Cambridge International
AS & A Level Further Mathematics
Further Probability & Statistics
STUDENT’S BOOK: Worked solutions
Yimeng Gu, Dr Patrick Wallace
Series Editor: Dr Adam Boddison
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1
Worked solutions
Worked solutions
1 Continuous random variables
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
1
y − 5 2
x=
 2 
6
∫1 k(x − 2)dx = 1
a
 2

k   x − 2x   = 1
2
 1

((
) ( ))
Exercise 1.1A
1
b P (X = 2.5) = 0
a P(X < 2) =
0
1
b P(−0.5 < X < 3) =
x 
41
∫1 7 x dx =  14 1 =
3
u = x2 − 4
(X 0.8) = 1 − ∫
∫2
x3
x2 − 4
dx =
2
−4=0
1 5 − 12
u (u + 4) du
2 ∫0
32
a
1
1
3
3
−4=5
1
(
41
dx = 1 − 1 = 14
15 15
1
∫1 6 dx = 2
)
2
⌠
 5 m 2 + a dm +
⌡−1
2
∫1 a dm = 1
 2 3
1
2
 15 m + am  + [am ]1 = 1
−1
a=
11
45
f(m)
b
5
1   2 u 23 + 8u 21  
= 
2 3

−0.8 1
−1
2
22
31
P(X > −0.8 ) = 1 − P(X  −0.8)
(1614 ) − (141 ) = 1141
du = 2x, x 2 = u − 4
dx
3
c
d P(X > 1) =
x
u
1
0
∫−0.5 3 dx + ∫0 6 dx = 6 + 2
=2
3
6
2  x3
44
2
2
∫1 x 15 (x − 2)dx = 15   3 − x 1  = 9
6
2
2
= 1 x  + 1 x  = 1 + 1 = 2
 3  −1  6 0 3 3 3
= 2 (4) = 8
15
15
2 4
21
∫−1 3 dx + ∫0 6 dx
0
5
2 (x − 2) dx = 2   x 2 − 2x  
∫3 15
 
15   2
3 
5
d
1
2
Range: f –1(x)  0
k 36 − 2(6) − 1 − 2 = 1
2
2
2
k=
15
c
( )
Therefore f −1( x ) = x − 5
2
Domain: x  5
6
29
45
 
0 
3
1

= 1  2 × 52 + 8 × 52 
23

= 5 5+4 5
3
17
=
5
3
4
y = 2x2 + 5
x2 =
y−5
2
11
45
–1
0
1
2
m
1
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 CONTINUOUS RANDOM VARIABLES
1
c
3
2
1
a
)
c
0
k cos x
−π
3
∫
P(−0.2 y 1.2)
=∫
0
=∫
0
1 × 8 (2 − y ) dy + 1 1 × 8 dy
∫0 2 15
15
1 .2 1
× 8 y dy
+∫
15
1 2
−0.2 4
1
dx + ∫ (1 − x ) dx = 1
0
1
[k sin x ]0− π3 +  x − 12 x 2 0 = 1
= 7 + 4 + 22 = 143
125 15 375 375
( )


 

 0 −  − 3 k   +  1 − 1 − 0 = 1
2


 2   
5
a
3
1
k + =1
2
2
25
∫0
1 a dt + ∞ at − 23dt = 1
∫25
50
∞
25
1
 1 at  +  −2at − 2  = 1


 50 0

 25
10
a=
9
3
1
k=
2
2
3
3
k=
1 m(2 − y ) dy + 1 1 m dy + 1.2 1 my dy
∫0 2
∫1 2
−0.2 4
11 dm 1
=
45
2
+∫
3
(
⌠ 2 2 11
P(0 < M  3 ) =  5 m + 45 dm +
⌡0
2
b
b
f(t)
1
45
f(x)
1
√3
3
0
–π
3
c
.
0.008
( )
⌠
P (x < π ) = 
3
⌡
1
x
π
P x= 4 =0
i
ii
0
−π
3
(
3 cosx dx + 1(1 − x ) dx = 1
∫0
3
)
π
0
iii P − π x π = ∫ π 3 cosx dx + ∫ 6 (1 − x )dx
6
6
3
0
−
6
= 0.675 (3 s.f.)
4
a
f(y)
)
a
∫0 3 t dt + ∫1
k −1 1
11
3
2
2
2
2
2
2
1 1
2 1
1
1
1
1
1
+ k − + k2 − k2 + k + k2 − k + = 1
6 3
3 6
3
3
6
3
6
0
1
2
y
−
1 1
1
+ k + =1
2 3
6
1
1 1
k =1− +
3
6 2
1
0 
1 
2
my − 1 my 2  +  1 my  +  1 my 2  = 1
8
 2
 −1  2
0  4
1
8
m = 15
k=4
1
1 (k − t ) dt = 1
k −1 3
 1 2  1  1  k −1  1
k
t
+ t
+  kt − 1 t 2 
=1
6  k −1
 6  0  3  1
3
∫−1 4 m(2 − y )dy + ∫0 2 m dy + ∫1 2 my dy = 1
0
k
dt + ∫
( 16 − 0) + ( 13 (k − 1) − 13 ) +  ( 13 k − 16 k ) − ( 13 k (k − 1) − 16 (k − 1)
( 16 − 0) + ( 13 (k − 1) − 13 ) +  ( 13 k − 16 k ) − ( 13 k (k − 1) − 16 (k − 1) ) = 1
1
1
2
1
1
1
1
1
1
+ k − ) +  k − ( k − k − k + k − ) = 1
6 (3
3 6
3
3
6
3
6 
1m
2
b
(
3
⌠ 1 × 10 dt + 30 10 t − 2dt = 0.150 (3 s.f.)

∫
25 9
⌡20 50 9
2
3m
4
–1
6
25
c
2
m
t
25
11
21
2
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions
b P(T > 2) = 1 − P (T  2)
=1−
3
2
t dt + ∫ 1 dt
1 3
(
1
11
0
=1− 1 + 1
6 3
1
=
2
c
(∫
1
2. 5 1
∫0.5 3 t dt + ∫1
3
Or
)
P(T  3)
)
= 1 – P(T > 3)
a
dt = 5
8
1
6
1
8
7
11
(
) (
)
75 10
27 6 
=1−
+
−
+
97 97 
 97 97
4
2
5
f(t)
0
5
∫3 97 (3t + 1)dt
3
2 
= 1 −  t2 +
t
97 3
 97
7
=1−
=1−
52
97
45
=
97
Alternative method, using the graph to
work out the area:
t
(
)
2 1
14 20
28 17
Using the graph from part a
P (T 3) = 14 ×
+ ×1×
+
=
+
=
97
2
97
97
97
97
4 1
1
2
P ( 2 T < 7 ) = ∫
t dt + ( 7 − 4 )
6 P (T 3) = 14 × 2 + 1 × 1 × 14 + 20 = 28 + 17 = 45
2 128
97 2 97 97
97 97 97
7
1
=
+ = 31
48 2 48
5 2
69
ii P ( 2 T 5 ) = ∫
(3t + 1) dt =
97
2 97
ii P (T > 7 ) = 1 × 1 × (11 − 7 ) = 1
2 6
3
Alternative method, using the graph:
b i
(
c
P (T < 2 ) =
2
1
∫0 128 t
2
P (2 T 5) =
dt = 1
48
Alternative method, using the previous answers:
9
P(T < 2) = 1 – P(2  T < 7) – P(T > 7)
31 1 1
= 1−
− =
48 3 48
(q) p – 4q
10
∫5
0
2
t
5
b Using the graph,
1
2 × 7k + ( 5 − 2 )( 7k + 16k ) = 1
2
14k +
k=
P (T 3 ) = 14 ×
2
4
10
x
From the graph, (10 – 5) × q = 0.5 q = 0.1
or q can be found using integration:
7k
i
0
b P(X > 5) = 0.5
f(t)
16k
c
)
p – 2q
1
× 100% = 2.08% appointments
48
were delayed by less than 2 minutes.
a
(
1
14 32
3
46 69
× 3×
+
= ×
=
2
97 97
2
97 97
f(x)
a
Therefore,
8
)
[qx ]105 = 0.5
10q – 5q = 0.5
q = 0.1
Total area under PDF = 1
69
k =1
2
4
∫2 p − 0.1x dx + 6 × 0.1 = 1
2
97
P (T 3 ) = 14 ×
q dx = 0.5
4
∫2 p − 0.1x dx = 0.4
3 2
2
28 17 45
+
3t + 1) dt =
+
=
97 ∫2 97 (
97 97 97
3 2
2
28 17 45
+∫
+
=
( 3t + 1) dt = 97
97 97 97
2 97
4
px − 0.05x 2  = 0.4
2
(4p – 0.8) – (2p – 0.2) = 0.4
2p = 0.4 + 0.8 – 0.2 = 1
p = 0.5
3
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
10
a
0. 5
∫0
1
1. 5
0. 5
1
kx dx + ∫ 1 dx + ∫ 3 − kx dx = 1
0.5
a For x ∈ ( 0, 1) , F ( x ) =
2
1.5
 kx 2 

1
kx 2 
 2  + [ x ]0.5 +  3x − 2  = 1

0

1
x
∫0 x
3
4
dx = x
4
x
For x ∈ (1, 1.5 ) , F ( x ) = F (1) + ∫ 1 dx = 1 + [ x ]1x
4
1
3
=x−
4
For x ∈ (1.5, 2.5 ) ,
( k8 − 0) + (1 − 0.5) +  ( 29 − 98k ) − (3 − k2 ) = 1
x
F ( x ) = F (1.5 ) + ∫ − ( x − 2.5 ) dx
4k
=1
8
3
1. 5
x
k=2
b P ( X < 1.2 ) =
0. 5
∫0
2x dx +
1. 2
1
∫0.51 dx + ∫1
=
3 − 2x dx
4
( x − 2.5)4
3  ( x − 2.5 ) 
+ −
 =1−
4 
4
4

1.5
Therefore
F( x ) = 
0
Alternatively:
x4
1. 5
4
P ( X < 1.2 ) = 1 − P ( X > 1.2 ) = 1 − ∫ 3 − 2x dx = 1 − 9 = 0.91

100
1.2
2
x − 3
4
1. 5

9
P ( X < 1.2 ) = 1 − P ( X > 1.2 ) = 1 − ∫ 3 − 2x dx = 1 −
= 0.91

100
1.22
( x − 2.5 ) 4
1 −
4

1
= 1 + 0.5 + ( (3.6 − 1.44) − (3 − 1) ) = 0.91
4
Exercise 1.2A
1
a For x ∈ (1, 6 ) , F ( x ) =
x
x
1
x

0

1
x

−
10 10

Therefore F ( x ) = 
2
 4x − x − 7
25
200
25


1
b For x ∈( 0, 2 ) , M( x ) =
6 x < 16,
3
x
For x ∈ ( 2, 8 ) , M ( x ) = M ( 2 ) +
2 x
P(0.5 < X < 2) = F(2) – F(0.5) = 0.969
a
∫−1 5 (x + 1)dx = 1
∫2


= 1 +  1 x − x  = −x + 1 x − 1
4 3
48 
48 3
3
2
2
0
 2
x
 16
Therefore M ( x ) =  2
 −x + 1 x − 1
3
 48 3
1
x <0
0x <2
2x 8
x > 8.
k
1
1 (k + 1)2 − 0 = 1
10
k = 10 − 1, reject k = − 10 − 1

0

2
b F(x) =  ( x + 1)
 10
1

2
1 (8 − x ) dx
24
)
k
 1
2
 10 (x + 1)  = 1
−1
∫0 8 x dx = 16
x
x 2.5.
x 16.
x1
1.5 x < 2.5
c
x < 1,
1 x < 6,
1 x 1.5
(
1
2
2


= 1 +  4x − x  = 4x − x − 7
2  25 200 
25 200 25
6
0x <1
b P(X > 1.2) = 1 – P(X  1.2) = 1 – F(1.2)
= 1 − 1.2 − 3 = 0.55
4
∫1 10 dx = 10 − 10
x
For x ∈( 6, 16 ) , F ( x ) = F ( 6 ) + ∫ 4 − 1 x dx
6 25 100
x <0
4
a
2
∫0 a(3 − z)
2
x < −1
−1 x 10 − 1
x > 10 − 1.
dz + ∫
6
2
1
(3z + 2) d z = 1
80
2
(
)
6
9az − 3az 2 + a z 3  +  1 3 z 2 + 2 z  = 1

3 0  80 2
2
(18a − 12a + 83 a ) + 107 = 1
26
3
a=
3
10
a= 9
260
4
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions
b For 0 z < 2 ,
z
∫−∞ a (3 − z)
2
z
dz = F(0) + ∫ a(3 − z) dz
72a =
a=
2
0
z
3
1
= 0 + − a (3 − z ) 
 3
0
=
81 3 ( 3 − z )
−
260
260
s
s 1
For 2 s < 8, F (2) + ∫
(s − 8)2 ds = 12 +
2 144
z 1
1
∫−∞ 80 (3 z + 2) dz = F(2) + ∫2 80 (3z + 2) dz
=
Therefore,
0

3
 81 − 3 ( 3 − z )
260
F(x) =  260
2
3
2)
z
+
(
1 +
6
480
1

5
1
144
2
s
b For 0 s < 2, F(0) + ∫ 1 s ds = 0 +  1 s 2  = s
0 4
 8 0 8
3
For 2 z < 6,
1
2
z
3  1
1 ( 3z + 2 )
+
(3 z + 2)2  = +
480
10  480
2 6
2
z
2 z<6
c
z 6.
a
P ( X > 8) = 1 − P ( X 8) = 1 −
1
2
41
8 − 5) =
50 (
50
8
 1 ( x − 5 ) 5 x 10
 25

f (x ) =  1
10 x 12
4
0
otherwise.

0s < 2
2s < 8
s 8.
P(1.5 < s 2.5) = F (2.5) − F (1.5)
b
P ( X > 1.5 ) = 1 − F (1.5 ) = 1 −
c
F (2 ) =
1
, therefore y > 2
3
F( y ) =
2
3
1
1 2
y− =
3
3 3
9
s <0
= 1153 or 0.334 (3 s.f.)
3456
1
1
a F(a) = 1, therefore a − = 1
3
3
Therefore the graph of f(x) is:
f(x)
3
s

0
s2

F(s) =  8
 ( s − 8 )3
 432 + 1

1
0 z < 2
c
(s − 8 ) + 1
1  1
+
(s − 8)3  =
2  432
432
2
Therefore,
z<0
83
P(1 < Z 3) = F(3) − F(1) = 416
b
=
a
3
1
13
2
1.5 =
12 ( ) 16
y=3
k
ln 2 1 −3t
∫0
a=4
e dt + ∫
1 dt = 1
ln 2 24
k
1
4
1
5
0
6
5


a f(t) =  cos t
0

(
x
10 12
t
) () ()
b P π <t < π = F π −F π = 2 − 1 =
6
4
4
6
2 2
7
a
21
= 0.207 (3 s.f.)
8
∫0 4 s ds + ∫2 a(s − 8)
2
2
ds = 1
k ln 2 65
−
=
24 24
72
65
k = 3 + ln 2
t
b For 0 t < ln 2, F ( 0 ) + ∫ 1 e −3t dt = 0 +  − 1 e −3t 
 9
0
03
0 t π ,
2
otherwise.
7 1 
+
t
=1
72  24  ln 2
= 1 (1 − e −3t )
9
For ln 2 t < ln 2 + 65 ,
3
t
F(ln 2) + ∫
t
1 dt = 7 +  t  = 7 − 3 ln 2 + t
72  24  ln 2
72
24
ln 2 24
8
 1 s 2  +  1 a(s − 8)3  = 1
 8 0  3
2
5
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
Therefore,

0
1
−3t
9 (1 − e )
F(t) = 
7 − 3 ln 2 + t
24
 72
1

ln 2 1 −3t
e dt
0. 5 3
∫
c
2
t <0
ln 2 t < ln 2 + 65
3
65
t ln 2 + .
3
+ ∫ 1 dt
ln 2 24
X
0
10
F(1) − F (0.5)
or
= 0.0237 (3 s.f.)
10
(
b
(
)
)
M
0
5
For m ∈( 0, 5 ) , FM (m ) = P
(
)
(
(
31
)
)
)
(
3
Y =X3
 0


F(x) =  1 x − 3
4
2
1
33

1
3
60 × P (t > 3 ) = 60 ×⌠
 − 5 t − 3 3 dt = 60 × 0.03 = 2 members  1
⌡3
3 13
Change limits:
1
3
×⌠
 − 5 t − 3 3 dt = 60 × 0.03 = 2 members
⌡3
X
(
(
)
x
∫0 0.08x dx = 0.04x
2
X
0
5
0x 5
x > 5.
4

For y ∈( 0,25 ) , FY ( y ) = P X 2 y = FX 

2
 21 
 21 
2
X y = FX  y  = 0.04 ×  y  = 0.04 y
 
 

y <0
 0
F ( y ) =  0.04y
0 y 25

y > 25 .
1
)
1

y2
)
6

Hence, f ( y ) =  0.04
 0
0 y 25
otherwise.
x > 10.
Y = X3
0
0
6
216
10
1000

2
1 −
f(y) = 12 y 3
0

Y
0
25
(
6 x 10
1
1 1 3
3
F(y) = P(X y) = P(X y 3 ) = 4 y 3 − 2
Therefore,
x <0
0

Therefore F ( x ) =  0.04x 2
1

Limits of Y :
x <6
)
Exercise 1.3A
1
X
)
3
2 3
⌠ 3 − 3 t − 3 1 dt =  − 3 t − 3 1  = 5 (or 0.208 to 3. s.f.)
=

3
3  1 24
⌡2 1 5
 10
2
2
2
c 60 members, therefore
(
( 12 x m) = F (2m) = 0.2(2m) − 0.01(2m)
1
2
m ∈ ( 0, 5 ) , FM (m ) = P x m = FX ( 2m ) = 0.2 ( 2m ) − 0.01( 2m ) = 0.4m − 0.04m 2
31
2
7, ⌠ 3 3
1
8
=
− t − 3 dt =
∫

15 
3
15
⌡2 5
m <0
0
7
8
F (m ) = 0.4m − 0.04m 2
0 m 5
+
= 1 , therefore a = 2
15 15

1
m
> 5.

1
31
2 33
3
 3
1
3
1
1 
5
P t 2 = ⌠
− t −3
dt =  −
t −3
=
(or 0.208 to 3. s.f.)
0m5
2 
3
10
3  1 24 Hence, f (m ) = 0.4 − 0.08m
⌡2 1 5

2
2
0
otherwise.
2

21 2
t dt
1 5
1
(
a Substitute a = 2,
2

x <0
 0
Therefore F ( x ) =  0.2x − 0.01x 2 0 x 10

x > 10.
1
Limits of M:
0 t < ln 2
1
x
∫0 0.02 (10 − x)dx = 0.2x − 0.01x

 = 0.04 × 
a
1 2

y2
10
∫0
216 y 1000
otherwise.
kx dx = 1
10
 = 0.04 y 1 kx 2  = 1
 2
0
50k = 1
k=
b
1
50

1
0 x 10
f ( x ) =  50 x
0
otherwise.

x 1
1
2
∫0 50 x dx = 100 x
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions

0
Therefore F ( x ) =  1 2
x
100
1
Change limits:
) , FY ( y ) = P
( )
Y
0
10
0
2
1
1
= F ( 5y ) =
5y = y
( 15 x y ) 100 ( )
4
2
X

0
F ( y ) =  1 2
y
4
1
a
)
T
(
r ∈ (1, 3 ) , FR (r ) = P
)
1
1
1
3
= F (3r ) =
+
+ r
3r =
( 13 t r ) 10 10 ( ) 10 10
T
( 15 x y ) = F (5y ) = 1001 (5y ) = 14 y
2
X
2
2
y <0
0y 2
0 y 2
otherwise.
9
∫1 0.4 − kt dt + ∫3 k dt = 1
3
0.4t − 1 kt 2  + [ kt ]9 = 1
3
2

1

0

F (r ) =  6 r − 9 r 2 − 7
5
20
20
1
3
10 + 10 r

1
r<1
3
1 r 1
3
1r 3
r > 3.

