S2 MS Jan 2006

January 2006

6684 Statistics S2 Mark Scheme

Question Number 1.(a)

Scheme

Marks

Let X be the random variable the number of heads.

X ~ Bin (4, 0.5)

P(X  2)  C24 0.520.52

Use of Binomial including nCr

=0.375

A1

or equivalent

(2)

P(X  4) or P(X  0)

(b)

M1

B1

 2  0.54  0.125

(0.5)4

M1

or equivalent

A1 (3)

(c)

P(HHT)  0.53

no nCr

 0.125

or equivalent

M1 A1 (2)

or P(HHTT) + P (HHTH)  2  0.54  0.125

Total 7 marks 1a) 2,4,6 acceptable as use of binomial.

1

January 2006

6684 Statistics S2 Mark Scheme

Question Number 2.(a)

Scheme

Let X be the random variable the no. of accidents per week X ~Po(1.5)

(b)

Marks

P(X  2) 



need poisson and B1 must be in part (a)

(1)

e  2 or P(X  2) - P(X  1) 2 M1

e1.51.52 2

 0.2510

awrt 0.251

A1 (2)

(c)

P(X  1)  1  P(X  0)  1  e1.5

correct exp awrt 0.777

B1

 0.77693

(p)3

M1

= 0.4689

awrt 0.469

A1

 0.7769 P(at least 1 accident per week for 3 weeks)

(3) (d)

X ~ Po(3)

may be implied P(X  4)  1  P(X  4)  0.1847

B1 M1

awrt 0.1847

A1 (3)

Total 9 marks

c) The 0.7769 may be implied

2

January 2006

6684 Statistics S2 Mark Scheme

3.(a) f ( x) 1 6

-1,5

B1

1 6

B1

B1 -1

(b)

5

x

E(X )  2 by symmetry

(3) B1 (1)

(c)

Var(X ) 

1 (5  1)2 12

or



5

x  x dx  4     4 6  18  1 2

3

M1

=3

A1 (2)

(d)

3.6 P(  0.3  X  3.3)  6

3.3

1 x dx    or   0.3 6  6   0.3 3.3

full correct method for the correct area

M1

A1 (2)

 0.6

Total 8 marks

3

January 2006

6684 Statistics S2 Mark Scheme

Question Number 4.

Scheme

Marks

X = Po (150  0.02) = Po (3) po,3

B1,B1(dep)

P(X  7)  1  P(X  7)

M1

= 0.0119

awrt 0.0119

A1

Use of normal approximation max awards B0 B0 M1 A0 in the use 1- p(x < 7.5)

7.5  3 =2.62 2.94 p  x  7   1  p  x  7.5  z

Total 4 marks

 1  0.9953  0.0047

5.(a)

3

 f  x  1

 kx( x  2)dx  1 2

3

1 3 2  3 kx  kx   1  2

attempt



M1

need either x3 or x2 M1 correct

(9k  9k )  ( k



A1

8k  4k )  1 3

3  0.75 4

*

cso

4

A1 (4)

January 2006

6684 Statistics S2 Mark Scheme

Question Number (b)

Scheme

E(X )  

3

2

Marks

3 2 x ( x  2)dx 4

attempt

 x f  x

3

1  3   x 4  x3  2 2 16

 2.6875  2 F(x)  

(c)

x

2

correct

11  2.69 (3sf) 16

3 2 (t  2t )dt 4



awrt 2.69

M1

A1

A1 (3)

 f ( x) with variable limit or +C

M1

x

3  1     t 3  t 2   2 4  3

correct integral A1 lower limit of 2 or F(2) = 0 or F(3) =1 A1

1  ( x3  3x 2  4) 4

A1

x2

0 1 F(x)  ( x3  3x 2  4) 4 1

2 x3

middle, ends

(6)

x3

1 2 1 3 1 ( x  3x 2  4)  4 2 F(x) 

(d)

B1,B1

M1 their F(x) =1/2

x3  3x 2  2  0 x = 2.75, x3 – 3 x2 + 2 > 0 x = 2.70, x3 – 3 x2 + 2 < 0  root between 2.70 and 2.75

