S1 Jan 06 Ms

January 2006

6683 Statistics 1 Mark Scheme

Question Number

Scheme

1. (a)

Mode is 56

(b)

Q1  35,Q 2  52,Q3  60

Marks B1 (1)

(c)

x

B1,B1,B1 (3)

1335 4  49.4 or 49 27 9

exact or awrt 49.4

B1 M1A1ft

2

71801  1335       214.5432... 27  27    14.6 or 14.9 2

awrt 14.6(5) or 14.9

A1 (4)

(d) (e)

49.4-56  0.448 14.6 For negative skew; Mean<median<mode (49.4<52<56 not required)

awrt range -0.44 to -0.46

M1A1 (2)

2 compared correctly 3 compared correctly

Q3 -Q 2
M1 A1 M1 A1 ft (4) Total 14 marks

2. (a)

(b)

(c)

(d)

p  q  0.4 2 p  4q  1.3

B1 Consider with (b).

M1A1 (3) M1

Attempt to solve

p  0.15, q  0.25

If both seen, award 3.

E(X 2 )  12  0.10  22  0.15  .....  52  0.30  14 Var( X )  14  3.52  1.75

A1A1 (3) M1A1ft M1A1 (4) M1A1ft

Var(3  2 X )  4Var( X )  7.00

(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)

(3) (b) (c)

B1

Points lie close to a straight line

(1)

106  704 S xy  8354   891.6 10 106 2 S xx  1352   228.4 10 891.6 b  3.903677... 228.4 704 106 a b  29.021015... 10 10

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)

(7) B1 (1) B1B1

(ii) 165.52

(2)

(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 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 (2)

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

(3)

exact or awrt 0.182

M1A

1

Total 7 marks

Any 2

(ii) score on a face after tossing a fair die

B1B1 (2) B1B1 Total 4 marks

3

(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)

(3) (3)

(c)

(d)

M1A1

awrt 0.478

(2) M1A1ft

2 =P( A B )  P(A)=0.54 for M1A1ft) 3

A1ft (3) Total 11 marks

Let H be rv height of athletes, so H
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 A1 A1 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 (3) 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 (3) B1 Total 9 marks

4

(1)