6 − 9 r
5 10
Hence, f (r ) = 
3
10
0

y > 2.
3
R
1
1
1
For r ∈ (1, 3 ) , FR (r ) = P 3 t r = FT (3r ) = 10 + 10 ( 3r ) =
 1
 y
Hence, f ( y ) =  2
 0
5
(
( 13 ,1), F (r ) = P ( 13 t r ) = F (3r ) = 25 (3r ) − 201 (3
1
1
2
1
2
7 6
9
7
r0∈ x,1 ,10
FR (r ) = P t r = FT ( 3r ) = ( 3r ) − ( 3r ) −
= r − r2 −
3
3
5
20
20 5
20
20
x > 10.
X
For y ∈ ( 0, 2 ) , FY ( y ) = P
For r ∈
x <0
6
c
P ( R < 1.5 ) = F (1.5 ) =
a
1 2
a = 1 so a2 = 16
16
1 r 1
3
1r 3
otherwise.
1
3
11
+
1.5 =
10 10 ( ) 20
Therefore, a = 4, a = –4 (reject as a > 0)
0.8 – 4k + 6k = 1
Change limits:
k = 0.1
t
2 t
X
0
4
⌠


b For t ∈ (1, 3 ) ,  0.4 − 0.1t dt = 0.4t − 0.1t  = 2 t − 1 t 2 − 7
2  5
20
20

⌡1
1
t
t
⌠

0.1t 2  2
1 2 7
 0.4 − 0.1t dt = 0.4t − 2  = 5 t − 20 t − 20
 1
⌡1
(
1t 3
3 t 9
b
t > 9..
7
R
1
1
3
3
9
1
3
y <0
0 y 16
y > 16.

1
Hence, f ( y ) =  16
0

t <1
T
)

0
F ( y ) =  1
y
16
1
Therefore
Change limits:
2
 1 1  1
1
FY ( y ) = P x 2 y = FX  y 2  =  y 2  =
y
16
  16  
t
For t ∈ (3, 9), F ( 3 ) + ∫ 0.1t dt = 2 + [0.1t ]t3 = 1 + 1 t
10 10
5
3
t
t
1
2
1
+ t
F ( 3 ) + ∫ 0.1t dt = + [0.1t ]3 =
5
10 10
3

0
2
1 2 7
F (t ) =  5 t − 20 t − 20
1 + 1t
10 10
1

Y
0
16
a
0 y 16
otherwise.
1 15
P (Y > 1) = 1 − F (1) = 1 −
=
16 16
x
∫−∞ 2 dx = F(−1) +  2 x −1 = 2 x + 2
x
1

0
F(x ) =  1
x+1
2
2
1
1
1
1
x < −1
−1 x 1
x > 1.
7
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
Therefore
Change limits:
Y = eX
X
−1
e−1
1
e
9
F(Y ) = P(Y y ) = P(e X y ) = P(x ln y ) =
=



F( y ) = 


a = −2
y <e
0
1
1
ln y +
2
2
1
1

Hence f(y) =  2 y
0

−1

0
F ( x ) =  1 3 1
x +
2
16
1
Change limits:
e −1 y e
y > e.
e −1 y e
otherwise.
b P(Y  k) = 0.25 is the same as
P(Y  k) = 0.75
1 ln k + 1 = 0.75
2
2
k = e0.5
1
∫02 c x
3
x < −2
−2 x 2
x > 2.
X
Y = X2
−2
4
0
0
2
4
1
1
1 3
F(y) = P(X 2 y) = P(− y 2 X y 2 ) = y 2
8
Therefore

1
3
0y 4
f(y) = 16 y 2
0
otherwise.

1 ln k = 0.25
2
ln k = 0.5
8
0 y 17
4
otherwise.
 3x 3  2

 =1
 48  a
1
1
ln y +
2
2
Therefore,

1024 3
f(y) =  83521 y
0

2 3 2
∫a 16 x dx = 1
dx = 1
x
10
1
4 2

 cx
 4  =1

0
2
3
3
x
⌠ ( x − 10 ) dx = F ( 0 ) +  ( x − 10 )  = ( x − 10 ) + 1



9000
9000
9
⌡−∞ 3000


0

0
F ( x ) = ( x − 10 )3 1
 9000 + 9
1

c = 64

3

f(x) =  64x
0

0x 1
2
otherwise.

0
F(x ) = 16x 4

1

x 0
0 < x < 30
x 30.
Since X + T = 30, T = 30 − X
x <0
Change limits:
0x 1
2
1
x> .
2
Change limits:
X
Y = 8.5X
0
0
1
2
17
4
y 
256 4
F(y) = P(8.5X y) = P  X =
y
8.5  83521

X
T
0
30
30
0
FT (t ) = P(T < t ) = P(30 − X < t ) = P(X > 30 − t ) =
= P(X > 30 − t ) = 1 − P(X 30 − t )
 (( 30 − t ) − 10 )3 1 
+ 
= 1 − F X (30 − t) = 1 − 
9000
9

(20 − t )
=8−
9000
9
3
8
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions
Therefore

0
F (t ) =  8 ( 20 − t )3
 9 − 9000
1

t 0
0 < t < 30
1
0
∫−1 3 x
b E(Z) =
∫−1 3 x
E(P) =
∫−1 3 x
c
2
4
33
dx + ∫ 1 x 2dx = 9
0 6
b
b
0
1
4
4
dx + ∫ 1 x 4dx = 513
15
0 6
1
f ( x ) =  8 x
0

10 1
4
57
4
a
x 3 dx = 544 cm3
4
E( X ) =
∫0 x
b Y = X3
b
=
c
10 1
∫6
× 3x 6dx =
4
7 290048
7
7
3 4
0
( )
4
( )
0x 4
otherwise.
x 
1 
∫ (2x ) 8 x  dx =  16 
4
4
2
0
4
= 16
0
31
2
∫−1k dy +∫2 5 (7 − 2y ) dy = 1
3
2


[ky ] 2−1+  75 y − 210y  = 1

2
5 2
E (Y ) =
∫0 x
( 252 x )dx = 1002 x  = 252
4
5
2k + k +
0
E(2X + Y) = 2E(X) + E(Y)
( )
k=
5
5
= 2∫ x 2 x dx + 25 = 2 ×  2 x 3  + 25
25
2
 75 0 2
2
a
a
2
25
=2 ×
( 18 x ) dx =  24x  = 83
4
⌠
 4
E X 2 =  x 2 1 x dx =  x  = 8
8
 32 0
⌡0
E(Y ) =
∫0 kx dx = 1
5
c
otherwise.
1
Y = 2X 2 and f ( x ) =  8 x
0

5
 1 kx 2  = 25 k = 1
 2
0 2
b
0x 4
Alternatively,
E(3Y 2 + Y + 2) = E(3Y 2 ) + E (Y ) + E(2)
k=
13
8
+3=5
9
9
E(Y) = E(2X2) = 2E(X2) = 16
7 290048
=
+ 544 + 2
7
7 293870
=
7
a
18 = 13
5
3
1
1 13 13
ET = ×
=
3 ( ) 3 3
9
E ( 2R + 3 ) = 2E(R) + 3 = 2 ×
E(3Y 2 ) = E(3X 6 )
4
E(R) =
4
dx + ∫ 1 x 3dx = 127
12
0 6
6
11
9
3
3Y 2 = 3X 6
3
3
10 5
− =5
3 3
∫1 t (0.4 − 0.1t )dt + ∫3 0.1t dt = 15 +
1
∫6
a E(Y) =
a E (T ) =
0
d E(Y + Z) = E(Y) + E(Z) =
2
10 5 15
+ =
=5
3 3 3
b E ( 2 X − M ) = 2E ( X ) − E ( M ) = 2 ×
t 30.
5
a E(Y) =
5
5
Therefore E ( X ) + E ( M ) =
Exercise 1.4A
1
∫0 m (0.4 − 0.08m )dm = 3
E(M ) =
b
1
5
f(y)
10 25 115
+
=
3
2
6
E( X ) =
10
∫0
3
5
2
5
0.02x (10 − x )dx = 10
3

f (m ) = 0.4 − 0.08m
0
4
=1
10
0 m 5
otherwise.
–1
0
2
3
y
9
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
2 2
2
3 3Therefore,
1 y dy + 3 1 y (7 − 2y ) dy =  y  +  7y − 2y  = 3 + 29 = 19
 10 

∫−1 5
∫2 5
 015
15  10 30
  −1  10
2

2
3
 1 w3
31
 7y 2 2y 3 
y2 
3
29
19
y dy + ∫ y (7 − 2y ) dy =   + 
−
=
+
=
 5

10 
10
15  10 30 15
2 5


 27
27
1
2
−
F (w ) =  40 w − 80

 189 + e −1.9 − e −w
d X = 1 Y , therefore X 2 = 1 Y 2
2
4
 200
 1
2
3

1
1
1
1
E (Y ) = ⌠
 4 y 2 5 dy + ⌠
 4 y 2 5 ( 7 − 2y ) dy
⌡−1
⌡−2
Y = 2.2W
c
E (Y ) =
2
( )( )
( )(
2
w<0
0 w < 1.5
1.5 w < 1.9
1.9 w < 2.36
)
3
 x3 
7
1 4
=   +  y3 −
y
40  2
 60  −1  60
w 2.36.
y 
For y ∈ ( 0, 3.3) , F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W =
2.2 

3
y  1 y 
25 3
∈ ( 0, 3.3) , F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W y
=
=
3 y 71
2.2  5  2.2 
1331

=
+
20 120
For y ∈(3.3, 4.18),
89
=
y 
27  y  27
120
F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W =
−
2.2 
40  2.2  80

21
8
1
2
2
8 ∫ x(x ) dx + ∫ (8 − x )(x ) dx = 82
0 4
2 4
y 
27  y  27 27
27
y−
F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W =
−
=
2.2
40
2.2
80
88
80




6
43
1
9 E(R 2) = ∫ r 2 dr = 3
1 5
For y ∈ (4.18, 5.192),
−5
y  189
E(A) = πE(R 2) = 43 π cm 2 or 45.0 (3 s.f.)
3
F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W =
+ e −1.9 − e 1
2.2  200

E 1 ( A + 1)2 = 1 E ( A + 1)2
2
2
−5 y
y  189
F ( y ) = P (Y y ) = P ( 2.2W y ) = P  W =
+ e −1.9 − e 11
2.2
200


1
2
= 2 (E( A ) + 2E( A) + E(1)) = 1059
10 To find the value of k,
Therefore, f ( y ) = d F( y)
dy
k
1.5
1.9
2
−w
∫0 0.6w dw + ∫1.5 0.675 dw + ∫1.9 e dw = 1
 75 2
k
1.5
0 < y < 3.3
 1331 y
0.2w 3  + [ 0.675w ]1.9 +  −e −w  = 1

0

1.9
1.5
 27

3.3 y < 4.18
0.675 + 0.27 + (– e–k + e–1.9) = 1
f ( y ) =  88
5
 5 −11 y
4.18 y < 5.192
e–k = e– 1.9 – 0.055
 11 e
 0
otherwise.
–k = ln(e–1.9 – 0.055)

4.18
3. 3
75 2
⌠
k = 2.36
y × 27 dy +
E(Y ) = ∫0 (y × 1331 y ) dy + 
88
⌡3.3
Change limits:
5.192
5y
⌠


5 −11

 y × 11 e
 d y = 2.94 (3 s.f.)
W
Y = 2.2 W
⌡4.18
0
0
Alternatively:
1.5
3.3
Y = 2.2W so E(Y) = 2.2E(W)
1.9
4.18
k
1.5
1.9
E(W ) = ∫ 0.6w 3dw + ∫ 0.675w dw + ∫ we −w dw
2.36
5.192
( )
( )
(
)
(
)
(
0
For W ∈(0,1.5),
w
∫0 0.6w
2
dw =
1.5
1 3
w
5
1.9
1.5
1.5
= 0.15w 4 
0
)
1.9
1.9
+ 0.3375w 2 
1.5
+  −e −w (w + 1) 
k
1.9
k
−w
0.3375
w)4 + w+ 0.675
w 2= 27+w −−e27
(w + 1)  where k = − ln(e −1.9 − 0.055) ≈ 2.36
For W ∈(1.5,1.9),0.15
F (1.5
dw
0 ∫1.5 
1.5
1.9
40
80
= 0.759375 + 0.459 + 0.116147 = 1.335
w
For W ∈ (1.9,2.36 ), F (1.9) + ∫ e −w dw = 189 + e −1.9 − e −w
200
1.9
E(Y) = 2.2 × 1.335 = 2.94 (3 s.f.)
10
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
where k = − ln(
Worked solutions
Exercise 1.5A
1
– 0.05x2 + 0.4x – 0.55 = 0
a F(x) = 0.5
1 x 2 = 0.5
25
x = 3.54
x = 1.76, x = 6.24 (reject, as x < 3 from part b)
Therefore the 20th percentile is 1.76.
3
x = −3.54, reject
a y = 1, 5 (y − 4)2 = 5
72
8
f(y)
b Q1: 1 x 2 = 0.25
25
x = 2.5
x = −2.5, reject
Q3: 1 x 2 = 0.75
25
x = 4.33
x = −4.33, reject
2
a
1
3
4
5
8
f(x)
0.3
0.1
0
1
9
3
1
x
Therefore the median value lies between 3
and 9.
Area of trapezium + area of rectangle = 0.5
c
Area of rectangle = 0.1
(x – 3) × 0.1 = 0.1
Therefore the median is 4.
Alternative method:
3
4
a
y
y <0
0y <1
1 y < 4
y > 4.
3
F(m) = 1 + 5(m − 4) = 0.5
216
m = 1.22 (3 s.f.)
31
3
1
∫0 7 dw = 7 < 2
Therefore, the median value lies between 3
and 5.
Therefore, the median value lies between 3
and 9.
0.4 + ∫ 0.1 dx = 0.5
3 + w 2 dw = 1
7 ∫3 7
2
[0.1x ]3x = 0.1
2w = 1 − 3
 7  3 2 7
x
3
w
0.1x – 0.3 = 0.1
w=3
x=4
F(x) = 0.2
x
∫1 0.4 − 0.1x dx = 0.2
x
0.4x − 0.1 x 2  = 0.2

1
2
1
4
Therefore, the median is 3
Therefore the median is 4.
c
4
d The 20th percentile is when y is between 0 and 1,
3 y 2 = 0.2
8
y = 0.730, y = −0.730, reject
x=4
∫1 0.4 − 0.1x dx = 0.4 < 0.5
3

0
3 2
8 y
b F(y) = 
3
 5( y − 4 )
1
+

216

1

b Using the graph, area of trapezium
1
= (0.1 + 0.3) × 2 = 0.4 < 0.5
2
Using calculus,
2
b
1
.
4
3 + w 2 dw = 7
7 ∫3 7
10
w
2w = 7 − 3
 7  3 10 7
w=3
19
20
11
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 CONTINUOUS RANDOM VARIABLES
5
a
3
F(Q3) = 0.75
5
∫1 k ( x − 1)dx + ∫3 k (5 − x )dx = 1
3
1
x − 1 = 0.75
2
2 5
 kx
 
kx 
 2 − kx  + 5kx − 2  = 1

1 
3
2
(
)( )
25k
9
+  ( 25k −
− 15k − k ) = 1
2 ) (
2 

 9k − 3k − k − k 
 2

2
b
7
a
f(x)
1− e
c
6
a
1
3
8
− 1m
3
=
1
2
x 0.
m = 3ln 2 or 2.08 (3 s.f.)
−1x
3
= 0.8
x = 3ln 5 or 4.83 (3 s.f.)
a
y
1
2
a
4
∫1 k dx + ∫3 2k dx = 1
[kx ] + [2kx ]
3
1
4
3
–1
k=
1
x
=1
a=1
2
(3k – k) + (8k – 6k) = 1
1
4
b F(t) = 0 +
x1

0
1
1
F ( x ) =  4 x − 4
1 x − 1
2
1

1
∫3 2 dx = 2 x − 1
For x ∈ (3, 4), F(x) =
c
x <1
x > 4.
1
1
Q1= −0.5
Q3 = 0.5
IQR = Q3 − Q1 = 1
1 x < 3
3x 4
t
∫−1 2 dt = 2 (t + 1)
Alternative method
1
(m + 1) = 0.5
2
m=0
x1
1
1
b For x ∈ (1, 3), F(x) = ∫ 4 dx = 4 x − 4
1
c
x <0
F(x) = 0.8
1− e
x
5
The graph is an isosceles triangle. It has the
line of symmetry x = 3. Therefore, the median
is 3.
3
x
−1x
 −1x 
dx =  −e 3  = 1 − e 3

0
1
b F(m) = 2
1
2
c
x 1 −1x
3
∫0 3 e
 0
Therefore F ( x ) = 
−1x
 1 − e 3
1
4
0
7
2
Therefore IQR = 7 − 2 = 1 1
2
2
4k = 1
k=
x=
9
a
4
k
1
∫1 0.25 dr + ∫4 − 8 (r − k ) dr
=1
k
2
 1 4  1
 4 r  +  − 16 (r − k )  = 1
1
4
F(x)
(1 − 14 ) +  0 − − 161 ( 4 − k )  = 1
2
1
1
2
0
1
3
4
x
d F(Q1) = 0.25. From the graph, LQ value must
be between 1 and 3.
1
1
x − = 0.25
4
4
x=2
1
1
(4 − k )2 =
16
4
4 – k = ±2
k = 2 (reject because k > 4) or k = 6
b For r ∈(1, 4),
F (r ) = F (1) + ∫
r
1
r
1
1
1
dr = 0 +  r  = (r − 1)
4
 4 1 4
12
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
1
Exam-style questions
For r ∈ (4, 6),
r
r
⌠ − 1 ( r − 6 )dr = 3 +  − 1 (r − 6)2  = 1 − 1 (r − 6)2
F( r ) = F( 4 ) + 
1
 4
16
4  16
⌡4 8
1 a F(7) = a(7)3 = 343a = 1, so a = 343
r
r
1 3
1 (r − 6)2  = 1 − 1 (r − 6)2
t = 0.2
) = F( 4 ) + ⌠⌡4 − 18 (r − 6 )dr = 34 +  − 16
343
 4
16
t = 4.0936…

t = 4 days
r <1
0
1

1r < 4
 3 2
 4 (r − 1)
0 t 7
b f(t) =  343 t
F(r) = 
2
0
 (r − 6 )
otherwise.

