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
e1.51.52 2
0.2510
awrt 0.251
A1 (2)
(c)
P(X 1) 1 P(X 0) 1 e1.5
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
x2
0 1 F(x) ( x3 3x 2 4) 4 1
2 x3
middle, ends
(6)
x3
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