M1 (2)

(or F(2.7)=0.453, F(2.75)=0.527  median between 2.70 and 2.75 Total 15 marks

5

January 2006

6.(a)

6684 Statistics S2 Mark Scheme

X

P(X  x)

1 1 2

2 1 3

5 1 6

1 1 1 Mean  1  2   5   2 2 3 6

 x . p  x  need ½ and 1/3

or 0.02

M1A1

For M M1A1

1 1 1 Variance=  12   22   52   22  2 or 0.0002 2 3 6

(b)

(c)

x 2 . p  x    2 (1,1) (1,2) and (2,1) (1,5) and (5,1) e.e.

(2,2) (2,5) and (5,2) (5,5)

x P(X  x )

1 1 1 1   2 2 4

(4)

B2 LHS

-1

B1 (3)

repeat of “theirs” on RHS

1.5 1 3

2 1 1 1   3 3 9

3 1 6

3.5 1 1 1 2   3 6 9

5 1 36

B1

¼

M1A1

1.5+,-1ee

M1A2 (6)

Total 13 marks

Two tail 6

January 2006

6684 Statistics S2 Mark Scheme

7.(a)(i) H0 : p  0.2, H1 : p  0.2 P(X  9)  1  P(X  8)

p=

or

attempt critical value/region

M1

CR X  9

= 1 – 0.9900 = 0.01

0.01 < 0.025 or 9  9 or 0.99 > 0.975 or 0.02 < 0.05 or lies in interval with correct interval stated.

Evidence that the percentage of pupils that read Deano is not 20%

(ii)

B1B1

X ~ Bin (20, 0.2)

A1 A1

may be implied or seen in (i) or (ii) B1 0,9,between “their 9” and 20 B1B1B1

So 0 or [9,20] make test significant.

(9)

(b)

H0 : p  0.2, H1 : p  0.2

B1

W ~ Bin (100, 0.2) W ~ N ( 20, 16)

P(X  18)  P(Z 

normal; 20 and 16

18.5  20 ) 4

=P(Z  0.375)

= 0.352 – 0.354

or

B1; B1

1 x(  )  20 M1M1A1 2  1.96  cc, standardise 4 or use z value, standardise

CR X < 12.16 or 11.66 for ½

A1

[0.352 >0.025 or 18 > 12.16 therefore insufficient evidence to reject H 0 ] Combined numbers of Deano readers suggests 20% of pupils read Deano

A1 (8)

Conclusion that they are different.

B1

Either large sample size gives better result Or Looks as though they are not all drawn from the same population.

B1

(c)

(2) Total 19 marks

7(a)(i)

One tail H0 : p = 0.2, H1 : p > 0.2

B1B0 7

January 2006

6684 Statistics S2 Mark Scheme

P(X  9)  1  P(X  8)

or

= 1 – 0.9900 = 0.01

attempt critical value/region

M1

CR X  8

A0

0.01 < 0.05 or 9  8 (therefore Reject H 0 , )evidence that the percentage of pupils that read Deano is not 20% X ~ Bin (20, 0.2)

A1

may be implied or seen in (i) or (ii) B1

(ii) 0,9,between “their 8” and 20 B1B0B1

So 0 or [8,20] make test significant.

(9) B1 √

H0 : p = 0.2, H1 : p < 0.2 (b) W ~ Bin (100, 0.2) W ~ N ( 20, 16)

P(X  18)  P(Z 

normal; 20 and 16 18.5  20 ) 4

=P(Z  0.375)

= 0.3520

or

B1; B1

x  20  1.6449  cc, standardise M1M1A1 4 or standardise, use z value

CR X < 13.4 or 12.9

awrt 0.352

A1

[0.352 >0.05 or 18 > 13.4 therefore insufficient evidence to reject H 0 ] Combined numbers of Deano readers suggests 20% of pupils read Deano

A1 (8)

Conclusion that they are different.

B1

Either large sample size gives better result Or Looks as though they are not all drawn from the same population.

B1

(c)

(2)

Total 19 marks

8