4 r 6
1 − 16
1
c F(t) = 0.25
r > 6.

c
1−
(r − 6)2 = 0.8
16
r = 4.21
10
−x

a f(x) = x
0

Lower quartile F(t) = 0.25
r = 7.79 reject
1 3
t = 0.25
343
−1 x < 0
0 x 1
otherwise.
0
1 1 2
 2 − 2 x
F(x) = 
1 + 1 x2
2 2
1
For t ∈(0, 7),
t 3
1 3
F (t ) = ∫
t 2 dt =
t
343
0 343
For the 80th percentile, R is between 4 and 6.
t3 = 85.75
t = 4.41
x < −1
2
−1 x 0
a
a
x > 1.
−
Change limits:
X
Y
−1
4
0
0
1
4
FY (y) = P(4 X 2 y) = P − 1 y X 1 y
2
2
1
1
1
= FX 2 y − FX − 2 y = 4 y
Therefore,

y <0
0

1
F(y) =  y
0y 4
4
1
y > 4.
1

Hence f(y) =  4
0

1
b
y = 0.5
4
M: y = 2
c
)
0 y 4
otherwise.
For Q1, 1 y = 0.25,
0.5 so y = 1
4
For Q3, 1 y = 0.75,
0.5 so y = 3
4
IQR = 2
1
 − 1 (4 − x)2  + 1 (5 − a) = 1
 2
 1 6
0 x 1
(
( ) (
a1
∫1 6 (4 − x)dx + 6 (5 − a) = 1
{
}
1
1
(4 − a)2 − 9 + (5 − a) = 1
12
6
(4 − a )2 − 9 − 2(5 − a ) = −12
a 2 − 6a + 9 = 0
2
(a − 3) = 0
a=3
)
3
4. 5 1
1
dx = 11
16
b
∫1.5 6 (4 − x)dx + ∫3
c
The 95th percentile lies in the region (3, 5).
1 1
F(x) = 6 + 6 x
6
3x 5
1 + 1 x = 0.95
6 6
x = 4.7
3
a
1
 1 x 4  + ( k − 1) x k +  − ( x − 2 )2 2 = 1
1 
k
 4 0 
8k 2 − 24k + 17 = 0
6+ 2
6− 2
or
4
4
13
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
b




f(x) = 




x3
0x <1
2+ 2
4
1 x <
− 2( x − 2)
0




or f ( x ) = 




Or change limits:
6+ 2
x < 2
4
otherwise
x3
0x <1
2− 2
4
6− 2
1 x <
4
6− 2
x <2
4
otherwise.
−2( x − 2)
0
x <0
0
4
x
4
2+ 2
2 +1
x−
4
4
1 − ( x − 2)
2
1

0
x4
4

or F(x) =  2 − 2 x − 1 − 2
4
 4

2
1 − ( x − 2 )

1
c
Y
0
1
6− 2
4
126 − 55 2
32
2
8
 1
1 y 3
3

−2
2 − 2 3
f(y) =  12 y
  1
 −2
− 2  y 3 − 2  y 3
3

 
0

Therefore,





F(x) = 





X
0
1
6+ 2
4
0x <1
4
6+ 2
1 x <
4
6+ 2
x 2
4
x >2
Q 3 : 1 − e −0.1t = 0.75
c
For median:
1 − e −0.1t = 0.5
t = 10 ln 2
1t
 1 − 10
 e
f (t ) = 10
 0





f(y) = 





126 − 55 2
32
2
8
1 13
y
3
0 y <1
2 + 2 − 23
y
12
1 y <
2
otherwise.
∞
∞
∞
∞
6+ 2
4
t 0
⌠ 
 − 1t
∞ − 1t
− 1t
Mean = E(t ) =  t  1 e 10  dt =  −te 10  − ∫ e 10 dt
0

⌡0  10

0
Change limits:
1
otherwisse.
Q 3 − Q1 = 10 ln 3
1 x < 6 − 2
4
6− 2 x 2
4
x > 2.
1
126 − 55 2 y < 8
32
t = 10 ln 4
3
t = 10 ln 4
b Q1 : 1 − e −0.1t = 0.25
0x <1
Y
0
1 y < 126 − 55 2
32
a F(15) − F(10) = 0.145
x <0
X
0
0y <1
⌠ 
 − 1t
∞ − 1t
− 1t
E(t ) =  t  1 e 10  dt =  −te 10  − ∫ e 10 dt
0

⌡0  10

0
126 + 55 2
32
 −3
2 1
−  y 3 − 2 y
3

126 + 55 2
y <8
32
0
otherwise.
∞
 − 1t
= 0 − 10e 10  = 0 − ( 0 − 10 ) = 10

0
P(mean  T  median) = F(10 ln 2) – F(10)
5
a
= 0.5 − e −1
iTwo points (−2, 0) and (0, 0.2) on the
first piece:
0.2 − 0 = 1
a=
0 + 2 10
Two points (1, 0.4) and (4, 0) on the third
piece:
0 − 0.4
2
b = 4 − 1 = − 15
14
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions
1
1
10 x + 5

1
f ( x ) = 5
− 2 x + 8
 15
15
0

 9 x − 1 x2  = 3
14 1 8
 14
0x <1
4x2 – 36x + 53 = 0
1 x < 4
x = 1.85, x = 7.15 reject
otherwise.
0

 1 x 2 + 4x + 4
20

ii F(x) = 1 (1 + x )
5

− 1 x 2 − 8x + 1
 15
1

b i Change limits:
(
−2 x < 0
)
1 x < 4
7
2
( ) = 0.803
919
239
− 1
192
240
2
e−2
1
4
e4
k = 0.25
b F(4) = 0.5, therefore the value of Q1 must lie
between 2 and 4; the value of Q3 must lie
between 4 and 5.
e
0.25(x – 2) = 0.25 x = 3
0.5(x – 3) = 0.75 x = 4.5
c
For y ∈ (e, e4), F ( y ) = 1 (ln y 2 − 8ln y + 1)
15
Therefore,
e −2 y < 1
E( X ) =
=1− F
8
a
3
5
(
4
otherwise.
4
3
9
∫2 0.25x dx + ∫4 0.5xdx = 2 + 4 = 3 4
P( X > µ) = P X >
ii
1 y <e
ey <e
IQR = 4.5 – 3 = 1.5

2x 4
 0.25
f ( x ) =  0.5
4x 5

otherwise.
0
d i
)
(
)
15
15
=1− P X .
4
4
(154 ) = 1 – 0.4375 = 0.5625
f(x)
e
⌠  1
 3 
ii ⌠
 y  10y  d y +  y  5y  d y
⌡e−2
⌡1
1 or 1
15
k
e4
 −2 
+⌠
 y  5y  d y = 21.4
⌡e
a P ( 0.5 < x < 3 ) =
1
1
31
∫0.5 2 x dx + ∫1 7 ( 4.5 − x )dx
3
1
1
∫0 2 x dx = 8 , therefore the median value
3
0
3
1
9
1
15 5 745
=  x 4  +  x − x2  =
+ =
14 1 128 7 896
 8 0.5  14
11
)
so k(4 – 2) = 0.5
1
(ln y 2 + 4 ln y + 4)
For y ∈ (e–2, 1), F ( y ) =
20
1
For y ∈ (1, e), F ( y ) = (1 + ln y)
5
b
4. 5
1
a F(4) = 0.5(4 – 3) = 0.5
F(y) = P(Y  y) = P(eX  y) = P(X  ln y)
6
5
2k = 0.5
−2
0
1
1
(
x 2 9 − 1 x dx
14 7
1
301
919
=
=
+
12 64 192
1
0
( )
x > 4.
Y

 3
10y
1

f(y) =  5y
 −2
 5y

0
( ) ∫ 12 x dx + ∫
E X2 =
Var ( X ) = E X 2 − ( E ( X )) =
0x <1
X
c
x < −2
)
(
x
−2 x < 0
b
x
∫0
15
0
1 dx = 1 x F x =  1
( ) 15 x
15
15

1
x
x <0
0 x 15
x > 15.
must lie between 1 and 4.5.
1 + x 1 4 . 5 − x dx = 1
)
2
8 ∫1 7 (
15
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
c
Change limits:
X
0
15
(16k − 8k ) − (8k − 2k ) = 53
Y
0
225
k=
b
1
 1  1  1  1 21


FY ( y ) = P X 2 y = P  X y 2  = FX  y 2  =  y 2  =
y
  15   15


(

 = FX 
1

y2
1

y2
1
 = 15 
)
1

y2
1
 = 15
0
0 y 225
y > 225.
0 y 225
E (T ) =
otherwise.
2
∫0 0.15t
 3X 4 
E(Z) = E  2  = E(3X 2)
 X 
= 3 × E(X2) = 3 × 75 = 225
a For X ∈(–∞, 0),
x
∫−∞
2
Mean = E (T ) =
c
t
4
2
∫0 0.15t
4
dt + ∫ 3 t ( 4 − t )dt = 3 + 8 = 2.2 hours
5 5
2 10
3
4
d t + ∫ 3 t ( 4 − t )dt = 3 + 8 = 2.2 hours
5 5
2 10
3
d For t ∈ ( 0, 2 ) , F (t ) =
d E(Y) = E(X 2) = 75
F(x) =
0.6
y <0

−1
Therefore, f ( y ) =  1 y 2
30
 0

9
f(t)
1
y2

0
F ( y ) =  1 21
y
 15
1

3
10
t
∫0 0.15t
2
dt = 0.05t 3
t
For t ∈ ( 2, 4 ) , F (t ) = F ( 2 ) + ∫
3 4 − t dt = 0 . 4 +  6 t − 3 t 2  =
(
)
20 2
2 10
 5
t
t
t
3 t 2 + 6t − 7
t ∈ ( 2, 4 ) , F (t ) = F ( 2 ) + ∫ 3 ( 4 − t )dt = 0.4 +  65 t − 230 t 2  = − 20
5
5
2 10

2
Therefore,
0 dx = 0

t <0
0
0.05t 3
For X ∈(0, ∞ ),
0 t 2
F (t ) = 
x
x
1x 
1x
1x
1x
3
6
7

2
−
−
−
−
1
2 t 4
F(x) = F( 0 ) + ⌠
 4 e 4 dx = 0 +  − e 4  = − e 4 − ( −1) = 1 − e 4 − 20 t + 5 t − 5
⌡0

0
t > 4.
1
x
1
1
1
1


1 e − 4 xdx = 0 + − e − 4 x = − e − 4 x − −1 = 1 − e − 4 x
17
3
( )
e P(T > 3) = 1 – F(3) = 1 − 20 = 20


4

0
Therefore,

0
F( x ) = 
1
1 − e − 4 x

b
x < 0,
2
∫0 0.15t
dt + ∫
2
4
2
6
2
 1 kx 2  + [ 2kx ]6 = 1
2
 2
0
x 0.
(2k – 0) + (12k – 4k) = 1
−1 
e 4
k=
 = 0.779
1 = e − 14 x
4
1
1
− x = ln
4
4
x = 5.55
a
2
∫0 kx dx + ∫2 2k dx = 1
a
−1 

P ( X > 1) = 1 − P ( X 1) = 1 − F (1) = 1 −  1 − e 4  = 0.779



1) = 1 − P ( X 1) = 1 − F (1) = 1 −  1 −

1
− x
3
c 1−e 4 =
4
10
11
1
10
2
31
1
3 1 7
x dx + ∫ dx =
+ =
10
20 5 20
2 5
b
P (1 < X < 3 ) = ∫
c
For x ∈ ( 0, 2 ) , F ( x ) =
1
x
1
1
∫0 10 x dx = 20 x
2
x
x
x1
1
1
1
1
For x ∈( 2, 6 ) , F ( x ) = F ( 2 ) + ∫ 5 dx = 5 +  5 x  = 5 x − 5
2
 2
x
x ∈( 2, 6 ) , F ( x ) = F ( 2 ) + ∫ 1 dx = 1 +  1 x  = 1 x − 1
5  5 2 5
5
2 5
k ( 4 − t ) dt = 1
4
16
2 
1
+ 4kt − kt 2  = 1
5 
2
2
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1
Worked solutions
Therefore,

0
1 2
F ( x ) =  20 x
1 x − 1
5
5
1

d
x <0
b
0x 2
2x 6
x > 4.
For a ∈ ( 4, 9 ) , F (a ) =
1
F ( 2 ) = , therefore the median value must lie
5
between 2 and 6.
a
9
a
(
)
b i
=−
( )
0x π
otherwise.
5
1
2
c
1
2
0
–5
π
5
a
1
1
4
∫4 10 da = 10 a − 10
(
)
1 2 19
161
a +
a−
200
100
200
Therefore,

0
1
4
F (a ) = 10 a − 10
− 1 a 2 + 19 a − 161
100
200
 200
1

f(x)
ii
otherwise.
(
1
1
1
P − π < x < π = 0 + F π = 0.383
4
4
4

1
1
f ( x ) =  2 cos 2 x
0

9 a < 19
1 19 − a da = 1 + 19 a − 1 a 2 − 261
)
100 ( 2 100
200
200
x = 3.5
12
4a < 9
a 1
1
19
For a ∈ ( 9, 19 ) , F (a ) = F ( 9 ) + ∫ 100 (19 − a ) da = 2 + 10
9
a ∈ ( 9, 19 ) , F (a ) = F ( 9 ) + ∫
1
1 1
x− =
5
5 2

1
10
f (a ) = 
1
100 (19 − a )
0

a<4
4 a 9
9 a 19
a > 19.
1
S = A2
Limits:
1
S = A2
2
3
A
4
9
x
19
–5
19
( 12 x ) dx


1
s
For s ∈ ( 2,3) , F ( s ) = P  A s  = P ( A s ) =
10


1
1
1
= 2x sin ( x )  + ∫ 2sin ( x ) dx = 2π −  −4cos ( x )  = 2.28
2 
2 s ∈ ( 2,3) , F (s ) = P 2A 
 s = P ( A s ) = 1 s − 4


10
10
1
1
1
sin ( x )  + ∫ 2sin ( x ) dx = 2π −  −4cos ( x )  = 2.28
P (S < 2.5) = F ( 2.5) = 1 (2.5) − 4 = 9
2 
2
2 

10
10 40
c
E ( 2X ) =
i
π
1
∫0 2x  2 cos
π
π
0
0
1
2
2
π
1
2
π
π
π
0
0
0
2
0
2
2
ii
13
a
E( X ) =
9
19
∫4 k da + ∫9
or 0.225
1
E 2 X = 1.14
2 ( )
14
1 19 − a da = 1
)
100 (
19
19
1 2
a−
a
[ka ]94 +  100
200 
9
=1
 192 192  19 × 9 81  
5k + 
−
−
−
=1
200  
 100 200  100
5k +
k=
a
x
∫0 cx
2
dx = 1
2
c x3 = 1
 3 0
8c
−0=1
3
c=
3
8
1
=1
2
1
10
17
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
−
1
1 Continuous Random Variables
b
0

F ( x ) =  1 x 3
8
1
0

F ( y ) =  1 y 4 − 1
80
 327680
1
x <0
0x 2
x > 2.
Q1 :
1 3 1
x =
8
4
x = 1.260
Q3 :
1 3 3
x =
8
4
x = 1.817
E (10Y ) = ∫
c
Limits of Y:
X
0
2
8 y 24
y > 24.
 1 y 3 8 y 24
Therefore, f ( y ) =  81920
0
otherwise.

b P(Y > 6) = 1 – F(6) = 1 – 0 = 1
IQR = 1.817 – 1.260 = 0.557
c
y <8
24
8
1
1
y 3  d y = 
y5
(10y )  81920
40960 


24
8
= 193.6
Y
0
16
16
( ) ( )( )
a
4k
∫−k
i
1 3
2
x + k dx = 1
5k
 1 21 
4k
2
1  1
1 3
FY ( y ) = F 4 X 2 y = FX  y  =
 4 y  = 64 y 2  x + 1 x  = 1
4
8




 10k 5 

 −k
1
1 3
1 2
1  1 2
1 23
y =
y  =
y

64
4
8  4



8
4
1
1
 k + k −
k − k  =1
5
10
5 
 5
0
y <0

3
F ( y ) =  1 y 2
12
1
0 y 16
k + k = 1
64
5
10

y > 16.
1

5
k = 1
2

1
3 2
0 y 16
Therefore, f ( y ) = 128 y
2
k =
5
0
otherwise.


3
16
x < −2
0
5
d E ( y ) = ∫ 3 y 2 dy = 9.60

128
0
( 5x + 2 )2
ii F(x) = 
−2 x < 8
5
5
 100
x
x 1 3
8
1
1
1
1
4
4

x .
15 a ∫
x dx =
x
= x −

5
80
 80 1 80
1 20
( ) ( )( )
(
)
(
0

F ( x ) =  1 x 4 − 1
80
80
1
x > 3.
x = 1.44 (3 s.f) −2.24, reject
b i
Limits of Y:
1
1
=
y4 −
327680
80
(5x + 2)2 = 0.85
100
1 x 3
FY ( y ) = F ( 8X y ) = FX
)
iii P(X  p) = 0.85
x <1
X
1
3
)(
Change limits:
X
Y
8
24
−
( 18 y ) = ( 801 )( 18 y ) − 801
4
Y
8
125
2
5
−
8
5
512
125
1
P (Y y) = P (X 3 y) = P (X y 3 ) =
 13

 5y + 2 
=
100
2
18
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
T) =
1
Worked solutions

0

2
  13

5
2
y
+
F(y) =  


100

1

0
 2
t
 24
F(t) = 
2
 1 − (12 − t )
192

1

y <− 8
125
− 8 y < 512
125
125
y 512 .
125
 1
2
 5y 3 + 2 

 = 0.5
ii
100
y = 1.01 −1.81, reject
17
t
−0.1t
)
1
a E(T) = ⌠
 t 10 e −0.1t dt = 10 minutes
⌡0
b i
∫0 0.1e
(
∞
t
F (T ) =
∫0 0.1e
−0.1t
(
t
t
)
)
1−
F(m) = P(2T  m) = P(T  0.5m)
= 1 – e–0.1(0.5m) = 1 – e– 0.05m

−0.05m

m 0
Therefore, f (m ) =  0.05 e
otherwise.
0
–
0.05m
= 0.5
To find median, 1 – e
ii
– 0.05m = ln 0.5
m = 13.9 minutes
= P − n T n
n
0.1 n
n
 0
FN ( n ) =  0.1
 e
n
18
n
)
x
x
2
2
t
n<0
−e
2 t
t 
1
∫0 12t dt =  24 
−0.1 n
t
4
=
0
=
x
 x2 
1
∫0 11 x dx =  22 
x
=
0
x2
22
x
∫2 11 dx = 11 + 11 x  2 = 11 +
x
2
2
2
2
(121 x −
5
n 0.
=1−
2
t
24
( 8 − x )2
33
Change limits:
2 t
 (12 − t ) 
1
(12 − t ) dt = 32 +  − 192 
96
 4

2
2  (12 − t )
1
+ −
+ 
3 
192
3
=1−
ottherwise.
2
8  (8 − x )
3
=
+ −
+ 
11 
33
11 
For t ∈ [4, 12],
F (t ) = F ( 4 ) + ∫
5x < 8
For x ∈ (5, 8)
x
x 2
8  (8 − x)2 
F ( x ) = F ( 5) + ∫
8 − x ) dx =
+ −
(

11 
33 
5 33
a For t ∈ (0 , 4)
F (t ) =
2x <5
2 x = 2 + 2 x − 4 = 2 x −1
(
)
) − (1 − e F(x ))=) F(2) + ∫ 112 dx = 112 + 
11 
11 ( 11
11 ) 11
− e −0.1
0.1
= e
F( x ) =
For x ∈ (2, 5)
F ( x ) = F ( 2) +
 n)
0x < 2
For x ∈ (0 , 2)
((
t = 5.07, 18.9, reject

1 x
11
2
f(x) =  11
2
 33 ( 8 − x )

0
b i
= 1 − e −0.1
(12 − 5)2 − 32 = 71
192
24 192
1 × 2 ×k + 5 − 2 × k + 1 × 8 − 5 × k = 1
(
)
)
2
2 (
2
k =
11
e– 0.05m = 0.5
(
ii 1 −
(12 − t )2 = 0.75
192
M = 2T
ii FN(n) =
t 12.
dt =  − e −0.1t  = − e −0.1t − ( −1)19= 1 −ae −0i.1t
0
dt =  − e −0.1t  = − e −0.1t − ( −1) = 1 − e −0.1t
0
P(T 2
4 t < 12
Q 3 when 4 t < 12,
(
0t < 4
2.52 = 25
24 96
b i
c
t <0
X
0
2
5
8
Y = X2
0
4
25
64
(12 − t )2
192
19
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
1 Continuous Random Variables
2
For y ∈ (0, 4)
1
1
1.5
 1
c y 2P( 0.5 < x < 1.5 ) = ∫ 0.2x 2 dx + ∫ 26 xdx = 0.08619... + 0.36111
1
1
45
.
0
5
1

  
1
F ( y ) = P (Y y ) = P X 2 y = P  − y 2 X y 2 =
y
−0=
1
22
22
1
1.5 
P ( 0.5 < x < 1.5 ) = ∫ 0.2x 2 dx + ∫ 26 xdx = 0.08619... + 0.36111... = 0.447
0.5
1 45
1 2
 2
 y 
1
1


1 ,
Mathematics in life and work
Y y ) = P X 2 y = P  −y 2 X y 2  =
y
−0=
22
22


(
(
)
)
because X cannot take negative values.
1
a
f(a)
For y ∈ (4, 25)
1
1
1
F ( y ) = 2  y 2 − 1 − 2  − y 2 − 1 = 4 y 2
11 
 11 
 11
For y ∈ (25, 64)


F( y ) = 1 −



f (y) =
1
∫0






1
22
1
11 y
1
0.2x 2
25 y 64
0
otherwise.
2 26
c
21
26
a
Mean is calculated from the integral of x f(x).
Mean is 22.1 years, so in the age group
21  a  26.
d Players between 21 and 26.
2
a The probability will be zero since t is the
continuous random variable measuring the
number of hours in one day.
c
dx + ∫
1
45
x dx = 1
1
Median > mean, negative skew.
Reduce the difficulty level in order to reduce
the median of the play time.
 1 2 23   13 2  b
 5 3 x  +  45 x  = 1
1

