S1 MS Jun 2005

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