Question 10:
--- 10(i) ---
10(i) | x = (y – 1.1664) / 0.4604 = 11.02 or Σ x = 110.2 | M1 | Find x or Σ x using eqn. of regression line of y on x
S = 1419.98 – 110.22/10 = 205.576
xx
S = 439.68 – 62.42/10 = 50.304 (AEF)
yy | M1 | Find S and S
xx yy
(may be earned in part (ii))
EITHER: b = S / S = (S / S ) × (S / S )
xy yy xy xx xx yy
= 0.4604 × 205.576 / 50.304 = 1.88[15] | M1A1 | Find gradient b to 3 s.f. in x – x = b (y – y)
OR: S = 0.4604 × S = 94.647
xy xx
b = S / S = 94.647 / 50.304 = 1.88[15]
xy yy | (M1A1)
(x – 11.02) = b (y – 6.24), x = 1.88y – 0.721 | M1A1 | and hence eqn. of regression line of x on y
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--- 10(ii) ---
10(ii) | r = S / √(S S ) = 94.647 / √(205.576 × 50.304)
xy xx yy
or √(0.4604 × b) = √(0.4604 × 1.8815) (AEF) | M1 | Find correlation coefficient r
= 0.931 | *A1
2
--- 10(iii) ---
10(iii) | H : ρ= 0, H : ρ> 0
0 1 | B1 | State both hypotheses (B0 for r …)
EITHER: r = 0.549
10, 5% | *B1 | State or use correct tabular one-tail r-value
Reject H if r > tab. r-value (AEF)
0 | M1 | State or imply valid method for conclusion
OR: t = r√((n–2) / (1 – r2)) = 7.21, t = 1.86
r 8,0.95 | (*B1 | (Rarely seen)
Reject H if | t | > tab. t-value (AEF)
0 r | M1)
Evidence of positive correlation (AEF) | A1 | Correct conclusion (dep *A1, *B1)
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Question | Answer | Marks | Guidance
11A(i) | ½mv 2 = ½mu2 + mga (1 – cos α) [v 2 = u2 + (2/5) ag]
C C | M1A1 | Find speed v at C by conservation of energy (A0 if no m)
C
R = mg cos α + mu2/a
A | B1 | Find reaction R at A by using F = ma
A
R = mg + mv 2/a
C C | B1 | Find reaction R at C by using F = ma
C
8 {ag + u2 + (2/5) ag} = 9 {(4/5) ag + u2) | M1 | Verify u2 using R / R = 9/8
C A
u2 = (8 + 16/5 – 36/5) ag = 4ag AG | A1
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11A(ii) | ½mv 2 = ½mu2 – 2mga cos α
B
or ½mv 2 – mga (1 + cos α) [v 2 = (22/5) ag]
C C | M1 | Find speed v at B by conservation of energy (A0 if no m)
B
v 2 = (4 – 16/5) ag or (22/5 – 18/5) ag = (4/5) ag
B | *A1
⩾ 0 so P reaches height of B (AEF; dep *A1) | A1 | (or valid argument based on R )
A
R = m v 2/a – mg cos α = 4mg/5 – 4mg/5 = 0
A B | M1**A1 | Find reaction R at B by using F = ma
B
⩾ 0 so P still [just] in contact at B (AEF; dep **A1) | A1
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11B(i) | x = (1/200) Σ x f(x) = 30 064 / 200 = 150.3[2] | M1A1 | Find sample mean to 4 s.f. using mid-interval values
(B1 for 29 964 / 200 = 149.8[2] or 30 164 / 200 = 150.8[2])
2
Question | Answer | Marks | Guidance
11B(ii) | P(151 ⩽ X < 152)
= P((151 – 150)/1.2 ⩽ Z < (152 – 150)/1.2)
= P(0.833 ⩽ Z < 1.667)
= 0.9522 – 0.7975 [= 0.1547] (to 4 s.f.) | M1A1 | Find P(151 ⩽ X < 152)
E = 190.44 – 159.5 or 200 × 0.1547 = 30.94 AG
151 | A1 | and hence expected frequency for 151 ⩽ x < 152
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11B(iii) | H : [Normal] distribution fits data (AEF)
0 | B1 | State (at least) null hypothesis in full
O: 3 23 52 69 36 17
i
E: 9.56 30.94 59.50 59.50 30.94 9.56
i | M1 | Combine values consistent with all exp. values ⩾ 5
X2 = 4.501 + 2.038 + 0.945 + 1.517 + 0.828 + 5.790 | M1 | Find value of X2 from Σ (E – O)2 / E [or Σ O2/E – n ]
i i i i i
= 15.6 (to 3 s.f.) | A1
No. n of cells: 8 7 6 5
χ 2: 14.07 12.59 11.07 9.488
n-1, 0.95 | B1 | State or use consistent tabular value χ 2 (to 3 s.f.)
n–1, 0..95
[FT on number, n, of cells used to find X2]
Accept H if X2 > tabular value (AEF)
1 | M1 | State or imply valid method for conclusion
15.6 [± 0.1] > 11.1 so distn. doesn’t fit [data] (AEF) | A1 | Conclusion (requires both values correct)
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