0
Put in extra help functions to support players
to complete each level, so that the mean of
play time will increase.
2 13 2 13
+ b −
=1
15 45
45
d Let daily playing time on the weekend be
Y, Y = 2X
E(Y) = 2 E(X)
b = 2, b = –2 reject
3
16
b Any increasing function, i.e. x 2, 0.2x 3.e x
b2 = 4
1
10
b A
ge is a non-negative continuous random
variable. However, we are not expecting a
baby to play computer games. The model can
be modified to take this into account.
4 y 25
8− y
33 y
b 26
( )( )
b
0
0 y 4
383
22
ii
a
2
d
F y , therefore:
dy ( )





f(y) = 





20
1


2
 8 − y 
33
E(x ) =
3
1
0.2x 2
0
∫
2
M(Y) = 2 M(X)
2
dx + ∫ 26 x 2dx = 2 + 182 = 1.443
25 135
1 45
182
x dx =
+
= 1.443
∫0 0.2x 2 dx + ∫ 25 135
1 45
2
20
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
2
2 Inference using normal and t -distributions
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in
this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Where values from the Cambridge International Education statistical tables are used, the same level of accuracy has
been used in workings unless stated otherwise.
Prerequisite knowledge
1
P(W < 53) = P(Z < 1.5) = 0.9332
2
a
P(W  3.0) = P(Z  −0.3529) = 1 − P(Z < 0.35)
= 0.3621. Expected number = 100 × 0.3621 = 36
b P(3.2  W  3.5) = P(−0.1176  Z  0.2353)
= 0.1398
c
3
4
Expected number = 100 × 0.1398 = 14
0.85
gives the interval [ 3.13,3.47 ]
3.3 ± 1.96
100
The test statistic T = 7.2 − 9 = − 4.495
1.2829
8
One-tailed test to the left, with p = 0.95 and v = 7;
the critical value of t = −1.895.
As −4.495 < −1.895, the test statistic T does not lie
in the acceptance region, so you should reject H0.
There is significant evidence to suggest that the
mean value of the random variable has decreased
from 9.
2
15.2
gives the interval [ 42.8,49.8 ]
50
b You are 90% confident that on average,
applicants can score between 42.8 and
49.8. Since the lower limit of the confidence
interval is greater than 42, that means
you are 90% confident that applicants could
achieve a mean score of 42 or higher.
a 46.3 ± 1.645
H0: m = 1.26
x=
a
H0 : m = 10 cm. The average length of leaves is
10 cm.
b H1 : m > 10 cm. The average length of leaves is
greater than 10 cm.
c
d Assume that the length of leaves is normally
distributed.
H1: m ≠ 1.26
x=
13.35
= 1.335
10
The test statistic z = 1.335 − 1.26 = 0.988
0.24
10
Two-tailed test, with p = 0.975, the critical values
are ± 1.96.
1
H0 : m = 9
57.6
x = 8 = 7.2
1
57.6 2 
s 2 = 7  423.7 − 8  = 1.2829


1
96 2 
s 2 =  1080 −
= 17.6
9
10 
9.6 − 10
= −0.302
17.6
10
A one-tailed test to the right, with p = 0.95
and m = 9. The critical value of t is 1.833.
The test statistic –0.302 < 1.833, lies within
the acceptance region. So you should accept
H0. There is no evidence to suggest that the
average length of leaves in Gemma’s garden is
greater than 10 cm.
3
H1: m < 9
96
= 9.6
10
The test statistic T =
As 0.988 < 1.96, the test statistic Z lies inside the
acceptance region, so you should accept H0.
There is no evidence to suggest a change in mean
growth of tomato plants.
Exercise 2.1A
Gemma should use the test statistic T,
because the population standard deviation is
unknown (and the sample size is small).
a H
0 : m = 165 cm. The average height of students
is 165 cm.
H1 : m > 165 cm. The average height of students
is greater than 165 cm.
b One-tailed test to the right.
c
Assume the height of students is normally
distributed.
21
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
x=
972
= 162
6
1
972 2 
s 2 =  157754.6 −
= 58.12
5
6 
The test statistic T =
162 − 165
= −0.964 .
58.12
6
One-tailed test to the right, with p = 0.99
and v = 5. The critical value of t is 3.365.
The test statistic –0.964 < 3.365, lies within the
acceptance region, so you should accept H0.
There is no evidence to suggest that the
average height of the students is greater than
165 cm.
4
Assuming the sample is normally distributed.
H0 : m = 10.5
H1: m > 10.5
2


127.3
= 12.73 s 2 = 1  2219.6 − 127.3  = 66.563
10
9
10


12.73 − 10.5
The test statistic T =
= 0.864
66.563
10
One-tailed test to the right, with p = 0.90 and
v = 9; the critical value of t = 1.383.
x=
As 0.864 < 1.383, the test statistic T lies inside the
acceptance region, so you should accept H0.
There is no evidence to suggest that the new
technology has increased the television lifetime.
207.7
1
207.7 2 
= 20.77 s 2 =  4361.33 −
= 5.267
10
9
10 
2
207.7
1
207.7 
x=
= 20.77 s 2 =  4361.33 −
= 5.267
10
9
10 
b H0 : m = 20
5
a x=
H1 : m > 20
The test statistic T =
20.77 − 20
= 1.061
5.267
10
One-tailed test to the right, with p = 0.975 and v =
9. The critical value of t is 2.262.
The test statistic 1.061 < 2.262, lies within the
acceptance region. So you should accept H0.
There is no evidence to suggest that the mean is
greater than 20.
6
H0: m = 138 ml
H1: m ≠ 138 ml
∑ x = 682 → x = 136.4
∑ x 2 = 93 246 → s 2 = 55.3
The test statistic T = 136.4 − 138 = − 0.481
55.3
5
Two-tailed test at the 5% significance level, with
p = 0.975 and v = 4; the critical values of t = ± 2.776.
As −2.776 < −0.481 < 2.776, the test statistic T lies
inside the acceptance region, so you should
accept H0. There is no evidence to suggest that
the mean volume is not as expected.
7
H0: m = 20 g
H1: m < 20 g
∑ x = 115 → x = 19.17
2
2
∑ x = 2231.26 → s = 5.419
The test statistic T = 19.17 − 20 = − 0.8734
5.419
6
One-tailed test to the left, with p = 0.05 and v = 5;
the critical value of t = −2.015 .
As −0.8734 > −2.015, the test statistic T lies inside
the acceptance region, so you should accept H0.
There is no evidence to suggest that the mean
weight of a pack of Brand A’s raisins is less than 20 g.
8
H0: m = 12 minutes
H1: m < 12 minutes
∑ x = 26 → x = 3.25
2
2
∑ x = 312 → s = 32.5
3.25 − 12
= − 4.341
The test statistic T =
32.5
8
One-tailed test at the 1% significance level,
with p = 0.01 and v = 7; the critical value of
t is −2.998.
As −4.341 < −2.998, the test statistic T does not lie
in the acceptance region, so you should reject H0.
There is evidence to suggest that the new mean
is less than 12 minutes. That means the new
schedule introduced by the control room works
better than the previous one during the
peak time.
9
1 H1 :Hm1 :≠m1≠ 1 kg
H0 H
: m0 :=m1=kg ∑ x = 10.54 → x = 1.054
∑ x 2 = 11.2202 → s 2 = 0.012 34
1.054 − 1 = 1.537
The test statistic T =
0.012 34
10
Two-tailed test at the 10% significance level, with
p = 0.95 and v = 9; the critical values of t = ±1.833.
As 1.537 < 1.833, the test statistic T lies within the
acceptance region, so you accept H0. There is no
evidence to suggest that the mean weight of bags
of vegetables has changed from 1 kg.
10
H0 : m = 50 grams
597
x=
= 49.75
12
H1 : m = 50 grams
1
597 2 
s 2 =  29 767 −
= 6.023
11 
12 
The test statistic T =
49.75 − 50
= −0.353
6.023
12
22
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
A two-tailed test, with p = 0.975 or p = 0.025;
v = 11. The critical values of t are ± 2.201.
lifespan of candles made at the two factories
are different. As the sample mean of factory X
is smaller than the sample mean from factory
Y, it also suggests that the mean lifespan of
candles produced by factory X is shorter.
The test statistic –0.353 is between –2.201 and
2.201, and lies within the acceptance region. So you
should accept H0. There is no evidence to suggest
that the mean weight differs from 50 grams.
b It is not necessary as both samples are larger
than 30. The Central Limit Theorem can be
applied and the unbiased estimators can be
used as the population variances.
Exercise 2.2A
1
a
Normal distribution test is suitable as the
population variances are known.
b H0 : m1 − m 2 = 0
H1: m1 − m 2 ≠ 0
The test statistic Z is
=
4
(4.06 − 3.91)
0.062 + 0.047
10
10
z = 1.437
Two-tailed test with p = 0.95; the critical values of
z = ±1.645.
Two-tailed test with p = 0.975; the critical
values of z = ±1.960.
2
As 1.437 < 1.645, the test statistic Z does lie
in the acceptance region, so accept H0. There is
no significant evidence to suggest that the mean
amounts of milk dispensed by the two machines
are different.
5
H0 : m M − m F = 0 H1: m M − m F > 0
xM =
256
= 3.2
80
2
sM
=
H1 : m1 – m2 > 0
1 
208 
208
s F2 =  731 −
= 2.408
= 2.6
79 
80 
80
The test statistic Z is
(3.2 − 2.6)
=
z = 2.564
1.972 + 2.408
80
80
One-tailed test to the right with p = 0.95; the
critical value of z = 1.645.
2

x1 = 2859 = 81.69 s12 = 1  235 425 − 2859  = 55.46
35 
34 
35
2
2

x 2 = 3052 = 87.20 s 2 2 = 1  268 450 − 3052  = 68.11
35 
34 
35
The test statistic Z =
As 2.564 > 1.645, the test statistic Z does not lie in
the acceptance region. So, you reject H0. There is
significant evidence to suggest that the mean
number of sales of the new cereal to men is
higher than to woman on that day.
3
The test statistic –2.93 < 1.65, lies within the
acceptance region. So accept H0. There is no
evidence to show that the warning signs reduce
road accidents.
H1: m X − mY ≠ 0
(305 − 309)
Z = −2.170
82.35 + 123.61
55
65
Two-tailed test with p = 0.975; the critical
values of z = ±1.960.
As −2.170 < −1.960, the test statistic Z does not
lie in the acceptance region. So, you reject H0.
There is evidence to suggest that the mean
81.69 − 87.20
= −2.93
55.46 + 68.11
35
35
A one-tailed test to the right with p = 0.95. The
critical value from the normal distribution table
is 1.645.
a H0 : m X − mY = 0
=
The test statistic Z is
Assume each accident that happened
was independent.
H0 : m1 – m2 = 0
1 
256 2 
975 −
= 1.972

79 
80 
xF =
H0 : m1 − m 2 = 0
H1: m1 − m 2 ≠ 0
The test statistic Z =
(23.4 − 19.8)
z = 0.4238
12 2 + 17 2
6
6
As 0.4238 < 1.960, the test statistic Z lies
within the acceptance region, so you accept
H0. There is no evidence to suggest that the
two random variables have different means.
2
6
Let the time spent on the internet by families
with children be X1 and the time spent by
families without children be X2.
H0 : m1 − m 2 60
H1: m1 − m 2 < 60
x1 = 728
s12 = 11 970.59
x 2 = 635
s12 = 49 929.66
23
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
(728 − 635) − 60
11970.59 + 49 929.66
35
30
Z
=
0.7367
One-tailed test to the left with p = 0.05; the
critical value of z = −1.645.
The test statistic Z =
9
H1 : mB – mG < 2
2

x B = 2474 = 82.47 s B 2 = 1  206 044 − 2474  = 69.71
30 
29 
30
2

xG = 2380 = 79.33 sG 2 = 1  191094 − 2380  = 78.64
30 
29 
30
0.7367 > −1.645, the test statistic Z lies within the
acceptance region. So, you accept H0. The
evidence supports the claim that families with
children spend at least sixty extra hours on the
internet than those without children in a year.
7
The test statistic Z =
Let the time taken by the old system be O.
H0 : mO − m N = 2
H1: mO − m N < 2
2
n = 4.3 s N = 2.56
The test statistic Z = (7 − 4.3) − 2
2.52 + 2.56
45
45
z
=
1
.
582
10
One-tailed test to the left with p = 0.05; the
critical value of z = –1.645.
As 1.582 > –1.645, the test statistic Z does lie in
the acceptance region, so accept H0. The hotel
manager’s claim is supported; there is at least
2 minutes improvement from the new
computer system.
8
(82.47 − 79.33) − 2 = 0.510
69.71 + 78.64
30
30
A one-tailed test to the left with p = 0.05. The
critical value from the normal distribution table
is –1.645.
The test statistic 0.510 > –1.645, lies within the
acceptance region, so accept H0. The evidence
supports the claim that boys scored at least 2
marks more than girls in the mock science exam.
The time taken by the new system is denoted by N.
H0 : mB – mG = 2
Let X1 be the size of flowers grown in high
nitrogen compost, and X2 be the size of flowers
grown in normal compost.
H0 : m1 – m2 = 1
a
H0 : m X − mY = 0
H1 : m X − mY ≠ 0
y = 19.99
(20.32 − 19.99)
0.36 2 + 0.36 2
35
40
z = 3.960
Two-tailed test with p = 0.95, the critical
values of z = ±1.645.
The test statistic Z =
3.960 > 1.645. The test statistic does not lie in
the acceptance region, so reject H0. There is
significant evidence to suggest that the mean
length has changed. After the machine has
been serviced, the mean length of flat-pack
components has decreased.
b H0 : m X − mY = 0
H1 : m X − mY ≠ 0
H1 : m1 – m2 < 1
1 
285.92 
=
= 0.6575
2357.75 −

34 
35 
285.9
x1 =
= 8.169
35
s12
244.9
x2 =
= 6.997
35
s2
2
The test statistic Z =
(8.169 − 6.997 ) − 1 = 0.764
1 
244.92 
=
= 1.118
1751.61 −
34 
35 
0.6575 + 1.118
35
35
A one-tailed test to the left with p = 0.10. The
critical value from the normal distribution table
is –1.282.
The test statistic, 0.764 > –1.282, lies within the
acceptance region, so accept H0. The evidence
supports the garden centre’s claim that flowers
grown in the high nitrogen compost are at least
1 cm bigger.
y = 19.99
s y2 = 0.0784
The test statistic Z =
(20.32 − 19.99)
0.36 2 + 0.0784
35
40
z = 4.39
Two-tailed test with p = 0.95, the critical
values of z = ±1.645.
4.39 > 1.645. The test statistic Z lies outside
the acceptance region, so reject H0. There is
significant evidence to suggest that the mean
length has changed after the service.
The second test is more reliable as the sample
variance is used.
24
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
Worked solutions
Exercise 2.2B
1
(8 − 1) × 3.7 + (15 − 1) × 2.9
= 3.167
8 + 15 − 2
b H0 : m X − mY = 0
a sp 2 =
H1 : m X − mY ≠ 0
The test statistic T =
As 0.8560 < 1.761, the test statistic T does lie in
the acceptance region. So, you accept H0. There
is no evidence to suggest that trains from
manufacturer A have greater fuel efficiency on
average than those from manufacturer B.
4
26.1 − 24.8
1
3.167 18 + 15
(
)
= 1.669
a
Two-tailed test with p = 0.975, v = 8 + 15 – 2 =
21; the critical values of t = ± 2.080.
x = 6.967
s x2 = 1.1
y = 6.556
s 2y = 0.8778
s p2 =
s p2 =
The test statistic T =
H1 : m X − mY > 0
(
)
One-tailed test with p = 0.95, v = 9 + 9 – 2 =16;
the critical value of t = 1.746.
As 0.877 < 1.746, the test statistic T lies within
the acceptance region. So, you accept H0.
There is no evidence to suggest that the
sunflowers planted in soil X are larger than
those planted in soil Y.
H0 : m A − m B = 0
sp 2 =
)
The five records are independent between Adam
and Bob. The population variances are the same.
H0 : m x A − m xB = 0
H1: m x A − m xB ≠ 0
x A = 49.3
s x A 2 = 0.0375
x B = 48.7
s xB 2 = 0.0625
s p2 =
(5 − 1) 0.0375 + (5 − 1) 0.0625 = 0.05
5+5−2
The test statistic T =
49.3 − 48.7 = 4.243
0.05 1 + 1
5 5
( )
Two-tailed test with p = 0.975, v = 8; the critical
values of t = ± 2.306.
H1: m A − m B > 0
(
= 0.499
As 0.0499 < 0.700, the test statistic T does lie
in the acceptance region, so accept H0. There
is no evidence to suggest that the new
fertiliser gives an increase in growth.
5
The test statistic T = 6.967 − 6.556 = 0.877
0.989 1 + 1
9 9
7.55 − 7.533
0.3488 1 + 1
6 6
One-tailed test to the right, with p = 0.75,
v = 10; the critical value of t = 0.700.
9+9−2
H0 : m X − mY = 0
x B = 3.8
(6 − 1) 0.619 + (6 − 1) 0.078 67
= 0.3488
6+6−2
H1 : m X − mY > 0
(9 − 1)1.1 + (9 − 1)0.8778 = 0.989
x A = 4.2
s y 2 = 0.07867
b H0 : m X − mY = 0
b Assume that both samples are independent
and randomly selected. They have the same
population variance.
3
s x 2 = 0.619
y = 7.533
a ∑ x = 62.7 ∑ x 2 = 445.61 ∑ y = 59 ∑ x 2 = 393.8
∑ x = 45.3 ∑ x 2 = 345.11 ∑ y = 45.2 ∑ y 2 = 340.9
x = 7.55
As −2.080 < 1.669 < 2.080, the test statistic t
lies within the acceptance region, so accept
H0. There is not enough evidence to suggest
that the two random variables X and Y have
different means.
2
Let X denote the height of young plants receiving
the new fertiliser. Y denotes the height of young
plants receiving the usual fertiliser.
s 2A = 0.49
s B2 = 1.257
(8 − 1) 0.43 + (8 − 1)1.257
= 0.8735
8+8−2
4.2 − 3 .8
The test statistic T =
= 0.8560
0.8735 1 + 1
8 8
One-tailed test with p = 0.95, v = 8 + 8 – 2 = 14; the
critical value of t = 1.761.
( )
As 4.243 > 2.306, the test statistic T does not lie in
the acceptance region, so you reject H0. There is
evidence to suggest that the average swimming
times of Adam and Bob are different.
6
H0 : md = 0 , there is no difference between the
population mean leaf widths.
H1 : md ≠ 0 , there is a difference between the
population mean leaf widths.
25
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
42
X d = 10 = 4.2
2
1  381.96 − ( 42 )  = 22.84
s d2 =

