Question 10:
--- 10(i) ---
10(i) | F(x) = ∫ f(x) dx = (1/30) (– 8/x + x3 – 14x) [+ c] | M1 | Find or state distribution function F(x) for 2 ⩽ x ⩽ 4
F(x) = (1/30) (– 8/x + x3 – 14x + 24) | M1 | Using F(2) = 0 or F(4) = 1 to find c if necessary. AEF
F(x) = 0 (x < or ⩽ 2), F(x) = 1 ( x > or ⩾ 4) | A1 | State F(x) for other values of x
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--- 10(ii) ---
10(ii) | G(y) = P(Y < y) = P(X2 < y)
G(y) = P(X < √y) = F(√y)
( )
G(y) = ( 1/30 ) −8/y 1 2 + y2 3 −14y 1 2 +24 | M1
A1 | Find or state G(y) for 2 ⩽ x ⩽ 4 from Y = X2
(allow < or ⩽ throughout)
Alternative method for question 10(ii)
Use x= y 1 2 to find
f ( x )=( 1/30 )( 8/y+3y−14 ) , dx =− 1 y−1 2
dy 2 | (M1
A1) | dx ( )=f ( )× dy
Find f(x) and for use in g y x
dy dx
( )
g ( y ) [=G′ ( y ) ]=( 1/30 ) 4/y 3 2 +( 3/2 ) y 1 2 −7/y 1 2
for 4 ⩽ y ⩽ 16 [g(y) = 0 otherwise] | A1
A1 | Find g(y). AEF
State corresponding range of y for G(y) or g(y)
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--- 10(iii) ---
10(iii) | ( )
( 1/30 ) −8/y 1 2 + y2 3 −14y 1 2 +24 =0.8 | M1 | Set G(y) = 0⋅8
– 8 + y2 – 14 y = 0, y = 7 + √57 or 14⋅5[5]
[rejecting 7 – √57; allow 14⋅6] | M1 A1 | Rearrange to give quadratic in y and solve to find value of y
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Question | Answer | Marks | Guidance
11A(i) | 10 (AP – 0⋅6) / 0⋅6 = 20 (1⋅2 – AP – 0⋅4) / 0⋅4 | M1 A1 | Verify AP by equating equilibrium tensions. AEF
4 AP – 2⋅4 = 9⋅6 – 12 AP
AP = 0⋅75 [m] | A1 | AG
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11A(ii) | m d2x/dt2 = – 10 (0⋅15 + x) / 0⋅6 + 20 (0⋅05 – x) / 0⋅4
or m d2x/dt2 = + 10 (0⋅15 – x) / 0⋅6 – 20 (0⋅05 + x) / 0⋅4 | M1
A1 A1 | Apply Newton’s law at 0⋅75 + x or 0⋅75 – x from A
(M1 requires LHS and 2 tensions: A1 for each correct tension)
⅔ d2x/dt2 = – (80 / 1⋅2) x , d2x/dt2 = – 100 x | M1 A1 | Simplify to give SHM eqn. in standard form
T = 2π/ω = 2π/10 = π/5 or 0⋅628 [s] | DB1 | State the period T with FT on ω from SHM eqn.
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11A(iii) | (a = 0⋅75 – 0⋅7 = 0⋅05)
v = ω × a
max | M1 | Find speed at equilibrium position from ωa
v = 0⋅5 [m s–1]
max | A1
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11A(iv) | x = a/2 = 0⋅025 | M1 | Find value of x giving half max. acceln.
v = ω √ (a2 – x2) = 10 √ (0⋅052 – 0⋅0252) | M1
v = 0⋅433 [m s-1] | A1 | Find corresponding speed
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Question | Answer | Marks | Guidance
11B(i) | x = (1/40) Σ x f(x) = 68/40 = 1⋅7 | B1 | Find mean of sample
(1/40) Σ x2 f(x) = 178/40 = 4⋅45, Var = 4⋅45 – 1⋅72 = 1⋅56 | B1 | Find variance of sample
Mean and variance are similar so Poisson may be suitable | B1 | State valid comment
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11B(ii) | a = 40 × 1⋅65 e–1⋅6 /5! = 40 × 0⋅01764 | M1 | AG
a = 0⋅706 | A1 | Verify a from Poisson term
b = 40 – 39⋅758 = 0⋅242 | B1* | Find b
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11B(iii) | H : Distribution fits/models data
0 | B1 | State (at least) null hypothesis in full
O: 6 15 9 10
i
E: 8⋅076 12⋅921 10⋅337 8⋅666
i | DM1
A1 | Combine values consistent with all exp. values ⩾ 5
(FT on b, dep B1*)
X2 = 0⋅5337 + 0⋅3345 + 0⋅1729 + 0⋅2053 = 1⋅25 | M1 | Find value of X 2 from Σ (E – O)2 / E [or Σ O2/E – n ]
i i i i i
X2 = 1⋅25 | A1
No. n of cells: 7 6 5 4 3
χ 2: 10⋅64 9⋅236 7.779 6⋅251 4⋅605
n-1, 0.9 | DB1 | State or use consistent tabular value χ 2 (to 3 s.f.)
n–1, 0.9
[FT on number, n, of cells used to find X2]
Accept H if X 2 < tabular value (using their values)
0 | M1 | AEF
1⋅25 [± 0⋅1] < 6⋅25 so distn. fits [data]
or distn. is a suitable model | A1 | Conclusion (requires both values approx. correct). AEF
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PMT
9231/21 Cambridge International A Level – Mark Scheme October/November 2019
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