Question 10:
--- 10(i) ---
10(i) | Σ x = 25, Σ y = 25 + q, Σ xy = 135 + 4q,
Σ x2 = 141, Σ y2 = 159 + q2 | B1 | Find required values
S = 135 + 4q – 25(25 + q)/5 = [10 – q] (AEF)
xy
[ S = 141 – 252/5 = 16 ]
xx
S = (159 + q2) – (25 + q)2/5 [= 34 – 10q + 4q2/5 ]
yy | M1 A1 | (S , S , S may be scaled by the same constant)
xy xx yy
40 – 4q = 170 – 50q + 4q2, 2 q2 – 23q + 65 = 0
(2q – 13)(q – 5) = 0, q = 5 | M1 A1 | Equate gradient 5/4 in line of x on y to S / S
xy yy
and solve quadratic to find integer value of q
5
--- 10(ii) ---
10(ii) | c = 25/5 – (5/4)(25 + q)/5 = – (5 + q) / 4 | M1 A1 | Find c from x – (5/4) y
= – 5/2 or – 2⋅5 | A1
3
--- 10(iii) ---
10(iii) | r = S / √(S S ) = 5 / √(16 × 4)
xy xx yy | M1 A1 | Find correlation coefficient r
= 5/8 or 0⋅625 | A1
3
Question | Answer | Marks | Guidance
11E(i) | ½mu 2 = ½m (21ag/2) + mga [u 2 = (25/2) ag]
P P | M1 A1 | u is speed of P at lowest point, v is speed of Q immediately after
P Q
collision.
Apply conservation of energy at lowest point (A0 if no m)
½ 4mv 2 = 4mag
Q | M1 | Find speed v at lowest point by conservation of energy
Q
(A0 if no m)
v = √(2ag) or 1⋅41√(ag) or 4⋅47√a
Q | A1
mu = [±] mv + 4mv
P P Q | M1 | Find v using conservation of momentum (m may be omitted)
P
v = [±] (– 5/√2 + 4√2) √(ag)
P | A1
v  = (3/√2) √(ag) or 2⋅12 √(ag) or 0⋅671√a (AEF)
P | A1 | Hence find speed of P
7
11E(ii) | V is speed of P when it loses contact
P
½mV 2 = ½mv 2 – mga (1 + cos α)
P P
[V 2 = (9/2)ag – 2ga (1 + cos α) = (5/2 – 2 cos α) ag]
P | M1 A1 | Apply conservation of energy at D (A0 if no m)
[R =] mV 2/a – mg cos α = 0 [V 2 = ag cos α]
D P P | M1 A1 | Apply F = ma radially at D with reaction = 0
(5/2 – 2 cos α) ag = ag cos α , cos α = 5/6 or 0⋅833 | A1 | Combine to find cos α
5
Question | Answer | Marks | Guidance
11O(i) | ts / √8 = ½ (16⋅7 – 13⋅5) [= 1⋅6]
A | M1 | Relate s to semi-width of confidence interval
A
t = 2⋅365 (to 3 s.f.)
7, 0.975 | A1 | State or use correct tabular t value
[s = √8 × 1⋅6 / 2⋅365 = 1⋅9135], s 2 = 3⋅66[16]
A A | A1 | Hence find unbiased estimate of A’s population variance
3
11O(ii) | H : µ = µ , H : µ >µ (AEF)
0 A B 1 A B | B1 | State hypotheses (B0 for x …)
[x =15.1], x =85.2/6=14.2
A B | B1 | Find sample mean for B
s 2 = (1221⋅06 – 85⋅22/6) / 5
B
= 561/250 or 2⋅244 or 1⋅4982 (all to 3 s.f.) | M1 | Estimate or imply population variance for B
(allow biased here: 1⋅87 or 1⋅3672)
s2 = (7 s 2 + 5 s 2) / 12 = 3⋅0709 or 1⋅7522
A B | M1 A1 | Estimate (pooled) common variance (s 2 not needed explicitly)
B
t = 1⋅782
12, 0.95 | B1 | State or use correct tabular t value
[-] t = (x – x ) / (s √(1/8 + 1/6)) = 0⋅951
A B
t < 1⋅78 so [accept H ]
0 | M1 A1 | Find value of t (or can compare x – x = 0⋅9 with 1⋅69)
A B
Correct conclusion
mean mass of B not less than mean mass of A (AEF) | B1
9
SC1: Implicitly taking s 2, s 2 as unequal population variances
A B
(may also earn first B1 B1 M1)
z = (x – x ) / √(s 2/8 + s 2/6)
A B A B
= 0⋅9 / √(0⋅8317) = 0⋅987
z < 1⋅645 so
DepSC1: mean mass of B not less than mean mass of A (AEF)
Comparison with z and conclusion (FT on z)
0.95
(can earn at most 5/9)