10 − 1 
10 
4.2
T =
= 2.779
22.84
10
Two-tailed test with p = 0.975, v = 9; the critical
values of t = ± 2.262.
One-tailed test to the right with p = 0.9, v = 9; the
critical value of t = 1.383.
As 2.78 > 2.262, the test statistic does not lie in the
acceptance region, so reject H0. There is evidence
to suggest that nitrogen affects leaf growth, as the
two samples indicate there is a difference in the
population mean leaf widths.
H1 : md ≠ 0
7
H0 : md 1
H1: md < 1
pH difference:
0.8
1.2
x d = 1.05
0.6
1.2
0.7
1.8
s d2 = 0.199
The test statistic T is
T =
1.05 − 1
= 0.2745
0.199
6
As 1.67 > 1.383, the test statistic does not lie in the
acceptance region, so you reject H0. There is
evidence to suggest that vitamins increase
attention span.
10
H0 : md = 0
Difference 7
H1: md ≠ 0
xd = −0.4571
1
a
A t-distribution with 15 degrees of freedom,
critical value = 2.131
c
A t-distribution with 22 degrees of freedom,
critical value = 2.819
d A normal distribution, critical value = 2.326
2
a Assuming sample is normally distributed.
xt = 5.3
−4.218 < −2.447, so the test statistic does not lie in
the acceptance region. So, you reject H0. There is
strong evidence to suggest that the wear on the
front and rear tyres is different.
H0 : md = 0
The test statistic T is T =
3.5
= 1.67
44.1
10
st2 = 2.161
A t-distribution with p = 0.975, v = 14, critical
value = 2.145
s d2 = 0.08219
H1: md > 0
3
b A t-distribution with 9 degrees of freedom,
critical value = 1.833
The test statistic T is
T = −0.4571 = −4.218
0.08219
7
Two-tailed test with p = 0.975, v = 6; the critical
values of t = ± 2.447.
9
−20
As −2.776 < −0.7567 < 2.776, the test statistic lies
within the acceptance region, so accept H0. There
is not sufficient evidence to suggest that the
percentages of bacteria from the two lakes
are different.
The sample is randomly selected; the differences
are normally distributed.
H0 : md = 0
−6
x = −2.4 s d2 = 50.3
d
−2.4
= −0.7567
The test statistic T is T =
50.3
5
Two-tailed test with p = 0.975, v = 4; the critical
values are t = ± 2.776.
0.2745 > −2.015, so the test statistic T lies within
the acceptance region. So, you accept H0. The
evidence suggests that the chemical reduces
pH value by at least 1.
−6
Exercise 2.3A
One-tailed test to the left with p = 0.05, v = 5; the
critical value of t = −2.015.
8
The differences are normally distributed.
5.3 ± 2.145
2.161
15
4.49 minutes  mean time  6.11 minutes
b You are 95% confident that the mean time of
completing the puzzle is between 4.49 and
6.11 minutes. That means the students can
complete the puzzle more quickly than the
time that the manufacturer suggested.
3
3.24 1.87
a (176 − 166) ± 1.645 40 + 32
9.39 cm  height difference  10.6 cm
2.5 + 2.9
b (80.3 − 67.6) ± 1.96
40 32
11.9 kg  mass difference  13.5 kg
26
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
c
(28.8 − 27.6) ± 2.326 4.6 + 5.9
40 32
−0.0727  BMI difference  2.47
9
a
Assume population variances of the times
spent watching TV by boys and girls are equal.
b
2.37 5.78
d (63 − 68) ± 2.576 40 + 32
−6.26 bpm heart rate difference −3.73bpm
4
Assume sugar content is normally distributed.
a
π = 3.13
b The confidence interval calculated is valid
because neither the distribution nor the
parameters used are approximated.
6
a
x = 501.7
s = 7.421
A t-distribution with p = 0.975, v = 5, critical
value = 2.571
7.421
501.7 ± 2.571
6
493.9 g  mean weight  509.5 g
6 × 2.2192 + 8 × 1.0142
7+9−2
( 17 + 19 )
c
10
You are 95% confident that, on average, the
difference between the times spent by boys
and girls watching TV each week lies in this
interval. As the interval contains the value
zero, there is no significant difference
between boys and girls.
Assume population variances of the study times
of first-year students and final-year students are
the same.
Xd = mean first-year study time – m
ean final-year
study time
4 × 1.3 + 4 × 0.98
The pooled estimate s p2 =
= 1.14
5+5−2
1 1
(3 − 2) ± 1.86 × 1.14 × +
5 5
( )
11
−0.256 xd 2.26
aAssume the pH values are normally
distributed.
x = 6.99
s = 0.4408
w = 5.5 sw2 = 1.692
A t-distribution with p = 0.975, v = 9, critical
value = 2.262
0.4408
6.99 ± 2.262
10
Let d = μM – μW
6.67  mean pH value  7.31
a m = 8.6
2
sm
= 7.575
(8.6 − 5.5) ± 1.96
b 0.95 × 60 = 57
7.575 1.692
+
600
800
12
2.86  d  3.34
b You are 95% confident that men spend more
money on breakfast at a café than women do
as both lower and upper limits are above the
value of zero.
8
sG = 1.014
−1.67 xd 1.88
b There is a 5% chance that the confidence
interval will not contain the population mean.
7
xG = 7.556
(7.657 − 7.556) ± 2.145 × 2.698 ×
s = 0.02915
A t-distribution with p = 0.975, v = 4, critical
value = 2.776
0.02915
3.13 ± 2.776
5
3.094  calculated value of π  3.166
s B = 2.219
The pooled estimate s p2 =
= 2.698
32.9 g  mean sugar content  39.1 g
5
x B = 7.657
Xd = μB − μG
s = 2.280
x = 36
A t-distribution with p = 0.99, v = 5, critical value
= 3.365
2.28
36 ± 3.365
6
2
2
2
a
The pooled estimate s p2 = 11 × 5.9 + 14 × 4.1
12 + 15 − 2
= 24.73
(
)
b (63 − 57) ± 1.708 × 24.73 × 1 + 1
12 15
2.71  difference in mean scores  9.29
Let Year 7 level of progress be the random
variable X, and Year 8 level of progress be the
random variable Y.
a
x = 2.28
s x2 = 0.907
y = 2.9 s y2 = 0.185
Xd = μX − μY
The pooled estimate
s p2 =
4 × 0.907 + 4 × 0.185 = 0.546
5+5−2
(2.28 − 2.9) ± 2.306 × 0.546 ×
( 15 + 15 )
−1.70 xd 0.458
27
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
b You are 95% confident that the difference in
mean levels of progress between Year 7 and
Year 8 is between −1.70 and 0.458. As the
interval contains the value 0, there is no
significant difference.
c
A normal distribution as populations are used.
Xd = μ7 – μ8
x 7 = 2.6
s 72 = 0.850
x 8 = 3.1
s 82 = 0.128
(2.6 − 3.1) ± 1.96 0.850 + 0.128
120
87
−0.681 xd −0.319
d 95% confidence interval found from the
population suggests that Year 7 students are
making less progress than Year 8, as the upper
and lower limits of the interval are both
negative values.
The interval calculated from part a does
suggest that there is no significant difference
between Year 7 and Year 8 level of progress.
However, since the upper limit of the interval
is only just above the value 0, this also
confirms the conclusion from part c.
Exam-style questions
1
H0 : m = 4.27
H1: m < 4.27
503.5 − 498
= 2.349
22.58 1 + 1
10 7
(
2.349 > 2.131. The test statistic T does not lie in
the acceptance region, so reject H0. There is
evidence to suggest that the population mean
weights of Bakery A’s white bread and brown
bread are different.
To test whether the mean weight of Bakery A’s
white bread is higher than the mean weight,
505 g, of Bakery B’s 50 bread:
50
H0 : µX = 505
H1 : µX > 505
x = 503.5
2
s x = 17.1
Test statistic T =
503.5 − 505
= −1.147
17.1
10
One-tailed test to the right with p = 0.95, v = 9
Critical value is t = 1.833
Since –1.147 <1.833, the test statistic T lies in the
acceptance region, so accept H0; there is no
evidence to suggest that Bakery A’s claim
is justified.
a A
ssuming the mobile phone signal strength is
normally distributed in city X and city Y.
4.023 − 4.27
= − 5.347
0.0256
12
One-tailed test to the left, with p = 0.01 and
v = 11; the critical value of t = −2.718.
x = −113.34
–5.347< −2.718. The test statistic T lies outside the
acceptance region, so reject H0. There is
significant evidence to suggest that the mean is
less than 4.27.
−0.835 m x − m y −0.625
The test statistic T =
2
Use a two-sample t-test, assuming both white
and brown bread weights are normally
distributed with the same variance.
Let X be the weight of white bread and Y the
weight of brown bread sold by Bakery A.
H0 : m X − mY = 0
H1: m X − mY ≠ 0
x = 503.5
s p2 =
s x2 = 17.1
y = 498
9 × 17.1 + 6 × 30.8
= 22.58
10 + 7 − 2
s 2y = 30.83
)
Two-tailed test with p = 0.975, v = 15; the critical
values of t = ±2.131
3
2
sm
= 0.02560
xm = 4.023
The test statistic T =
s x2 = 0.18
y = −112.61 s y2 = 0.052
(−113.34 − (−112.61)) ± 1.645 ×
( 060.18 + 0.50052 )
b H0 : m X − mY = 0
H1 : m X − mY ≠ 0
−113.34 − ( −112.61)
0.18 + 0.052
60
50
z = −11.49
The test statistic Z =
Two-tailed test with p = 0.99; the critical
values of z = ± 2.326.
As −11.49 < −2.326, the test statistic Z does not
lie in the acceptance region, so reject H0.
There is significant evidence to suggest that
the mean signal strengths in the two cities are
different.
28
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
4
A paired sample t-test
H0 : md = 0
H1 : md ≠ 0
y = 6.025
Difference
X d = −0.66
1
2
3
4
5
−3
0.4
−1
−1.5
1.8
b No ladybird with nine spots was found in
forest B. However, a few of them were found
in forest A. Therefore, forest A is more likely to
be located in the north-eastern United States.
s d = 1.835
7
sm = 0.2130
xw = 5.795
sw = 0.439
∑(x − x )2 = 0.67
)
(2.67 − 3.2) ± 1.812 ×
∑(y − y )2 = 1.84
( )
94
1 1
× +
375
3 9
−1.13 m x − m y 0.0748
8
H0 : mm − mw = 2.20
a H0 : mr = 36.5
H1 : mr ≠ 36.5
H1: mm − mw ≠ 2.20
The test statistic T = (7.934 − 5.795) − 2.2 = −0.395
0.119 1 + 1
10 10
Two-tailed test with p = 0.995, v = 18; the
critical values of t = ± 2.878.
(
)
−2.878 < –0.395 < 2.878. The test statistic lies
within the acceptance region, so accept H0.
There is evidence to suggest that the mean
distance of long jump of men is 2.20 m greater
than that of women.
a ∑ x = 0 × 8 + 2 × 5 + 7 × 24 + 9 × 7 + 13 × 6 = 319
∑ x 2 = 0 × 8 2 + 2 × 52 + 7 × 24 2 + 9 × 7 2 + 13 × 6 2 = 4991
2
5 + 7 × 24 + 9 × 7 +
13 × 6 2 = 4991
2
x = 6.38
y = 3.2
t-distribution with p = 0.95, v = 10, critical
value = 1.812
9 × 0.21302 + 9 × 0.4392
=
= 0.119
10 + 10 − 2
1.81 mm − mw 2.46
2
28.82
3n
b x = 2.67
(
6
∑(y − y )2 = 94 −
n = 3 or n = 113.2 (reject)
1
1
+
(7.934 − 5.795) ± 2.101 × 0.119 × 10
10
c
82
n
1128n 2 − 131064n + 383040 = 0
The pooled estimate
s p2
∑ ( x − x )2 = 22 −
94 ( 4n − 2 )
1021.44
= 116 −
375
3n
a Population variances are equal.
xm = 7.934
a

82  
28.82 
 22 − n  +  94 − 3n 
94
=
375
n + 3n − 2
−2.132 < −0.804 < 2.132. The test statistic T lies
within the acceptance region, so accept H0.
There is not sufficient evidence to suggest that
the mean running time over the two courses
is different.
b
 7.7672 9.0992 
 50 + 40 
−3.20 x − y 3.90
The test statistic T = −0.66 = − 0.804
1.835
5
Two-tailed test with p = 0.95, v = 4; the critical
values of t = ± 2.132.
5
s x = 9.099
(6.38 − 6.025) ± 1.96 ×
The differences, course 1 – course 2, are
Runner
2
s x = 7.767
xr = 34.8 sr 2 = 15.46
34.8 − 36.5
= −1.675
15.46
15
Two-tailed test with p = 0.975, v = 14, critical
values = ± 2.145.
The test statistic T =
−2.415 < −1.675 < 2.415. The test statistic T
lies within the acceptance region, so accept
H0. There is no evidence to suggest that the
mean is not 36.5.
b A t-distribution with p = 0.975, v = 14, critical
value = 2.145
15.46
15
32.6 mr 37.0
34.8 ± 2.145
∑ y = 0 × 4 + 2 × 7 + 7 × 25 + 9 × 0 + 13 × 4 = 241
2
2
2
2
2
2
∑ y = 0 × 4 + 2 × 7 + 7 × 25 + 9 × 0 + 13 × 4 = 4681
7 2 + 7 × 252 + 9 × 0 2 + 13 × 4 2 = 4681
29
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
9
One-tailed test to the left with p = 0.05; the
critical values of z = −1.645.
−1.104 > −1.645. The test statistic Z lies within
the acceptance region, so accept H0. There is
no evidence to suggest that college A students
took less time than college B students.
a A paired sample t-test
H0 : md = 0
H1 : md > 0
The differences before – after are:
Staff
Difference
A
4
X d = 3.5
s d = 7.672
B
1
C D E
4 −2 0
F G H
10 18 −7
3.5
The test statistic T = 7.672 = 1.29
8
One-tailed test with p = 0.90, v = 7, critical
value t = 1.415.
b Calculate value of Z =
12
H1: m < 28
s = 1.502
27.44 − 28
The test statistic T = 1.502 = –0.9133
6
One-tailed test to the left with p = 0.05, v = 5,
critical value t = −2.015.
−0.9133 > −2.015. The test statistic lies in the
acceptance region, so accept H0. There is no
evidence to suggest that the mean completion
time for the swimmers is less than 28 seconds.
b p = 0.975, v = 9; the critical value of t = 2.262
m − 2.262 s = 26.07
10
m + 2.262 s = 28.17
10
m = 27.12 s 2 = 1.468
H1: m > 15
x = 15.95
b 15.95 ± 2.571 1.629
6
14.2 mm m 17.7 mm
11
a H0 : m A − m B = 0
H1 : m A − m B < 0
t A = 5.7
s 2A = 0.7838
t B = 5.9
s B2 = 0.2669
The test statistic Z is
=
(5.7 − 5.9)
0.7838 + 0.2669
30
40
z = −1.104
a H0 : m = 28
x = 27.44
a H0 : m = 15
1.428 < 2.015. The test statistic lies in the
acceptance region, so accept H0. There is no
evidence to suggest that the mean tail length
of new-born mice is greater than 15 mm.
= 0.5522.
1 – 0.7095 = 0.2905, therefore β > 29.05%
b The mean number of absent hours is reduced
by 3.5 hours from eight staff. However, at the
10% significance level, the test statistic is not
significant enough to suggest that the number
of hours of absence is reduced. As it costs the
agency to run a kids’ club, it is recommended
not to do it in the coming year.
s = 1.629
The test statistic T = 15.95 − 15 = 1.428
1.629
6
One-tailed test to the right with p = 0.95, v = 5,
critical value t = 2.015.
0.7838 + 0.2669
30
40
Gives probability 0.7095.
1.29 < 1.41. The test statistic T lies within the
acceptance region, so accept H0. There is not
sufficient evidence to suggest that the holiday
kids, club reduced the absence rate.
10
(5.9 −5.7 ) − 0.1
13
Assume the numbers of people coming to the
gym on different days are independent.
Population variances before and after extended
opening hours are the same.
Let X1 be the number of people using the gym
each day when the gym is open for 12 hours, and
X2 be the number of people when the gym is
open for 18 hours.
H0 : m1 – m2 = 0
H1 : m1 – m2 < 0
x1 =
407
= 40.7
10
1
407 2 
s12 =  18125 −
= 173.34
9
10 
x2 =
511
= 51.1
10
1
5112 
s2 2 =  28109 −
= 221.88
9
10 
sp2 =
(9)(173.34 ) × (9)(221.88) = 197.61
18
The test statistic T =
40.7 − 51.1
= −1.65
1 + 1
197.61 10
10
(
)
30
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
Worked solutions
A one-tailed test to the left with p = 0.05, v = 18.
The critical value is –1.734. The test statistic,
–1.65 > –1.734, lies within the acceptance region,
so accept H0. There is not sufficient evidence to
suggest that more people use the gym when the
gym is open for 18 hours.
14
A one-tailed test to the right with p = 0.90. The
critical value is 1.282. The test statistic,
2.070 > 1.282, lies outside the acceptance region,
so reject H0. There is significant evidence to show
that the new menu increases sales.
16
a Let X1 be the time needed to solve the puzzle
without any training and X2 be the time
needed with training.
H0 : m1 – m2 = 0
b Let X1 be the height of the boys and X2 be the
height of the girls.
H1 : m1 – m2 > 0
x1 =
19
= 3.8
5
s12 =
17.4
x2 =
= 3.48
5
sp2 =
a A
suitable test would be a two-sample t-test.
Assume that the height of each group of
children is independent. Population variances
are the same.
s2
2
H0 : m1 – m2 = 2
1
192 
75.34 −
= 0.785
4 
5 
H1 : m1 – m2 < 2
( 4 )(0.785) × ( 4 )(0.112 ) = 0.4485
sp 2 =
8
The test statistic T =
3.8 − 3.48
= 0.7555
0.4485 1 + 1
5 5
( )
sp2 =
sp2 =
( 4 )(0.785) × ( 4 )(0.112 ) = 0.4485
c
(112.8 − 111) − 2 = −0.1255
(
6.35 15 + 15
)
For a 90% confidence interval, p = 0.95 and
v = 8. The critical values are ± 1.860.
( )
17
Assume the sales made on different days
are independent.
Let X be the random variable of new menu sales
and Y be the random variable of old menu sales.
H0 : mx – my = 0
a A
ssume the weight of each pack of potatoes is
independent and normally distributed.
Population variances are the same.
Let X1 be the weight of King Edward potatoes,
and X2 be the weight of salad potatoes.
H0 : m1 – m2 = 0
H1 : m1 – m2 ≠ 0
H1 : mx – my > 0
311
X2 =
= 10.37
30
8
–1.16 cm  μ1 – μ2  4.76 cm
– 0.468 minutes  μ1 – μ2  1.11 minutes
343.8
= 11.46
30
( 4 )(5.2 ) × ( 4 )( 7.5) = 6.35
(112.8 − 111) ± 1.860 × 6.35 15 + 15
( 3.8 − 3.48) ± 1.860 × 0.4485 15 + 15
X=
2

s 2 2 = 1  61 635 − 555  = 7.5
5 
4
The test statistic, –0.1255 < –1.860, lies within
the acceptance region, so accept H0. There is
sufficient evidence to suggest that the boys
are taller than the girls by at least 2 cm.
8
The confidence interval for the difference in
means:
15
= 0.002139
A one-tailed test to the left with p = 0.05,
v = 8. The critical value is –1.860.
19
17.4
= 3.8 x2 =
= 3.48
5
5
( )
14
The test statistic T =
b Assume the times spent completing the
puzzle are independent. Population variances
with and without training are the same.
From part a, x1 =
(7 ) (0.003 343) × (7 ) (0.000 9357 )
X 2 = 555 = 111
5
A one-tailed test to the right with p = 0.95,
v = 8. The critical value is 1.860. The test
statistic, 0.7555 < 1.860, lies within the
acceptance region, so accept H0. There is not
sufficient evidence to suggest that the training
improves the time taken to solve the puzzle.
c
2

s12 = 1  63 640 − 564  = 5.2
4
5 
X 1 = 564 = 112.8
5
1
17.42 
=  61 −
= 0.112
4
5 
sx2 =
sy2
The test statistic Z =
2
1
343.8
= 4.451
4069.04 −
29 
30 
1
3112 
=  3336.29 −
= 3.871
29 
30 
11.46 − 10.37
= 2.070
4.451 + 3.871
30
30
x1 = 8.4 = 1.05
8
2

s12 = 1  8.843 4 − 8.4  = 0.003 343
8 
7
2

x 2 = 8.18 = 1.0225 s 2 2 = 1  8.3706 − 8.18  = 0.0009357
8 
7
8
The test statistic
1.05 − 1.0225
= 1.189
0.002139 1 + 1
8 8
T =
( )
31
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2 Inference using normal and t-distributions
b Let X be the length of a piece before and Y the
length after the machine is serviced.
A two-tailed test with p = 0.05 and p = 0.95.
v = 14. The critical values are ± 1.761.
H0 : µ X − µY = 0
The test statistic –1.761 < 1.189 < 1.761 lies
within the acceptance region, so accept H0.
There is not sufficient evidence to suggest
that the weights of the two types of potatoes
are different.
H1 : µ X − µY ≠ 0
∑ x = 6.04 ∑ x 2 = 6.0864
b For a 95% confidence interval, p = 0.975 and
6.04
1
6.04 2 
x=
= 1.006667 s x2 =  6.0864 −
= 0.0012267
6
5
6 
v = 14. The critical values are ± 2.145.
2
6.04
1
6.04 
2
= 0.0012267
x=
= 1.006667
1 + 1s x = 5  6.0864 − 6
6 
.
−
.
.
.
1
05
1
0225
±
2
145
×
0
002139
(
)
8 8
∑ y = 6.01 ∑ y 2 = 6.0223
( )
–0.0221 kg  μ1 – μ2  0.0771 kg
18
x=
6.01
1
6.012 
= 1.001667 s x2 =  6.0223 −
= 0.0004567
6
5
6 
a A
paired sample t-test. Assume that the
6.01
1
6.012 
difference betweenx the
of pain
1.001667
= 0.0004567
= two=types
s x2 relief
=  6.0223 − 6
5
6 
tablets is normally distributed.
5 × 0.0012267 + 5 × 0.0004567
s p2 =
= 0.00084167
b Let d = Tablet A time – Tablet B time H0 : md = 0
6+6−2
H1 : md < 0
6.04 − 6.01
= 1.791
0.00084167 16 + 16
Test statistic T =
(
Differences: 4 1.5 0 –1.5 1.5 –2.5
xd =
3
= 0.5
6
1
32 
sd 2 =  29 −  = 5.5
5
6
Two-tailed test with p = 0.025, 0.975 and v = 10
Critical values are t = ±2.228
0.5 − 0
= 0.5222
The test statistic T =
5.5
6
Since −2.228 < 1.791 < 2.228, the test statistic
lies within the acceptance region, so accept
H0. There is no significant difference between
the mean lengths before and after the
machine’s service.
A one-tailed test to the left with p = 0.95.
v = 5. The critical value is –2.015.
The test statistic, 0.5225 > –0.2015, lies within
the acceptance region, so accept H0. There is
not sufficient evidence to suggest that
Tablet A is more efficient than Tablet B.
c
19
20
a L
et X1 be the 11-year-olds’ progress and
X2 be the 12-year-olds’ progress.
x1 = 0.84 s12 = 0.148
Tablet A, x A =
52.5
= 8.75
6
s A = 2.44
x2 = 1.16 s22 = 1.528
Tablet B, x B =
49.5
= 8.25
6
s B = 0.524
sp2 =
The test statistic T in part b can be used to
show that Tablet B is more efficient than
Tablet A at the 5% significance level. Also,
tablet B has a smaller sample mean and
standard deviation. Tablet B would be
recommended.
aA two-sample t-test. Assume that the samples
of length measurements before and after the
machine’s service are independent, that each
is taken from a normally distributed
population, and that the two populations
have the same variance.
)
( 4 )(0.148) × ( 4 )(1.528) = 0.838
8
The confidence interval:
( )
(0.84 − 1.16 ) ± 2.306 × 0.838 15 + 15
–1.655  μ1 – μ2  1.015
b The confidence interval contains the value 0,
so there is no significant difference between
the two groups of children.
32
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
2
Mathematics in life and work
1
A 2-sample t-test
Let H denote the goals scored in home matches
and A denote the goals scored in away matches
H0 : m H − m A = 0
H1 : m H − m A > 0
x H = 2.33
s H 2 = 3.87
x A = 1.83
s A 2 = 1.37
s p 2 = 5 × 3.87 + 5 × 1.37 = 2.62
6+6−2
2.33 − 1.83
The test statistic T =
= 0.535
2.62 1 + 1
6 6
One-tailed test to the right, p = 0.95, v = 10; the
critical value of t = 1.812.
(
)
0.535 < 1.812. The test statistic lies in the
acceptance region, so accept H0. There is not
enough evidence to show that Liverpool plays
better at a home match than an away match.
2
For home matches, ∑ x = 14 + 19 = 33
x H = 2.06
Pooled variance estimate of home matches
2


