GCE Edexcel GCE Statistics S1 (6683)
Summer 2005
Statistics S1 (6683)
Edexcel GCE
Mark Scheme (Results)
June 2005 6683 Statistics S1 Mark Scheme Question Number
1.
2. (a) (b) (c)
Scheme
Marks
Diagram A : y & x : r = - 0.79; As x increases, y decreases or most points lie in the 2nd and 4th quadrant.
B1;B1dep
Diagram B : v & u: r = 0.08; No real pattern. Several values of v for one value of u or points lie in all four quadrants, randomly scattered.
B1;B1dep
Diagram C : t & s : r = 0.68; As s increases, t increases or most points lie in the 1st and 3rd quadrants
B1;B1dep
Distance is a continuous.
B1
continuous
F.D = freq/class width ⇒ 0.8, 3.8, 5.3, 3.7, 0.75, 0.1 Q2
(67 − 23) × 10 = 50.5 + 53
or the same multiple of
(6)
(1) M1 A1 (2)
= 58.8
Q1 = 52.48; Q3 = 67.12
awrt 58.8/58.9
M1 A1
awrt 52.5/52.6 67.1/67.3
A1 A1 (4)
Special case : no working B1 B1 B1 ( ≡ A’s on the epen) (d)
x= s =
8379.5 = 62.5335... 134 557489.75 ⎛ 8379.5 ⎞ −⎜ ⎟ 134 ⎝ 134 ⎠
awrt 62.5 2
B1 M1 A1√
s = 15.8089…. ( Sn - 1 = 15.86825...)
awrt 15.8 (15.9)
A1 (4)
Special case : answer only B1 B1 ( ≡ A’s on the epen)
(e)
Q3 − 2Q2 + Q1 67.12 − 2 × 58.8 + 52.48 = 67.12 − 52.48 Q3 − Q1 = 0.1366 ⇒ ; +ve skew
subst their Q1,Q2 &Q3 need to show working for A1√ and have reasonable values for quartiles awrt 0.14
M1 A1√ A1; B1
(f)
For +ve skew Mean > Median & 62.53 > 58.80 or Q3 – Q2 (8.32) > Q2 – Q1(6.32) Therefore +ve skew
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
(4) B1 (1)
Question Number 3. (a)
Scheme
Sxy = 8880 −
Marks
130 × 48 = (8100) 8
may be implied
B1
Sxx = 20487.5
b=
s xy s xx
=
8100 = 0.395363... 20487.5
allow use of their Sxy for M
M1 A1
awrt 0.395
a=
48 130 − (0.395363...) = −0.424649... 8 8
allow use of their b for M
M1 A1
awrt -0.425
y = −0.425 + 0.395 x
3s.f.
B1 √ (6)
Special case answer only B0 M0 B1 M0 B1 B1(fully correct 3sf) ( ≡ to B0 M0 A1 M0 A1 B1 on the epen)
(b)
f − 100 = −0.424649... + 0.395...(m − 250) f = 0.735 + 0.395m
(c)
m = 235 ⇒ f = 93.64489....
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
subst f - 100 & m - 250 3 s.f. awrt 93.6/93.7
M1 A1√ A1 (3) B1 (1)
4(a)
1.5 (Q3 - Q1) = 1.5 (28 - 12) = 24 Q3 + 24 = 52 ⇒ 63 is an outlier
may be implied att Q3 +… or Q1 - …, 52 and -12or <0 or evidence of no lower outliers
Q1 - 24 < 0 ⇒ no outliers
63 is an outlier
B1 M1, A1 A1
M1 A1 A1
(7)
(b)
Distribution is +ve skew; Q2 - Q1 (5) < Q3 - Q2 (11);
B1; B1 (2)
(c)
Many delays are small so passengers should find these acceptable or sensible comment in the context of the question.
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
B1 (1)
5.(a)
k + 2k + 3k + 5k + 6k = 1
use of ∑ P ( X = x ) = 1
M1
17 k = 1
k=
(b)
(c)
1 = 0.0588 17
1 2 6 64 + 2 × + .... + 5 × = 17 17 17 17 13 =3 17
E(X) = 1 ×
E(X2) = 12 ×
A1 (2)
use of ∑ xP ( X = x ) and at least 2 prob correct
M1
Do not ignore subsequent working
A1
1 2 6 ⎛ 266 ⎞ + 2 2 × + .... + 5 2 × = ⎜ = 15.6 ⎟ 17 17 17 ⎝ 17 ⎠
2 use of ∑ x P ( X = x )
M1 A1
and at least 2 prob correct
266 ⎛ 64 ⎞ Var (X) = −⎜ ⎟ 17 ⎝ 17 ⎠
2
2 use of ∑ x P ( X = x ) -
A1
(E(X))2
= 1.4740… (d)
Var ( 4 - 3X) = 9 Var (X) = 9 × 1.47 = 13.23 ⇒ 13.2 or 9 × 1.4740…= 13.266 ⇒ 13.3
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
M1 (4)
awrt 1.47
cao
9 Var X
M1 A1 (2)
6(a)
M ~ N( 155, 3.52)
standardising
160 − 155 ⎞ ⎛ P(M > 160) = P⎜ z > ⎟ 3.5 ⎠ ⎝ = P( z > 1.43)
± (160 − 155 ), σ, σ2,√ σ
A1 A1
= 0.0764
(b)
M1
(3)
P(150 ≤ M ≤ 157) = P(−1.43 ≤ z ≤ 0.57) = 0.7157 - ( 1 - 0.9236) = 0.6393
awrt -1.43, 0.57
p>0.5 0.6393 - 0.6400 4dp
B1 B1 M1 A1 (4)
special case : answer only B0 B0 M1 A1 (c)
P( M ≤ m ) = 0.3 ⇒
-0.5244 att stand = z value
m − 155 = −0.5244 3.5
for A1 may use awrt to 0.52.
m = 153.2
cao
B1 M1 A1 A1 (4)
Glasses
No Glasses
Totals
Science Arts Humanities
18 27 44
12 23 24
30 50 68
Totals
89
59
148
7.
(a)
(b)
(c)
P(Arts) =
50 may be seen in (a) 23 may be seen in (b)
50 25 = = 0.338 148 74
23 23 P(No glasses / Arts) = 148 = = 0.46 50 50 148 P(Right Handed) = (
a number/148
B1 B1
M1 A1 (4)
prob number or their (a )prob their 50
30 50 68 × 0.8) + ( × 0.7) + ( × 0.75) 148 148 148
attempt add three prob
M1 A1 (2) M1 A1 √
A1 √ on their (a)
=
(d)
55 = 0.743 74
30 × 0.8 12 148 = P ( Science /Right handed) = = 0.218 (c) 55
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
awrt 0.743
√ on their (c)
A1 (3)
M1 A1√ A1 (3)
6683 Statistics S1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics