Mock Paper Mark Scheme
Advanced Subsidiary/Advanced GCE
General Certificate of Education
Subject STATISTICS
Paper No. Mock S2
Question
number
1. (a)
Scheme
Marks
x
42
f(x)
1
7
1
7
Labelled axes and < 0, > 10
B1
B1
B1
x
6
(b)
1
7
10
P(X  5) = 1 – P(X < 5)
=1–
5
42
=1–
25
84
 12  5
=
, 6, 10
B1
5 and  or area ∆
M1
full method
M1
(area of ∆)
A1
59
84
(c)
Probability it does not break down is
8459 2 =
(3)
M1
8459 2
 probability it does break down is 1 –
(4)
(awrt) 0.507
A1
(2)
B1
(1)
(9)
f(x)
2.
1
10
5.0 4.2 1.5
(a)
(b)
Question
number
P(X < 4.2) =
1.5
5.0
0 .8
= 0.08
10
P(X < 1.5) =
3
= 0.3
10
x
M1 A1 (2)
Scheme
UA009018 – AS/Advanced GCE Mathematics: Mock Papers with Mark Schemes (Part 1)
Marks
Statistics S2 Mark Scheme
(c)
Y = no. of lengths with X < 1.5
 Y ~ B(10, 0.3)
M1
P(Y > 5) = 1 – P(Y  5)
= 1 – 0.9527
(d)
M1
A1
= 0.0473
R = no. of lengths of piping rejected
R ~ B(60, 0.08)  R  ~Po(4.8)
4.8 or 60  (a)

4.82 
P(R  2) = e4.8 1  4.8 

2! 

= 17.32  e4.8 = 0.1425…
3. (a)
(b)
(c)
B1
Po and  2, formula M1, M1 A1
(ft for their  if full
expression seen)
(accept awrt 0.143)
A1 cao (5)
D is continuous
B1
(1)
Sampling Frame is the list of competitors or their results,
e.g. label the results 1—200 and randomly select 36 of them
B1
B1
(2)
X = no. of competitors with A = 2
X ~ B(36,
1
3
M1 A1

19.5  12 

P(X  20)  P  Z 
8 

 12 , ‘z’ M1, M1
A1
= P(Z  2.65…)
= 1 – 0.9960
(11)
)
X  ~N(12, 8)
(d)
(3)
A1
= 0.004
Probability is very low, so assumption of P(A = 2) =
(Suggests P(A = 2) might be higher.)
UA009018 – AS/Advanced GCE Mathematics: Mock Papers with Mark Schemes (Part 1)
1
3
is unlikely.
(6)
B1, B1 (2)
(11)
Statistics S2 Mark Scheme
Question
number
4. (a)
Scheme
X = no. of vases with defects
X ~ B(20, 0.15)
P (X  0) = 0.0388
P(X  6) = 0.9781
Marks
 P(X  7) = 0.0219
B1
Use of tables to
M1
find each tail
M1
 critical region is X  0, or X  7
A1, A1 (5)
(b)
Significance level = P(X  0) + P(X  7) = 0.0388 + 0.0219 = 0.0607
B1
(c)
H0:  = 2.5, H1:  > 2.5
[or H0:  = 10, H1:  > 10]
B1, B1
Y = no. sold in 4 weeks.
Under H0 Y ~ Po(10)
M1
P(Y  15) = 1 – P(Y  14) = , 1 – 0.9165 = 0.0835
M1, A1
More than 5% so not significant. Insufficient evidence of an increase in the
rate of sales.
A1
UA009018 – AS/Advanced GCE Mathematics: Mock Papers with Mark Schemes (Part 1)
(1)
(6)
(12)
Statistics S2 Mark Scheme
Question
number
5. (a)
Scheme
Marks
F(1.5) = 1  k(2  (1.5)3 – (1.5)4) = 1
M1
81
=1
i.e. k 2  278  16
81
 =1
i.e. k 108
16
(b)
(c)
P(T > 1) = 1 – F(1) ,
f(t) = F(t) = ,
16
27
= 1–
E(T) =

1.5
0
(2 – 1) =
16
27
16
27
(*)
11
27
t f (t ) dt
=
0  t  1.5
(2)
M1, A1
(2)
Full definition B1
otherwise
32
27
A1 cso
M1, A1
(6t 2 – 4t 3)
 32
(3t 2  2t 3 )
27
i.e. f(t) = 
0
(d)
k=

1.5
0
(3t 3  2t 4 ) dt

t f (t )
(3)
M1
3
 3t 4 2t 5  2

= 32

27 
5 0
 4
(
=
32
27
=
9
2

243
64
 185
A1
  (0)
 52  243
32
= 0.9
A1 cso
(*)
(3)
(e)
F(E(T)) =
16
27
(2  0.93 – 0.94) = 0.4752
(f )
P(T > 1T > 0.9) =
evidence seen
P(T  1)
part (b)
, =
, = 0.7763…
P (T  0.9)
1  part (e)
accept awrt 0.776
B1
M1, M1,
A1
(3)
(14)
UA009018 – AS/Advanced GCE Mathematics: Mock Papers with Mark Schemes (Part 1)
Statistics S2 Mark Scheme
Question
number
6. (a)
Scheme
X = no. of customers arriving in 10 minute period
P(X  4) = 1 – P(X  3) = , 1 – 0.6472 = 0.3528
X ~ Po(3)
(b)
(c)
Marks
M1 A1
Y = no. of customers in 30 minute period Y ~ Po(9)
B1
P(Y  7) = 0.3239
M1 A1
p = probability of no customers in 5 minute period = e1.5
B1
(2)
(3)
C = number of 5 minute periods with no customers
C ~ B(6, p)
M1
P(C  1), = (1 – p)6 + 6(1 – p)5p
M1, M1 A1
= 0.59866…
(d)
A1
(accept awrt 0.599)
(6)
W = no. of customers on Wednesday morning
3 12 hours = 210 minutes
 W ~ Po(63)
‘63’
M1 A1
Normal approximation W  ~N(63, (63)2)
P(W > 49)  P(W  49.5)

49.5  63 

= P  Z 
63 

 12
standardising
= P(Z  1.7008)
= 0.9554 (tables)
UA009018 – AS/Advanced GCE Mathematics: Mock Papers with Mark Schemes (Part 1)
B1
M1
M1
A1
(accept awrt 0.955 or 0.956)
A1
(7)
Statistics S2 Mark Scheme
6
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