5 × 1.97 +  49 − 19

10

=
= 1.625
6 + 10 − 2
For away matches, ∑ x = 11 + 11 = 22
x A = 1.375
Pooled variance estimate of away matches
2


5 × 1.17 +  21 − 11
10 

= 1.054
=
6 + 10 − 2
Pooled estimate of home and away matches
15 × 1.625 + 15 × 1.054
s p2 =
= 1.3395
16 + 16 − 2
95% confidence interval: p = 0.975, v = 30,
t = 2.042
(2.06 − 1.375) ± 2.042 × 1.3395 × 1 + 1
16 16
−0.15 m H − m A 1.52
From part b, the confidence interval contains 0,
so you are 95% confident that there is no
difference between how Liverpool play at home
or away; both parts suggest that there is not
enough statistical evidence to support the claim
that Liverpool plays better at home.
(
)
33
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
-Tests
3 χ2-tests
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in
this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
a The total frequency is 25 + 31 + ⋅⋅⋅ + 1 = 100
1
1 × 25 + 2 × 31 + ⋅ ⋅ ⋅ + 8 × 1) = 2.5
100 (
x=
{(
}
)
1
189
12 × 25 + 2 2 × 31 + ⋅ ⋅ ⋅ + 82 × 1 − 100 × 2.52 =
= 1.91
99
99
s2 =
b The total frequency is 13 + 25 + 32 + 5 = 75
1
566
10 × 13 + 30 × 25 + 50 × 32 + 70 × 5 ) =
= 37.7 (3 s.f.)
75 (
15
x=
s2 =
2
a
( )
2
1  2
566 
10 × 13 + 302 × 25 + 502 × 32 + 702 × 5 − 75 ×
= 291

74 
15 
(
)
 10
10 − 2
P ( X = 2 ) =   0.32 (1 − 0.3 )
= 0.233
 2
i
ii E(X) = 10 × 0.3 = 3
P (150 < Y 200 ) = Φ
b i
( 20050− 260 ) − Φ(15050− 260 )
= Φ(–1.2) – Φ(–2.2) = (1 – Φ(1.2)) – (1 – Φ(2.2))
= (1 – 0.8849) – (1 – 0.9861) = 0.101
ii
3
Var(Y) = 502 = 2500
Performing a paired t-test, assume the differences between sample A and sample B are normally
distributed. Take differences to be sample A minus sample B.
Differences 2, – 7, 6, 5, 3, 7, 4, 0
d=
1
2 − 7 + ⋅ ⋅ ⋅ + 0 ) = 2.5
8(
((
)
)
sd2 = 71 2 2 + (−7)2 + ⋅ ⋅ ⋅ + 02 − 8 × 2.52 = 138
7
H0: μd = 0
H1: μd ≠ 0
2.5 − 0
= 1.593. Critical value: t7 (2.5%) = 2.365.
19.71
8
As 1.593 < 2.365 there is no reason to doubt H0. There is insufficient evidence at the 5% significance level to
5% significance level, two-tailed test. The test-statistic T =
suggest the means of sample A and sample B are different.
34
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
3
Worked solutions
Exercise 3.2A
1
H0: Distribution of flights is as Yusuf claims.
H1: Distribution of flights is not as Yusuf claims.
5% significance level, degrees of freedom: 4 − 1 = 3, critical value 7.815
On time
Under 30 mins
Over 30 mins
Cancelled
Probability
0.5
0.2
0.2
0.1
Observed
35
10
3
2
Expected
25
10
10
5
4
0
4.9
1.8
(Ok − E k )
Ek
2
X 2 = 10.7 > 7.815. Therefore reject H0; the distribution of flight departure times is not as Yusuf claims.
2
H0: Number rolled on the dice can be modelled by a uniform distribution.
H1: Number rolled on the dice cannot be modelled by a uniform distribution.
10% significance level, degrees of freedom: 6 − 1 = 5, critical value 9.236
1
2
3
4
5
6
Probability
1
6
1
6
1
6
1
6
1
6
1
6
Observed
29
44
38
34
48
47
Expected
40
40
40
40
40
40
3.025
0.4
0.1
0.9
1.6
1.225
(Ok − E k )
Ek
2
X2 = 7.25 < 9.236. Therefore, no reason to doubt H0, the dice are fair.
3
a H
0: The distribution of ‘shiny’ stickers in packs can be modelled by B(8, 0.2).
H1: The distribution of ‘shiny’ stickers in packs cannot be modelled by B(8, 0.2).
5% significance level.
0
1
2
3
4
5
6
0.1678
0.3355
0.2936
0.1468
0.045 88
0.009 175
0.001 147
8.192
× 10–5
2.560
× 10–6
Observed
32
43
40
21
10
3
1
0
0
Expected
25.17
50.33
44.04
22.02
6.881
1.376
0.1720
0.012 29
0.000 384 0
Probability
7
8
Combining columns 4 to 8:
0
1
2
3
4 or more
Observed
32
43
40
21
14
Expected
25.17
50.33
44.04
22.02
8.442
1.856
1.068
0.3706
0.047 26
3.659
(Ok − E k )
Ek
2
Degrees of freedom 5 − 1 = 4, critical value 9.488
X 2 = 7.001 < 9.488. Therefore no reason to doubt H0, the distribution of shiny stickers in packs is B(8, 0.2).
b For the test statistic to be approximately described by the χ2-distribution, expected values must be
greater than 5.
35
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
4
-Tests
a A
geometric distribution would be suitable if the probability of success is fixed and if successes
occur independently.
b Expected frequencies, under Geo(0.4)
Probability
Expected
c
1
2
3
4
5
6
7 or more
0.4
0.24
0.144
0.0864
0.051 84
0.031 10
0.046 66
8.64
5.184
3.110
1.866
2.799
24
14.4
H0: First sale of the day can be modelled by Geo(0.4).
H1: First sale of the day cannot be modelled by Geo(0.4).
2.5% significance level. Combine the classes 5, 6 and 7 or more.
1
2
3
4
5 or more
Observed
10
20
16
7
7
Expected
24
14.4
8.64
5.184
7.776
(Ok − E k )2
Ek
8.167
2.178
6.270
0.6362
0.077 44
Degrees of freedom 5 − 1 = 4, critical value 11.14
X 2 = 17.33 > 11.14. Therefore reject H0; the distribution for the first sale of the day cannot be modelled
by Geo(0.4).
5
a H0: Defective parts can be modelled by Po(2.5).
H1: Defective parts cannot be modelled by Po(2.5).
1% significance level
0
1
2
3
4
5
6 or more
Prob
0.082 08
0.2052
0.2565
0.2138
0.1336
0.066 80
0.042 02
Obs
28
49
50
44
16
8
5
16.42
41.04
51.30
42.75
26.72
13.36
8.404
8.172
1.543
0.033 10
0.036 40
4.301
2.151
1.379
Exp
(Ok − E k )
Ek
2
Degrees of freedom 7 − 1 = 6, critical value 16.81
X 2 = 17.62 > 16.81. Therefore reject H0; defective parts cannot be modelled by Po(2.5).
b Defects should occur singularly and randomly.
6
H0: Sarah’s cat’s food preference can be modelled by a uniform distribution.
H1: Sarah’s cat’s food preference cannot be modelled by a uniform distribution.
5% significance level, degrees of freedom: 5 − 1 = 4, critical value 9.488
Turkey
Fish
Chicken
Lamb
Beef
Prob
0.2
0.2
0.2
0.2
0.2
Obs
4
7
3
6
10
6
6
6
6
6
2
3
1
6
3
2
0
8
3
Exp
(Ok − E k )
Ek
2
X 2 = 5 < 9.488. Therefore, no reason to doubt H0, the cat does not have a preference.
36
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
7
3
H0: The number of goals scored in a penalty shootout can be modelled by B(5, 0.7).
H1: The number of goals scored in a penalty shootout cannot be modelled by B(5, 0.7).
5% significance level
0
1
2
3
4
5
Prob
0.002 43
0.028 35
0.1323
0.3087
0.3602
0.1681
Exp
0.243
2.835
13.23
30.87
36.02
16.81
Combine 0, 1 and 2:
2 or fewer
Obs
Exp
(Ok − E k )
Ek
2
3
4
5
22
40
27
11
16.31
30.87
36.02
16.81
1.987
2.700
2.257
2.006
Degrees of freedom 4 − 1 = 3 , critical value 7.815
X 2 = 8.950 > 7.815. Therefore reject H0; goals scored in a penalty shootout cannot be modelled by B(5, 0.7).
8
(
)
12
25
1 1 1
+ +
= 1, therefore 12 k = 1 so k = 25 .
2 3 4
b Expected frequencies for sample of size 50.
a The sum of the probabilities is 1. k 1 +
r
P(X = r)
1
2
3
4
0.48
0.24
0.16
0.12
24
12
8
6
Er
c
H0: The data can be modelled by the random variable X.
H1: The data cannot be modelled by the random variable X.
10% significance level, Degrees of freedom 4 − 1 = 3, critical value 6.251
3 1 1 3 43
+
+ + =
2 12 2 2 12
X 2 = 3.583 < 6.251. Therefore, no reason to doubt H0, the random variable X is a good model for the data.
X2 =
9
a
b
c
(15050− 260 ) = Φ(−2.2) = 1 − 0.9861 = 0.0139
200 − 260
0.1012 using sum of probabilities.
P(150 X < 200) = Φ (
) − 0.0139 = Φ(−1.2) − 0.0139 = 0.1012
50
P(X < 150) = Φ
H0: the finishing times can be modelled by N(260, 50)
H1: the finishing times cannot be modelled by N(260, 50)
Time
under 150
150–200
200–250
250–300
over 300
Prob
0.0139
0.1012
0.3057
0.3674
0.2119
Obs
20
83
373
476
298
Exp
17.38
126.5
382.1
459.3
264.82
(Ok − E k )2
Ek
0.3952
14.93
0.2162
0.6105
4.158
5% significance level, degrees of freedom 5 − 1 = 4, critical value 9.488
X 2 = 20.31 > 9.488. Therefore reject H0; there is evidence that the finishing times cannot be modelled by
N(260, 50).
37
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
-Tests
Exercise 3.2B
1
a H0: The number of buses arriving before Nury’s can be modelled by a Poisson distribution.
H1: The number of buses arriving before Nury’s cannot be modelled by a Poisson distribution.
0 × 4 + 1 × 13 + 2 × 10 + 3 × 3
= 1.4, hence λ = 1.4
30
Expected frequencies under Po(1.4)
b Sample mean: x =
c
0
1
2
3
Probability
0.2466
0.3452
0.2417
0.1128
0.053 73
Expected
7.398
7.250
3.383
1.612
10.36
4 or more
d Groups 2, 3 and 4 must be combined in order to get expected frequencies greater than 5.
e
Three (combined) groups, two constraints, degrees of freedom 3 − 2 = 1.
f
5% significance level, critical value 3.841
0
Obs
Exp
(Ok − E k )
Ek
2
1
2 or more
4
13
13
7.398
10.36
12.25
1.561
0.6744
0.046 55
X 2 = 2.282 < 3.841. Therefore, no reason to doubt H0, the number of buses arriving before Nury’s bus can
be modelled by a Poisson distribution.
2
a Sample mean: x =
b p=1÷7= 3
3 7
c Probabilities
1 × 51 + 2 × 46 + 3 × 29 + … + 8 × 1 7
=
150
3
1
2
3
4
5
6
7
8
9 or more
0.4286
0.2449
0.1399
0.079 97
0.045 69
0.026 11
0.014 92
0.008 526
0.011 37
d Expected frequencies
1
2
3
4
5
6
7
8
9 or more
64.29
36.73
20.99
12.00
6.854
3.917
2.238
1.279
1.705
e
Combined groups 6, 7, 8 and 9 or more. Degrees of freedom: 6 combined groups minus 2 constraints, so 4.
f
H0: The number of darts taken to hit a double can be modelled by a geometric distribution.
H1: The number of darts taken to hit a double cannot be modelled by a geometric distribution.
5% significance level, critical value 9.488
X 2 = 2.746 + 2.337 + 3.056 + 0.082 54 + 0.1065 + 0.501 = 8.83
X 2 = 8.83 < 9.488. Therefore accept H0, the number of darts required to hit a double can be modelled by
a geometric distribution.
g
3
The geometric distribution requires the probability of success to remain constant. As the dart player
throws more darts, they are probably more likely to hit a double through practice. Therefore, it is
unlikely this condition would be met.
0 × 49 + 1 × 64 + 2 × 34 + 3 × 3
= 0.94, hence λ = 0.94
150
b H0: The visits by patients can be modelled by a Poisson distribution.
H1: The visits by patients cannot be modelled by a Poisson distribution.
a Sample mean: x =
38
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
c
3
Expected frequencies
Prob
Exp
0
1
2
3
4 or more
0.3906
0.3672
0.1726
0.054 07
0.015 53
8.111
2.329
58.59
55.08
25.89
d Combine groups 3 and 4 or more. 1% significance level, degrees of freedom 4 − 2 = 2, critical value 9.210
X 2 = 1.571 + 1.445 + 2.543 + 5.302 = 10.86
X 2 = 10.86 > 9.210. Therefore reject H0; the visits by patients cannot be modelled by a Poisson
distribution. As a generic Poisson distribution is not a good fit, it is not surprising that in the example
the null hypothesis is rejected as well.
4
H0: The number of goals scored in a penalty shootout can be modelled by a binomial distribution.
H1: The number of goals scored in a penalty shootout cannot be modelled by a binomial distribution.
Sample mean: x = 0 × 3 + 1 × 6 + … + 5 × 11 = 3.15, so an estimate for p is 3.15 ÷ 5 = 0.63
100
0
1
Prob
0.006 934 0.059 04
Exp
0.6934
5.904
2
3
0.2010
0.3423
20.10
34.23
2
3
4
5
0.2914
0.099 24
29.14
9.924
Combine 0 and 1:
1 or fewer
Obs
Exp
(Ok − E k )
Ek
2
4
5
9
13
40
27
11
6.597
20.10
34.23
29.14
9.924
0.8753
2.510
0.9721
0.1576
0.1166
5% significance level, degrees of freedom 5 − 2 = 3, critical value 7.815
X 2 = 4.632 < 7.815. Therefore, no reason to doubt H0, goals scored in a penalty shootout can be modelled by
a binomial distribution. This reverses the result from the previous question; a 70% chance of scoring was
too high, with this sample suggesting a 63% chance would be more appropriate.
5
a c + (c + d ) + (c + 2d ) + (c + 3d ) + (c + 4d ) = 1
5c + 10d = 1
c + 2d = 0.2
b E ( X ) = 0 × c + 1 × ( c + d ) + … + 4 × ( c + 4d ) = 10c + 30d
0 × 12 + 1 × 20 + … + 4 × 3
= 1.5
60
c
x=
d
c + 2d = 0.2 
 ⇒ c = 0.3, d = −0.05
10c + 30d = 1.5 
e
H0: The data can be modelled by the random variable X.
H1: The data cannot be modelled by the random variable X.
5% significance level, degrees of freedom 5 − 2 = 3, critical value 7.815
0
1
2
3
4
P(X = r)
0.3
0.25
0.2
0.15
0.1
Or
12
20
17
8
3
18
15
12
9
6
2
5
3
25
12
1
9
3
2
r
Er
(Or − E r )
Er
2
X 2 = 7.361 < 7.815. Therefore, no reason to doubt H0, the random variable X is a good fit for the data.
39
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
-Tests
6
a x = 38.68 (by symmetry), y = 160 − 2 × 0.9935 + 9.696 + 38.68) = 61.26
Alternative method:
( 5010− 45 ) − Φ ( 4010− 45 ) = 160 × (2 Φ (0.5) − 1) = 61.28 . Actual value using spreadsheet, y = 61.27.
y = 160 ×  Φ

(You get slightly different answers, due to the fact that normal tables round probability values to 4 d.p.)
b H0: Plant growth can be modelled by N(45, 102).
H1: Plant growth cannot be modelled by N(45, 102).
5% significance level, combine ‘20 or less’ and ‘20–30’, and combine ‘more than 70’ and ‘60–70’; degrees
of freedom 5 − 1 = 4, critical value 9.488
X 2 = 0.009 040 + 2.421 + 0.083 96 + 0.002701 + 11.97 = 14.49
X 2 = 14.49 > 9.488. Therefore reject H0; plant growth cannot be modelled by N(45, 102).
c
x=
25 × 11 + 35 × 29 + 45 × 59 + 55 × 39 + 65 × 22
= 47mm
160
d z = 160 − ( 0.5547 + 6.576 + … + 1.716 ) = 13.77
Alternative method: z = 160 ×  Φ

e
( 7010− 47 ) − Φ ( 6010− 47 ) = 160 × (Φ(2.3) − Φ(1.3)) = 13.78
H0: Plant growth can be modelled by N( μ, 102).
H1: Plant growth cannot be modelled by N( μ, 102).
5% significance level, combine ‘20 or less’ and ‘20–30’, and combine ‘more than 70’ and ‘60–70’; degrees
of freedom 5 − 2 = 3, critical value 7.815
X 2 = 2.102 + 0.2108 + 0.021 99 + 0.9677 + 2.738 = 6.0404
X 2 = 6.0404 < 7.815. Therefore, no reason to doubt H0, plant growth can be modelled by N(μ, 102) with μ
estimated to be 47.
7
a At x = 3, F(3) = 3a + 9b = 1
b f(x) = F ′(x) = a + bx2 for 0  x  3, 0 otherwise
3
E ( X ) = ⌠ x f ( x ) dx =
⌡0
3
 ax 2 bx 4 
9a 81b
3
∫0 ax + bx dx =  2 + 4 0 = 2 + 4
3
(0.5 × 3 + 1.5 × 16 + 2.5 × 21)
= 1.95
40
c
Average time =
d
3a + 9b = 1 

