Question 10:
--- 10(i) ---
10(i) | x = Σ x f(x) = 118/40 = 2.95 AG | B1 | Verify mean of sample data (B0 for x = 118/40 = 2.95)
σ2 = (454 – 1182/40) / 40 = 2.65
n
or σ 2 = (454 – 1182/40) / 39 = 2.72
n–1 | B1 | Find variance of sample data (accept either σ2 or σ 2 )
n n–1
2.95 ≈ 2.65 or 2.72 | B1 | Valid comment on correct values
3
Question | Answer | Marks | Guidance
--- 10(ii) ---
10(ii) | E = 40 λ4 e–λ /4! with λ = 2.95 [= 40 × 0.1652]
4
(allow (λ/4) × E = 0.7375 × 8.96)
3 | M1A1 | State expression for reqd. expected value E
4
2
--- 10(iii) ---
10(iii) | H : [Poisson] distribution fits data (AEF)
0 | B1 | State (at least) null hypothesis in full
O: 8 8 10 5 9
i
E: 8.27 9.11 8.96 6.61 7.05
i | M1 | Combine values consistent with all exp. values ⩾ 5
X2 = 0.0088 + 0.1352 + 0.1207 + 0.3921 + 0.5394 | M1 | Find value of X2 from Σ (E – O)2 / E [or Σ O2/E – n ]
i i i i i
= 1.20 (to 3 s.f.) | A1
No. n of cells: 8 7 6 5
χ 2: 12.59 11.07 9.488 7.815
n–2, 0.95 | B1 | State or use consistent tabular value χ 2 (to 3 s.f.)
n–2, 0.95
[FT on number, n, of cells used to find X2]
Accept H if X2 < tabular value (AEF)
0 | M1 | State or imply valid method for conclusion
1.20 [± 0.1] < 7.81[5] so [Poisson] distn. fits [data]
or distn. is a suitable model (AEF) | A1 | Conclusion (requires both values correct)
7
Question | Answer | Marks | Guidance
11A(i) | EITHER: 40 e / 0.8 = 2g, e = 0.4 [or OP = 1.2] [m]
0 0 0 | M1A1 | Find extension e [or OP ] at equilibrium position P
0 0 0
2 d2x/dt2 = – 40 (e + x) / 0.8 + 2g
0
or = + 40 (e – x) / 0.8 – 2g
0 | M1A1 | Use Newton’s law at general point (e.g. x below or above P )
0
(ignore LHS sign here only)
OR: 2 d2y/dt2 = – 40 (y – 0.8) / 0.8 + 2g
= – 50 y + 60 | (M1A1 | Use Newton’s law at general point in terms of y = OP
(ignore LHS sign here only)
Take x = y – 1.2 to give | M1A1) | Change variable to give standard form of SHM eqn
d2x/dt2 = – 25x | A1 | Hence SHM (A0 if wrong sign or LHS unclear)
(B1 only for stating SHM eqn. without proof)
T = 2π / √(25) = 2π/5 AG | A1 | Verify period T using T = 2π/ω with ω = √(25)
OP = 1.2 [m]
0 | B1 | State OP explicitly (may imply first M1 A1)
0
7
11A(ii) | 0.42 = 25 (a2 – 0.062) | M1A1 | Find amplitude a from v2 = ω2 (a2 – x2)
a = √(0.0064 + 0.0036) = 0.1 [m] | A1
3
11A(iii) | 40 e / 0.8 = (2 + M) g, [e = 0.4 + 0.2 M]
1 1
or 40(e – e ) / 0.8 = Mg [e – e = 0.2 M]
1 0 1 0 | M1A1 | Find extension e [or OP ] at first equilibrium position P
1 1 1
or equate additional extension to M by Hooke’s Law
e – e = a , M = a/0.2 = 0.5
1 0 | M1A1 | Find M by relating e , e and a
0 1
4
Question | Answer | Marks | Guidance
11B(i) | H : µ = µ , H : µ > µ (or in terms of µ , µ )
0 x y 1 x y A B | B1 | State hypotheses (B0 forx …)
s 2 = (14.1775 – 10.562/8) / 7 = 0.03404 and
x
s 2 = (15.894 – 12.392/10) / 9 = 0.06031 (to 3 s.f.)
y | M1A1 | Estimate both population variances
(allow biased here: 0.02979 and 0.05428)
s2 = (7 s 2 + 9 s 2) / 16
x y
or (14.1775 – 10.562/8 + 15.894 – 12.392/10) / 16
= 0.78 109 / 16 or 0.04882 or 0.220952 (to 3 s.f.) | M1A1 | Find pooled estimate of common variance
(M1 A1 for s 2 and s 2 may be implied here)
x y
t = 1.337 (to 3 s.f.)
16. 0.9 | *B1 | State or use correct tabular t value
t = (1.32 – 1.239) / s√(1/10 + 1/8) = 0.773 | M1A1 | Calculate value of t (or –t)
(or can comparey –x = 0.081 with 0.140)
t < tabular value
so claim not justified or A’s not heavier than B’s (AEF) | B1 | Correct conclusion (FT on t, dep *B1)
SC: z = (1.32 – 1.239) / √(s 2/8 + s 2/10)
x y
= 0.081 / √(0.078) = 0.799 | (B1 | SC: Implicitly taking s 2, s 2 as unequal popln. variances
x y
(may also earn first B1 M1 A1)
z < 1.282 so claim is not justified (AEF) | B1) | Comparison with z and conclusion (FT on z; max 5/9)
0.9
9
11B(ii) | x = 1.28 and s2 = 0.294 / 7 [= 0.042 (or 0.2052)] | M1 | Find sample mean & estimate popn. var (allow M1 if 0.294 / 8)
t = (1.28 – p) / √(0.042/8) (AEF) | M1A1 | Find value of t
t > 1.415 (< is A0), p < 1.28 – 0.1025, p = 1.18
max | M1A1 | Find p by comparison with tabular value, here t
max 7. 0.9
5