Exam (elaborations)
A-LEVEL EDEXCEL PAPER 31 STATISTICS June 2024 MARK SCHEME (9MA0/31)
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A-LEVEL EDEXCEL PAPER 31 STATISTICS June 2024 MARK SCHEME (9MA0/31)
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Mark Scheme (Results)Summer 2024
Pearson Edexcel GCE
In Mathematics (9MA0)
Paper 31 Statistics
, Scheme Marks AO
(a) X~B(10, ) [Allow 0.167 or better for 1
] M1 3.3
6
(i) [P(X = 3) = ] 0.155045... awrt 0.155 A1 1.1b
[P(X < 3) = P(X 2) =] 0.775226... awrt 0.775 A1 1.1b
(3)
[Let D = no. of days when X = 3] D~B(60, “0.155”) M1 3.3
M1 3.4
P(D 12) = 1 – P(D 11) [Allow 1 – P(D < 12)]
= 1 – 0.78819... awrt 0.212 A1 1.1b
(3)
[n = 600, p = ] estimate = 100 B1 3.4
(1)
(d) 2
5 M1A1 3.3,1.1b
[S = total no. of sixes over 60 days. ] S ≈ 𝑇 ∼ N ("100", √ × 100 )
6
95.5 "100"
P(S > 95) P(T 95.5) o r P Z o r P(Z M1 3.4
0.49..) "9.128..." A1 1.1b
(4)
= 0.688976… awrt 0.689 (11 marks)
Notes
If you see any attempt using an n-sided die with n not equal to 6 please send to review.
(a) M1 for sight or use of the correct distribution. Must have B, or Bin or Bpd or Bcd and the
correct value for n and p, just n = 10, p = is M0
3 7
10 1 5
Implied by one answer correct to 2dp or by sight of or one of :
3 6 6
[P(X = 0) =] 0.16(1…),[P(X = 1) =] 0.32(3…), [P(X = 2) =] 0.29(0…), [P(X 3) =] 0.93(0…)
(i) 1st A1 for awrt 0.155
(ii) 2nd A1 for awrt 0.775
(b) 1 M1 for selecting a correct model. Sight or use of correct binomial, ft their (a)(i)
st
May be implied by sight of [P(D 11) = ] 0.78… or 0.79 or [P(D 12) = ] 0.87…
2 nd
M1 for correct interpretation of “at least 12” and writing or using 1 – P(1D1)
We are not attempting to ft their incorrect 0.155 on our calculators here.
A1 for awrt 0.212 [Answer only 3/3]
(c) B1 for 100 but must be seen in part (c) i.e. between (b) and (d)
(d) 1st M1 for attempting normal with mean = 100 or ft their answer to (c)
May be implied by the correct mean and a correctly labelled s.d. ( σ ) or var (σ 2 )
1st A1 for correctly labelled standard deviation allow 250
= 83.3... = 9.1(28….) or
3
correctly labelled variance. Implied by N( μ, 250
) or correct answer.
3
2 M1 for attempt at continuity correction i.e. sight of 95 + 0.5
nd
2nd A1 for awrt 0.689 [Answer only 4/4]
NB If they don’t state the model for 1st M1 but just give probabilities with
probability statements (Y is any letter):
σ 3250 1st M1 implied by: P(Y > 94.5) = 0.52(63…), P(Y > 95) = 0.52(39…), P(Y > 95.5) = 0.52(15..)
No cc 1st M1 1st A1 implied by: P(T > 95) = 0.70(805…)
1st M1 1st A1 2nd M1 implied by: P(T > 94.5) = 0.72(657…)
Exact binomial gives 0.68567… and will likely score 0/4
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