January 2006
6683 Statistics 1 Mark Scheme
Question Number 1. (a) (b) (c)
Scheme
Marks
Mode is 56
B1 B1,B1,B1
Q1 35,Q2 52,Q3 60 x
1335 4 49.4 or 49 27 9
exact or awrt 49.4
(e)
B1
awrt 14.6(5) or 14.9
49.4-56 0.448 14.6 For negative skew; Mean<median<mode (49.4<52<56 not required)
(3)
M1A1ft
2
71801 1335 214.5432... 27 27 14.6 or 14.9 2
(d)
(1)
awrt range -0.44 to -0.46
2 compared correctly 3 compared correctly
Q3 -Q2 <Q2 -Q1 8 and 17 Accept other valid reason eg. 3(mean-median)/sd as alt for M1A1
A1 M1A1
(4) (2)
M1
A1 M1 A1 ft
(4)
Total 14 marks 2. (a)
(b)
(c)
p q 0.4 2 p 4q 1.3
Consider with (b).
p 0.15, q 0.25
If both seen, award 3.
E(X 2 ) 12 0.10 22 0.15 ..... 52 0.30 14
B1 (3)
M1
Attempt to solve
Var( X ) 14 3.52 1.75 (d)
M1A1
A1A1 M1A1ft M1A1 M1A1ft
Var(3 2 X ) 4Var( X ) 7.00
(3)
(4)
(2) Total 12 marks
1
January 2006 3. (a)
6683 Statistics 1 Mark Scheme
Sensible graph scales, labels, shape
B1,B1,B1 3(a) Scatter Diagram
120
100 18, 96
Evaporation Loss (y ml)
15, 90 16, 88 13, 82 12, 79
80 10, 69 8, 61
60 6, 53 5, 50 40 3, 36
20
0 0
2
4
6
8
10
12
14
16
18
20
Time (x weeks)
(b) (c)
106 704 S xy 8354 891.6 10 1062 S xx 1352 228.4 10 891.6 b 3.903677... 228.4 704 106 a b 29.021015... 10 10
(1) B1 B1
awrt 3.9
M1A1
awrt 29
M1A1 A1ft
29.02, 3.90 (d)
For every extra week in storage, another 3.90 ml of chemical evaporates
(e)
(i) 103.12
(f)
(3)
B1
Points lie close to a straight line
(7)
B1 B1B1
(ii) 165.52
(i) Close to range of x , so reasonably reliable (ii) Well outside range of x , could be unreliable since no evidence that model will continue to hold
2
B1,B1 B1
(1) (2)
B1
(4) Total 18 marks
January 2006
6683 Statistics 1 Mark Scheme
8 11
4. (a)
9 12
3 12
Blue
Blue
3 11
Red
9 11
(c)
5. (a)
(b)
P(Second ball is red)=
M1
Blue
9 3 , 12 12
A1
Red
Complete & labels
A1
Red
2 11 (b)
Tree
9 3 3 2 1 12 11 12 11 4
M1A1
3 2 2 P( Both are red Second ball is red)= 12 11 1 11 4 To simplify a real world problem To improve understanding / describe / analyse a real world problem Quicker and cheaper than using real thing To predict possible future outcomes Refine model / change parameters possible (i) e.g.s height, weight
exact or awrt 0.182
M1A
(2) 1
Total 7 marks
Any 2
(ii) score on a face after tossing a fair die
B1B1 B1B1 Total 4 marks
3
(3)
(2)
(2) (2)
January 2006
6683 Statistics 1 Mark Scheme
6. (a)
A
B
0.32
0.22
0.11
0.35 (b)
P(A) 0.32 0.22 0.54;P(B) 0.33
(c)
P( A B)
(d)
For independence P(A B) P(A)P(B) For these data 0.22 0.54 0.33 0.1782
NOT independent 7. (a)
Let H be rv height of athletes, so H
(c)
(d)
A1 A1
(3) (3)
M1A1
awrt 0.478
(2) M1A1ft
2 =P( A B) P(A)=0.54 for M1A1ft) 3
A1ft
(3) Total 11 marks
N(180,5.22 )
188 180 ) P(Z >1.54) 0.0618 ± stand. √, sq, awrt 0.062 5.2 Let W be rv weight of athletes, so W N(85,7.12 ) standardise, awrt 0.9545 P(W<97) P(Z 1.69) 0.9545 P(H >188) P(Z
(b)
M1
M1A1ft;A1ft
P(A B) 32 P(B) 67
(OR P( A B) P(A) for M1A1ft OR
Venn Diagram 0.32,0.11 & A,B 0.22,0.35 & box
P(H >188 & W 97) 0.0618(1 0.9545) 0.00281
M1A1A1 M1A1
(2)
allow (a)x(b) for M awrt 0.0028
Evidence suggests height and weight are positively correlated / linked Assumption of independence is not sensible
M1A1ft A1 B1 Total 9 marks
4
(3)
(3) (1)