AG so must be convincing]
σ 2 + k 2 × 3σ 2 (1 + 3k 2 ) 2
Var(W) =
=
σ
(1 + 2k ) 2
(1 + 2k ) 2
E(W) =
2
2
d
(Var(W ) ) = 6k (1 + 2k ) − 4(1 + 42k )(1 + 3k )
dk
(1 + 2k )
B1
M1A1
M1A1
M1A1
For minimum variance,
6k (1 + 2k ) 2 = 4(1 + 2k )(1 + 3k 2 )
M1
3k (1 + 2k ) = 2(1 + 3k 2 )
k=
A1
2
3
A1
[Award full marks if correct answer given with no/partial working]
GCE M/S Mathematics M1-M3 & S1-S3 (June 2006)/JD
23
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