2
1
9a 81b 39  ⇒ a = 15 , b = 15
+
=
2
4
20 
e
H0: The data can be modelled by the random variable X.
H1: The data cannot be modelled by the random variable X.
1% significance level, degrees of freedom 3 − 2 = 1, critical value 6.635
0t1
1<t2
2<t3
7
45
13
45
5
9
Ot
3
16
21
Et
56
9
104
9
200
9
1.669
1.709
0.067 22
Prob
(Ot − Et )2
Et
X 2 = 3.445 < 6.635. Therefore, no reason to doubt H0, the random variable X is a good fit for the data.
40
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
8
a
x=
3
5 × 0 + 15 × 12 + 25 × 48 + … + 55 × 7
= 35
200
s=

200  52 × 0 + 152 × 12 + 252 × 48 + … + 552 × 7
− 352  = 9.563 (4 s.f.)
199 
200

b H0: Income distribution can be modelled by N( μ, σ 2)
H1: Income distribution cannot be modelled by N( μ, σ 2)
Less than 10
Prob
0.004 472
Exp
0.8943
10–20
20–30
30–40
40–50
50–60
60 or more
0.053 91
0.2422
0.3989
0.2422
0.053 91
0.004 472
10.78
48.43
79.78
48.43
10.78
0.8943
Combine ‘less than 10’ and ‘10–20’ and combine ‘60 or more’ and ‘50–60’:
Less than 20
Obs
Exp
(Om − E m )
Em
2
20–30
30–40
40–50
50 or more
12
48
75
58
7
11.68
48.43
79.78
48.43
11.68
0.009 026
0.003 864
0.2869
1.890
1.872
5% significance level, degrees of freedom 5 − 3 = 2, critical value 5.991
X 2 = 4.062 < 5.991. Therefore no reason to doubt H0, income distribution can be modelled by N(μ, σ2).
c
P(M 22.5) = Φ
( 22.59.563− 35 ) = Φ(−1.307) = 0.095 59
200 × 0.095 56 = 19.12, therefore approximately 19 families.
Exercise 3.3A
1
60 × 80
a
= 32, row total multiplied by column total divided by grand total is expected frequency.
150
70 × 60
x=
= 28
150
b
c
2
(34 − 32)2 1
(Ok − E k )2 (26 − 28)2 1
= = 0.125
(4 s.f.). y =
0.1429
=
=
=
32
8
Ek
28
7
H0: Age group and clinic time attendance are independent.
H1: Age group and clinic time attendance are not independent.
5% significance level, degrees of freedom (4 − 1) × (2 − 1) = 3, critical value 7.815
X 2 = 12.84 > 7.815. Therefore reject H0; age group and clinic time attendance are not independent.
a 365 × 460 = 115, row total multiplied by column total divided by grand total is expected frequency. The
1460
column totals (365) are identical for each region, so expected frequency will be equal in each row.
b Expected frequencies: 115, 187.5, 62.5.
2
X England
=
(120 − 115)2 (199 − 187.5)2 (46 − 62.5)2
+
+
= 5.279
115
187.5
62.5
2
X Scotland
=
(103 − 115)2 (215 − 187.5)2 (47 − 62.5)2
+
+
= 9.130
115
187.5
62.5
2
X Wales
=
(103 − 115)2 (184 − 187.5)2 (78 − 62.5)2
+
+
= 5.162
115
187.5
62.5
X N2 .Ireland =
(134 − 115)2 (152 − 187.5)2 (79 − 62.5)2
+
+
= 14.22
115
187.5
62.5
Hence X2 = 33.79.
41
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
-Tests
c
Degrees of freedom (4 − 1) × (3 − 1) = 6
d H0: Region and rainfall are independent.
H1: Region and rainfall are not independent.
0.1% significance level, 6 degrees of freedom,
critical value 22.46
X 2 = 33.79 > 22.46. Therefore reject H0; region and rainfall are not independent. There is strong evidence
to justify this conclusion.
3
H0: Gender and political preference are independent.
H1: Gender and political preference are not independent.
5% significance level, degrees of freedom (5 − 1) × (2 − 1) = 4, critical value 9.488
Key
Observed
 (Ok − E k )2 


Ek

A
Male
B
27
Female
30.5 (0.4298) 28.5
34
30.5 (0.4298) 28.5
61
D
24
12
(0.5976)
20.5 (0.2857)
17
16
(0.5976)
20.5 (0.2857)
41
28
25
(0.4016)
Total
C
32
(0.4016)
Expected
57
E
Total
5
14
(0.3462)
100
6.5
8
14
(0.3462)
13
100
6.5
200
X 2 = 4.122 < 9.448. Therefore, no reason to doubt H0; there is no relationship between gender and
political preference.
4
H0: Consumption type and taste change are independent.
H1: Consumption type and taste change are not independent.
1% significance level, degrees of freedom (3 − 1) × (3 − 1) = 4, critical value 13.28
Key
Observed
 (Ok − E k )2 


Ek

Better
High
Medium
20.625
(2.133)
16.5
(1.5)
55
30
(0.8776)
24
(2.885)
(0.1538)
80
Total
75
24.375
12
28
17.875
Worse
29
30
13
(1.330)
Total
22
18
(0.1364)
Low
No Change
24
(0.5523)
Expected
60
19.5
24
26
(0.3913)
65
65
21.125
200
X 2 = 9.959 < 13.28. Therefore, no reason to doubt H0, volume of consumption is not related to taste change.
5
a
34 × 25
= 6.8, row total multiplied by column total divided by grand total is expected frequency.
125
42
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
3
b The expected frequencies are less than 5, and in order for the test statistic to be approximated by the
χ2-distribution, this cannot be the case.
c
H0: Advertising and sales are independent.
H1: Advertising and sales are not independent.
10% significance level, degrees of freedom (4 − 1) × (3 − 1) = 6 , critical value 10.64
Key
Observed
 (Ok − E k )2 


Ek

0a<5
None
Low
& High
6.8
(0.4765)
8
6.8
(0.5718)
8
(0.2042)
12
(0.3176)
10
8.16
(0.4099)
11
(0)
10.2
15  a  20
6
8
7
(1.004)
10  a < 15
5
5
(1.125)
Medium
5  a < 10
13
(5.653)
Expected
16
9.6
(0.1778)
13
10.2
(0.04719)
12.24
14.4
19
12.24
(0.02231)
18.36
X 2 = 10.01 < 10.64. Therefore, no reason to doubt H0; there is insufficient evidence at the 10%
significance level to state that advertising and sales are linked.
d H0: Advertising and sales are independent.
H1: Advertising and sales are not independent.
10% significance level, degrees of freedom (4 − 1) × (3 − 1) = 6, critical value 10.64
Key
Observed
 (Ok − E k )2 


Ek

None
0  a < 10
10  a < 15
18
6
(1.424)
Low
16
10
8.16
(0.2042)
11.6
(2.345)
9.6
(2.038)
12.24
(0.1778)
14.4
7
6.96
2
8.8
(0.4099)
16
11
8
(0.072 73)
(0.5718)
15  a  20
11
11
(0.031 03)
High
13.6
13
(0.5625)
Medium
Expected
(1.133)
10.44
12
5.28
(2.102)
7.92
X 2 = 11.07 > 10.64. Therefore reject H0; there is sufficient evidence at the 10% significance level to state
that advertising and sales are linked.
e
By combining different rows or columns, opposite conclusions are reached.
43
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
6
-Tests
a H0: Gender and subject preference are independent.
H1: Gender and subject preference are not independent.
5% significance level, degrees of freedom (3 − 1) × (2 − 1) = 2, critical value 5.991
Key
Observed
 (Ok − E k )2 


Ek

Maths & History
Male
Female
Science
14
(0.050 78)
Geography
5
13.18
6
(0.4848)
15
(0.042 32)
Expected
6.818
(0.2)
10
15.82
5
5
(0.4040)
8.182
(0.1667)
6
X 2 = 1.349 < 5.991. No reason to doubt H0, there is no relationship between gender and subject preference.
b H0: Gender and subject preference are independent.
H1: Gender and subject preference are not independent.
5% significance level, degrees of freedom (3 − 1) × (2 − 1) = 2, critical value 5.991
Key
Observed
 (Ok − E k )2 


Ek

History
Male
Female
Science
12
(0.9309)
Maths & Geography
5
9.091
8
(0.7758)
Expected
(0.4848)
8
6.818
(0.1309)
10
10.91
(0.4040)
9.091
12
8.182
(0.1091)
10.91
X 2 = 2.836 < 5.991. No reason to doubt H0, there is no relationship between gender and subject preference.
Both tests return the same result, despite different column groupings. However, the second test has a
p-value of approximately 24%, whereas the first has a p-value of approximately 51%, meaning the
grouping of columns (which here are arbitrary) could affect conclusions at weaker significance levels
(for example 25%).
Exam-style questions
1
a Geometric distribution
b H0: The first time a head is tossed can be modelled by Geo(0.4).
H1: The first time a head is tossed cannot be modelled by Geo(0.4).
5% significance level
44
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
3
Worked solutions
1
2
3
4
5
6
Prob
0.4
0.24
0.144
0.0864
0.051 84
Obs
70
42
33
21
20
Exp
80
48
28.8
17.28
10.37
1
2
3
4
7
8
9+
0.018 66
0.011 20
0.016 80
5
2
0
6.221
3.732
2.239
3.359
6
7+
0.031 10
7
5
Obs
70
42
33
21
20
Exp
80
48
28.8
17.28
10.37
6.221
0.75
0.6125
0.8008
8.948
0.097 60
 (Ok − E k )2 

 1.25
Ek

7
7
9.331
0.5824
Combine 7, 8 and 9+, degrees of freedom 7 − 1 = 6, critical value 12.59
X 2 = 13.04 > 12.59, therefore reject H0; there is evidence that the first time a head is tossed cannot be
modelled by Geo(0.4).
2
H0: The number of broken teacups in a pack of four can be modelled by B(4, 0.15).
H1: The number of broken teacups in a pack of four cannot be modelled by B(4, 0.15).
5% significance level
Prob
0
1
2
0.5220
0.3685
0.0975
3
4
0.011 48
0.000 506 3
Obs
42
16
5
2
0
Exp
33.93
23.95
6.340
0.7459
0.032 91
0
1
2 or more
Obs
42
16
7
Exp
33.93
23.95
7.119
 (Ok − E k )2 


Ek

1.919
2.639
0.001 980
Combine 2, 3 and 4, degrees of freedom 3 − 1 = 2, critical value 5.991
X 2 = 4.561 < 5.991, therefore no reason to doubt H0, the number of broken teacups in a pack of four can be
modelled by B(4, 0.15).
3
H0: Gender and vegetable preference are independent.
H1: Gender and vegetable preference are not independent at the 5% significance level.
Key
Observed
 (Ok − E k )2 


Ek

Male
Tomatoes
Carrots
24
22
10
(2.701)
Mushrooms
Female
26
(0.7273)
(0.5714)
28
28
16.72
19
(0.4544)
Expected
(2.122)
21.28
18
16.28
(0.3571)
20.72
45
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
-Tests
Degrees of freedom (3 − 1) × (2 − 1) = 2, critical value 5.991
X 2 = 6.933 > 5.991, therefore reject H0; there is a relationship between gender and vegetable choice.
4
a H0: House type and supermarket shopped at are independent.
H1: House type and supermarket shopped at are not independent.
b Total columns minus one multiplied by total combined rows minus one:
degrees of freedom (3 − 1) × (3 − 1) = 4
5
c
Critical values: 2.5% significance level 11.14, 1% significance level 13.28, 0.5% significance level 14.86.
13.28 < 13.95 < 14.86, so 0.5% is the largest significance level for which there would be no reason to
doubt H0. (Note: from calculator 0.75% is the solution.)
a
P(150 < l 170) = Φ
(17020− 160 ) − Φ(15020− 160 ) = Φ(0.5) − Φ(−0.5)
= 2 × 0.6915 − 1 = 0.3830
Expected value: 100 × 0.3830 = 38.30
100 × P(l 150) = 100 × Φ(−0.5) = 100 × 0.3085 = 30.85
100 × P(170 < l 190) = 100 × [ Φ(1.5) − Φ(0.5) ] = 24.17
100 × P (l > 190 ) = 100 × [1 − Φ(1.5 )] = 6.681
b H0: The length of Aesculapian snakes can be modelled by N(160, 202).
H1: The length of Aesculapian snakes cannot be modelled by N(160, 202).
2.5% significance level, degrees of freedom 4 − 1 = 3, critical value 9.348
X2 = 1.997 + 0.075 46 + 0.1386 + 2.794 = 5.005
X2 < 9.348, therefore no reason to doubt H0, the length of Aesculapian snakes can be modelled by
N(160, 202).
c
Confidence interval

20
20 
165 − 1.96 × 100 ,165 + 1.96 × 100 
[161, 169]
The proposed population mean of 160 lies outside the 95% confidence interval, and so it would seem
unlikely that it would make a suitable model to measure the lengths of the snakes.
6
a
3
∫0 ky
2
4
d y + ∫ k(12 − y) d y = 1
3
3
4
y3 

y2 
35
k   + k 12y −  = k (9 − 0) + k (40 − 31.5) = k = 1 therefore k = 2
3
2 
2
35
 0

3
3
2
∫0 35 y
2
3
2 3
18
dy = 
y
=
 105 0 35
b
P(Y 3) =
c
From data given in the table P(Y 2) =
6 + 10 16
=
105
105
18 16
38
−
=
35 105 105
18 17
and P(3 < Y 4) = 1 − P(Y 3) = 1 −
=
35 35
so the remaining two expected values are 38 and 51 respectively.
H0: Amount of chemical produced can be modelled by the random variable Y.
H1: Amount of chemical produced cannot be modelled by the random variable Y.
5% significance level, degrees of freedom 4 − 1 = 3, critical value 7.815
Therefore P(2 < Y 3) =
46
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
3
X 2 = 1.5 + 1.6 + 3.789 + 7.078 = 13.97
X 2 > 7.815, therefore reject H0, so the amount of chemical produced cannot be modelled by the random
variable Y.
7
a x = 0 × 29 + 1 × 38 + 2 × 33 + 3 × 13 + 4 × 8 + 5 × 4 = 1.56
125

125  02 × 29 + 12 × 38 + 22 × 33 + 32 × 13 + 42 × 8 + 52 × 4
2
s =
− 1.562  = 1.7
124 
125

b Poisson distribution has equal expectation and variance, and the sample mean and variance are close.
c
P(X = 2) =
e −1.56 × 1.56 2
= 0.2557, and expected value is 125 × 0.2557 = 31.96
2!
x = 125 − 26.27 − 40.98 − 31.96 − 16.62 − 6.482 − 2.022 = 0.6705
d Groups 4, 5 and 6 or more must be combined (to form 4 or more) so that expected frequencies are
greater than five.
e
H0: The number of A* gained can be modelled by a Poisson distribution.
H1: The number of A* gained cannot be modelled by a Poisson distribution.
10% significance level, degrees of freedom 5 − 2 = 3, critical value 6.251
X 2 = 0.2844 + 0.2162 + 0.033 73 + 0.7885 + 0.8701 = 2.193
X 2 < 6.251, therefore no reason to doubt H0, the number of A* gained can be modelled by a Poisson
distribution.
f
8
Expected frequencies would change. One more degree of freedom.
a H0: Opening day and level of demand are independent.
H1: Opening day and level of demand are not independent.
5% significance level
Key
Observed
 (Ok − E k )2 

 Expected
Ek

Low
Tuesday
Wednesday
Thursday
15
18
12
(0)
Normal
30
(0.6154)
High
15
15
25
26
5
(1.778)
(0.6)
(0.038 46)
(0.4444)
15
23
26
7
9
(0.6)
(0.3462)
26
15
9
(4)
9
Degrees of freedom (3 − 1) × (3 − 1) = 4, critical value 9.488
X 2 = 8.422 < 9.448, therefore no reason to doubt H0, there is not a relationship between level of demand
and the weekday of opening.
b The conclusion states there is no relationship between level of demand and opening day, meaning
no specific day gains a higher or lower demand. Therefore, it would be difficult to choose which one of
the days to open on. Also, the test does not give any information on whether the demand is sufficient for
the club to make a profit.
47
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
2
3
9
-Tests
a H0: The number of cases assigned can be modelled by a uniform distribution.
H1: The number of cases assigned cannot be modelled by a uniform distribution.
5% significance level
A
B
C
Observed
43
32
45
Expected
40
40
40
0.225
1.6
0.625
 (O − E )2 
k
 k

Ek


Degrees of freedom 3 − 1 = 2, critical value 5.991
X 2 = 2.45 < 5.991, therefore no reason to doubt H0, the number of cases assigned can be modelled by a
uniform distribution.
b H0: Cases solved and officer are independent.
H1: Cases solved and officer are not independent.
1% significance level
Key
Observed
 (Ok − E k )2 


Ek

A
Solved
B
31
21.5
16
(5.878)
14
(4.198)
21.5
60
22.5
34
(0.25)
43
Total
11
(0.25)
12
Total
C
18
(4.198)
Unsolved
Expected
16
(5.878)
32
45
60
22.5
120
Degrees of freedom (3 − 1) × (2 − 1) = 2, critical value 9.210
X 2 = 20.65 > 9.210, therefore reject H0, so there is a relationship between the proportion of cases solved
and assigned officer.
10
a ΣP(X = r) = a + 2a + 3a + 4b + 5b + 6b = 1 ⇒ 6a + 15b = 1
b E(X) = ΣrP(X = r) = a + 4a + 9a + 16b + 25b + 36b = 14a + 77b
c
1 × 3 + 2 × 13 + 3 × 11 + 4 × 10 + 5 × 16 + 6 × 7 56
=
60
15
6a + 15b = 1 

1
1
56  ⇒ a = 12 , b = 30
14a + 77b =

15 
x=
d Use these values to calculate expected frequencies:
1
Obs
Exp
(Or − E r )
Er
2
4
5
3
13
2
3
11
10
16
6
7
5
10
15
8
10
12
0.8
0.9
1.067
0.5
3.6
2.083
48
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
3
X2 = 8.95, degrees of freedom 6 − 2 = 4
critical values: 10% significance level 7.779, 5% significance level 9.488
7.779 < X2 < 9.488, therefore 5% significance level is the greatest at which there would be no reason to
doubt H0, the data is modelled by the random variable X.
11
a R
ow total multiplied by column total divided by grand total gives expected frequency. Here
48 × 70 32
=
= 10.67 to 2 d.p.
315
3
b Expected frequencies for the row x > 150 are all less than five, and so must be combined to get expected
frequencies of over five in order for the test statistic to be compared to the χ2-distribution.
c
X F2 =
(11 − 9.90)2 + (19 − 27.24 )2 + ( 35 − 27.86 )2 = 4.44 to 2 d.p.
9.90
27.24
27.86
d 1% significance level, degrees of freedom (5 − 1) × (3 − 1) = 8, critical value 20.09. X2 > 20.09, so she would
have rejected the null hypothesis under these conditions.
12
a x = 25 × 10 + 35 × 42 + 45 × 31 + 55 × 17 = 40.5
100

100  252 × 10 + 352 × 42 + 452 × 31 + 552 × 17
− 40.52  = 8.92 to 3 s.f.
99 
100

60 − 40.5
50 − 40.5 

−Φ
b 100 × P (50 length < 60) = 100 × Φ
8.92
8.92


s=
(
) (
)
= 100 × [Φ(2.186) − Φ(1.065)] = 100 × (0.9856 − 0.8566) = 12.90
c
H0: Algae length can be modelled by a normal distribution.
H1: Algae length cannot be modelled by a normal distribution.
1% significance level, combine first two and last two cells, three constraints, degrees of freedom 4 − 3 = 1,
critical value 6.635
X2 = 0.3202 + 1.070 + 1.253 + 0.4938 = 3.137
X2 < 6.635, therefore no reason to doubt H0, algae length can be modelled by a normal distribution.
d Expected values will change, and there will be two fewer constraints, so degrees of freedom will increase
by two.
13
a H0: Mobile phone signal strength and service provider are independent.
H1: Mobile phone signal strength and service provider are not independent.
1% significance level
Key
Observed
 (Ok − E k )2 


Ek

A
Good
Medium
55
56
(0.018 52)
B
54
(0.074 07)
36
(0.1111)
35
(0.027 78)
Expected
Bad
39
54
(0.2143)
36
(0.3214)
34
42
31
28
Degrees of freedom (3 − 1) × (2 − 1) = 2, critical value 9.210
X 2 = 0.7302 < 9.210, therefore no reason to doubt H0, there is no relationship between signal and provider.
b Increasing sample size by a factor of n and keeping observed frequencies in the same proportion
increases X 2 by a factor of n.
0.7302n > 9.210 ⇒ n > 12.61 therefore n = 13.
49
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
3
2
14
-Tests
a H0: Treatment and reaction are independent.
H1: Treatment and reaction are not independent.
b Column total multiplied by row total divided by grand total gives expected frequency,
here 41 × 41 = 11.21 to 2 d.p.
150
c
X S2 =
(25 − 21.05)2 + (7 − 11.21)2 + (9 − 8.75)2 = 2.33 to 2 d.p.
21.05
11.21
8.75
d Degrees of freedom (3 − 1) × (3 − 1) = 4
Critical value at 5% is 9.488 and at 2.5% is 11.14. Since 9.488 < X 2 < 11.14, the null hypothesis would be
rejected at the 5% significance level, but not at the 2.5% significance level. Therefore 5% is the smallest
significance level from the tables for which the null hypothesis would be rejected.
e
H0: Proportions of people given the treatments A, B and C are in the ratio 3:2:1.
H1: Proportions of people given the treatments A, B and C are not in the ratio 3:2:1.
5% significance level
A
B
C
77
41
32
75
50
25
0.053 33
1.62
1.96
Observed
Expected
 (Ok − E k ) 


Ek

2
Degrees of freedom 3 − 1 = 2, critical value 5.991
X 2 = 3.633 < 5.991, therefore no reason to doubt H0, the proportion of people given each treatment is in
the ratio 3:2:1.
Mathematics in life and work
H0: Age and demand are independent
H1: Age and demand are not independent
5% significance level, combine 40–49 and 50+ rows.
Key
Observed
 (Ok − E k )2 


Ek

Low
<20
12
(2)
20–29
40+
8
7.667
7
(1.093)
10.67
(0.7621)
6.708
(0.3487)
9
(3.350)
8.625
13
9.333
14
13.67
(0.4444)
14
12
16
(0.3983)
(0.5714)
7
4
7
(1.262)
High
5
5
(0.9277)
30–39
Medium
Expected
(0.083 33)
12
11
11.96
(1.247)
15.38
Degrees of freedom (4 − 1) × (3 − 1) = 6, critical value 12.59
X 2 = 12.49 < 12.59, therefore just about no reason to doubt H0, there is no relationship between age
and demand.
50
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
3
H0: The ages of customers attending the coffee shop can be modelled by a uniform distribution.
H1: The ages of customers attending the coffee shop cannot be modelled by a uniform distribution.
5% significance level
<20
Observed
Expected
 (Ok − E k ) 


Ek

20–29
30–39
40–49
50+
24
23
32
27
14
24
24V
24
24
24
0
0.041 67
2.667
0.375
4.167
2
Degrees of freedom 5 − 1 = 4, critical value 9.488
X 2 = 7.25 < 9.488, therefore no reason to doubt H0, the ages of customers attending the coffee shop can be
modelled by a uniform distribution.
51
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4 Non-parametric tests
4 Non-parametric tests
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in
this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Where values from the Cambridge International Education statistical tables are used, the same level of accuracy has
been used in workings unless stated otherwise.
Prerequisite knowledge
1
P(X  12) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
15
1
576
=
× { 455 + 105 + 15 + 1} =
= 0.0176
2
32768
2
1
156 + 46 + … + 175 ) = 121
10 (
1
12102  21350
= 2372.2…
s 2 =  167760 −
=
9
10 
9
()
x=
H0 : μ = 150
H1 : μ < 150
121 − 150
= −1.883. Critical value:
2372
10
t9 (5%) = –1.833. As –1.883 < –1.833, reject H0. There is (just) sufficient evidence at the 5% significance level
to reject the null hypothesis that the mean is 150 in favour of the alternative hypothesis that the mean is less
than 150.
5% significance level, one-tailed test. The test statistic T =
3
Perform a paired t-test, assuming the differences in the calorie intakes in January and June are normally
distributed. Take differences to be January minus June.
Differences: 126, 63, 189, – 92, – 49, 6, 93, 141
d=
1
126 + 63 + … + 141) = 59.625
8(
sd2 =
1
126 2 + 632 + … + 1412 − 8 × 59.6252 = 9508
7
((
)
)
H0 : md = 0
H 1 : md > 0
10% significance level, one-tailed test.
Test statistic T = 59.625 − 0 = 1.730.
9508
8
Critical value: t7 (10%) = 1.415. As 1.730 > 1.415 reject H0. There is sufficient evidence at the 10% significance
level to support Juliet’s hypothesis that people consume fewer calories in summer than winter.
52
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
4
Exercise 4.1A
1
a H0: median revision time is 30 hours a week
H1: median revision time is less than 30 hours a week
X ∼ B(9, 0.5), 5% significance level, one-tailed test
Signed differences: −14, −11, −16, −18, −1, 9, −9, −20, −21
One positive sign. P(X  1) = 0.01953 < 0.05
Reject H0. Sufficient evidence that students are doing less than 30 hours revision a week.
b H
0: median revision time is 20 hours a week
H1: median revision time is not 20 hours a week
X ∼ B(9, 0.5), 5% significance level, two-tailed test
Signed differences: −4, −1, −6, −8, 9, 19, 1, −10, −11
Three positive signs. P ( X 3 ) = 0.2539 > 0.025
No reason to doubt H0. Insufficient evidence to refute the claim that median revision time is 20 hours
per week.
2
H0: Standard and premium brand yield same total distance.
H1: Premium brand yields higher distance than standard.
X ∼ B(8, 0.5), 5% significance level, one-tailed test
Signed differences: 15, −7, 24, 23, −16, 21, 10, 33
Two negative signs. P ( X 2 ) = 0.1445 > 0.05.
No reason to doubt H0. Insufficient evidence to show premium is better than standard.
The cars with the lowest distance travelled with standard are the ones that travel less far with premium.
Perhaps premium works better with more efficient cars.
3
H0: Wheatees and Crunchos are equally popular.
H1: There is a difference in popularity between Wheatees and Crunchos.
X ∼ B(12, 0.5), 5% significance level, two-tailed test
Two signs for Crunchos. P(X  2) = 0.01929 < 0.025
Reject H0. Sufficient evidence to say there is a difference in preferences for the two breakfast cereals.
4
H0: Chesford and Amerston have the same median crime rate.
H1: Chesford and Amerston do not have the same median crime rate.
X ∼ B(12, 0.5), 5% sig level, two-tailed test
Signed differences: 0.78, 0.27, 0.48, −1.29, −0.15, 0.50, −0.04, 0.07, 2.09, 1.20, 1.04, 0.02
Three negative signs. P ( X 3 ) = 0.07300 > 0.025 No reason to doubt H0. Insufficient evidence to show that
Chesford and Amerston have different median crime rates.
5
H0: Insulation does not affect median heat loss.
H1: Homes with insulation have a lower median heat loss.
X ∼ B(10, 0.5), 2% significance level, one-tailed test
Signed differences: 0.9, 3.2, 1.4, 0.1, −1.3, 1.2, 1.7, 5.1, 0.4, −1.9
Two negative signs. P(X  2) = 0.05469 > 0.02
No reason to doubt H0. Insufficient evidence to show that insulation reduces median heat loss.
6
a 2.17, 3.50, 2.06, 0.55, 2.05
b 0.17, 1.50, 0.06, −1.45, 0.05
c
H0: Derivative A performs two percentage points better than derivative B.
H1: Derivative A performs more than two percentage points better than derivative B.
X ∼ B(5, 0.5), 10% significance level, one-tailed test
One negative sign. P ( X 1) = 0.1875 > 0.1
No reason to doubt H0. Insufficient evidence to show that derivative A has median performance more
than two percentage points better than derivative B.
53
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4 Non-parametric tests
7
H0: WBC count is normal, with a median of 7 million per 1 ml.
H1: WBC count is abnormal; median is not 7 million per 1 ml.
X ∼ B(25, 0.5), 10% significance level, two-tailed test
Signed differences: −0.27, –0.08, 0.21, −0.06, 0.45, 0.32, −0.49, −0.29, −0.09, −0.13, −0.05, 0.93, 0.35, −0.37,
−0.18, −0.39, −0.03, 0.23, 0.02, −0.24, −0.28, −0.10, −0.59, −0.34, −0.16
Seven positive signs
Normal approximation: X ∼ N(12.5, 6.25)
 7.5 − 12.5 
P ( X 7) = Φ 


6.25 
= Φ ( −2 ) = 1 − Φ ( 2 )
= 1 − 0.9772
= 0.0228 < 0.05
Reject H0. The patient has an abnormal white blood cell count.
8 X ∼ B(n, 0.5): P(X = 0) = (0.5)n
Note: 0 negative signs, implies one-tailed test
0.5n < 0.001
n ln 0.5 < ln 0.001
ln 0.001
n>
= 9.966
ln 0.5
Therefore n = 10.
X ∼ B(30, 0.5); therefore, by the normal approximation X ∼ N(15, 7.5)
 8.5 − 15 
P ( X 8) = Φ 
= Φ ( −2.373 )
 7.5 
= 1 − 0.9912 = 0.0088
Two-tailed test, therefore significance level is 2 × 0.0088 = 0.0176 , or 1.76%.
Exercise 4.2A
1
Signed diff
4
−7
−5
−13
1
3
Unsigned diff
4
7
5
13
1
3
Unsigned rank
3
6
4
10
1
2
Signed rank
3
−6
−4
−10
1
2
−20 −31
24
10
20
31
24
10
11
15
14
8
−11 −15
14
8
−12 −23
6
−21
−8
21
8
12
23
6
9
13
5
12
7
−9 −13
5
−12
−7
H0: Employees stay with the company for a median period of a year (52 weeks).
H1: Employees stay with the company for a median period of less than a year.
1% significance level, one-tailed test, n = 15, critical value from table: 19
P = 33, Q = 87, therefore T = 33
T > 19, so no reason to doubt H0. Employees stay with the company for a year on average.
Assumption: weeks worked at the company are distributed symmetrically.
2
H0: Median is the same.
H1: Median is different.
10% significance level, two-tailed test,n = 18, critical value from table: 47
P = 133, Q = 38, therefore T = 38
T < 47, so reject H0. The median is different.
3
H0: Chesford and Amerston have the same median crime rate.
H1: Chesford and Amerston do not have the same median crime rate.
5% significance level, two-tailed test, n = 12, critical value from table: 13
54
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4
Worked solutions
Signed diff
0.78 0.27 0.48
Unsigned diff
0.78 0.27 0.48
−1.29 −0.15 0.50
1.29
−0.04 0.07
2.09
1.20 1.04 0.02
0.04 0.07
2.09
1.20 1.04 0.02
0.15 0.50
Unsigned rank
8
5
6
11
4
7
2
3
12
10
9
1
Signed rank
8
5
6
−11
−4
7
−2
3
12
10
9
1
P = 61, Q = 17, therefore T = 17
T > 13, no reason to doubt H0. Median crime rates in Chesford and Amerston are equal.
4
H0: Literacy rates between genders (people aged 35–39) in South America are equal.
H1: Literacy rates between genders (people aged 35–39) in South America are not equal.
5% significance level, two-tailed test, n = 9, critical value from table: 5
Signed diff
1.59
−1.28
2.70
1.67
3.62
1.60
−0.19
0.46
1.79
Unsigned diff
1.59
1.28
0.19
2.70
1.67
3.62
1.60
0.46
1.79
Unsigned rank
4
3
8
6
9
5
1
2
7
Signed rank
4
−3
8
6
9
5
−1
2
7
P = 41, Q = 4, therefore T = 4
T < 5, reject H0. There is sufficient evidence at the 5% significance level that literacy rates between genders
(people aged 35–39) in South America are not equal.
5
a H0: Median consumption of refined petroleum products is 35 barrels a day.
H1: Median consumption of refined petroleum products is less than 35 barrels a day.
X ∼ B(12, 0.5), 5% significance level, one-tailed test
Signed differences: −9.99, 23.33, −6.24, −7.59, −5.66, −3.54, −11.06, 25.50, −10.40, −5.03, −2.37, −9.38
Two positive signs. P ( X 2 ) = 0.0193 < 0.05
Reject H0. Sufficient evidence to demonstrate median consumption of refined petroleum products is
less than 35 barrels a day.
b Same hypotheses. 5% significance level, one-tailed test, n = 12, critical value: 17
Sign diff
Unsign diff
Unsign rank
Sign rank
−9.99 23.33 −6.24
9.99 23.33
−7.59 −5.66 −3.54 −11.06 25.5 −10.4
6.24
7.59
5.66
3.54
11.06 25.5
−5.03 −2.37
−9.38
10.4
5.03
2.37
9.38
8
11
5
6
4
2
10
12
9
3
1
7
−8
11
−5
−6
−4
−2
−10
12
−9
−3
−1
−7
P = 23, Q = 55, therefore T = 23
T > 17, no reason to doubt H0. Median consumption of refined petroleum products is 35 barrels a day.
c T
he two biggest consumers are Netherlands and Belgium. They only count for two positive signs in the
sign test, but are the two largest deviations from the median in the Wilcoxon signed-rank test, so count
more significantly towards this test. Netherlands and Belgium could be considered outliers, so the sign
test would be more persuasive. Or, thinking of ‘average’ consumption, the fact that Netherlands and
Belgium consume so much means the Wilcoxon signed-rank test might be more appropriate.
6
H0: Median is as given.
H1: Median is less than the value given (also acceptable: greater than).
4% significance level, one-tailed test. Normal approximation required
µ = 1 × 50 × ( 50 + 1) = 637.5 and
4
σ 2 = 1 × 50 × ( 50 + 1) × ( 2 × 50 + 1) = 10 731.25
24
T = 423
55
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4 Non-parametric tests
 423.5 − 637.5 
P (T 423 ) = Φ 

 10 731.25 
= Φ ( −2.066 )
= 1 − 0.9806
= 0.0194 < 0.04
Therefore reject H0. Median is not as the given value.
7
a H
0: WBC count is normal, with a median of 7 million per 1 ml.
H1: WBC count is abnormal; median is not 7 million per 1 ml.
10% significance level, two-tailed test, n = 25. Normal approximation required
µ = 1 × 25 × 26 = 162.5 and
4
1 × 25 × 26 × 51 = 1381.25
σ 2 = 24
Signed ranks: −14, −5, 11, −4, 22, 17, −23, −16, −6, −8, −3, 25, 19, −20, −10, −21, −2, 12, 1, −13, −15, −7, −24,
−18, −9
P = 107, Q = 218, therefore T = 107
 107.5 − 162.5 
P (T 107 ) = Φ 


1381.25 
= Φ( −1.480 ) = 1 − 0.9306
= 0.0694 > 0.05
Therefore no reason to doubt H0. WBC count is normal.
b T
he Wilcoxon signed-rank test has a lower probability of a Type II error (incorrectly rejecting a true null
hypothesis). Given the data does not look asymmetric, the Wilcoxon signed-rank test would be more
appropriate here.
8
a H
0: Population median is as given
H1: Population is not as given
1% sig level, two-tailed test, n = 30. Normal approximation required
µ = 1 × 30 × 31 = 232.5 and
4
1 × 30 × 31 × 61 = 2363.75
σ 2 = 24
T = 105
 105.5 − 232.5 
P (T 105 ) = Φ 


2363.75 
= Φ( −2.612 ) = 1 − 0.9955
= 0.0045 < 0.005
Therefore reject H0. The population median is not as given.
b SN = 1 N ( N + 1)
2
The maximum number of positive ranks would occur if these were all the lowest ranks (because T is the
smaller of P and Q). If the lowest N ranks were all positive, then
1
N N + 1) = 105
2 (
N 2 + N = 210
N 2 + N − 210 = 0
( N − 14 )( N + 15) = 0
As N is positive, therefore N = 14, so at most 14 positive ranks.
56
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
4
c X ∼ B(30, 0.5), two-tailed test. Normal approximation required.
 14.5 − 15 
P ( X 14 ) = Φ 


7.5 
= Φ ( −0.1826 )
= 1 − 0.5726
= 0.4274
Therefore the probability of a Type I error (rejecting a true null hypothesis) is 2 × 0.4274 = 0.855.
9
a
b
1
2
3
P
Q
T −
−
−
0
6
0
+
−
−
1
5
1
−
+
−
2
4
2
+
+
−
3
3
3
−
−
+
3
3
3
+
−
+
4
2
2
−
+
+
5
1
1
+
+
+
6
0
0
1
2
3
4
P
Therefore, for a two-tailed test with rejection region T  2, P (T 2 ) = 6 = 63
8 2
T Q
+
+
+
+
10
0
0
−
+
+
+
9
1
1
+
−
+
+
8
2
2
+
+
−
+
7
3
3
−
−
+
+
7
3
+
+
+
−
6
4
−
+
−
+
6
4
3
+
c
−
−
+
5
Therefore, for a two-tailed test with rejection region T  2,
6
6
P (T 2 ) =
=
16 2 4
4
4
5
5
−
+
+
−
5
5
5
+
−
+
−
4
6
4
−
−
−
+
4
6
4
1
2
3
4
P
Q
T
+
+
−
−
3
7
3
−
−
+
−
3
7
3
−
+
−
−
2
8
2
+
−
−
−
1
9
1
−
−
−
−
0
10
0
1
2
3
4
5
−
−
−
−
−
0 15
0 +
−
−
−
−
1 14
1
−
+
−
−
−
2 13
2
+
−
+
+
+
13
2
2
−
+
+
+
+
14
1
1
+
+
+
+
+
15
0
0

P
Q

T
Each increase in n raises the total number of possible outcomes
by an additional power of two; however, the number of different
ways of getting signed-ranks of two or less remains constant at
six. Therefore, for a two-tailed test with rejection region T  2,
6
6
P (T 2 ) =
=
32 2 5
57
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4 Non-parametric tests
d For a sample of size n, P (T 2 ) = 6n
2
e P (T 2 ) = 6n < 0.001
2
6
< 0.001
2n
6 < 0.001 × 2 n
ln6 < ln0.001 + n ln2
ln6 − ln0.001 < n ln2
ln6 − ln0.001
<n
ln2
n > 12.55
Therefore, n = 13.
Exercise 4.3A
1
a β = 81
b γ = 55
c m
(n + m + 1) = 7 × ( 9 + 7 + 1) = 119. 119 − 81 = 38 and 119 − 55 = 64
d There are fewer boys, so should use 119 − 81 = 38 or 81. But as 38 is lower, the test statistic is W = 38.
e Critical value for 5% significance level one-tailed test: 43.
f
H0: Girls and boys raise the same amount of sponsorship.
H1: Girls raise more sponsorship than boys.
W < 43, so reject H0. There is sufficient. evidence at the 5% significance level that girls raise more
sponsorship than boys.
2
a
Rank
1
2
3
4
5
6
7
8
9
10
98A
95B
93A
90B
87B
84B
81A
78A
77A
67A
 10
b   = 210
 6
c 17
d H0: Quarry A and quarry B have the same purities of iron ore.
H1: Quarry A and quarry B have different purities of iron ore.
10% significance level, two-tailed test, m = 4, n = 6, critical value: 13, m (n + m + 1) − 17 = 27, therefore W = 17
As W > 13, no reason to doubt H0; the quarries have the same purity of iron ore.
e No assumptions on the underlying probability distribution are necessary to perform this test.
3
a H
0: Group 1 and Group 2 are drawn from identical populations.
H1: Group 2 has higher values than Group 1.
5% significance level, one-tailed test, m = 6, n = 8, critical value: 31, m is the second group with
summed rank
Rm = 62, m (n + m + 1) − 62 = 28 , therefore W = 28
As W < 31, reject H0, so Group 2 has higher values than Group 1.
b T
wo-tailed test, 2% significance level (from tables) critical value is 27, as W > 27, no reason to doubt H0.
At 5%, critical value is 29, so this would yield a rejection of H0. Answer: 2%.
58
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
4
a
2
3
4
5
6
7
8
Ball
C
C
C
G G
C
G G
C
9 10
G
Sum of ranks C = 17
Rank 1
4
Sum of ranks G = 38
b The two sample sizes are equal.
c H
0: Claxxon and Galway golf balls travel the same distance.
H1: Claxxon and Galway golf balls do not travel the same distance.
10% significance level, two-tailed test, m = 5, n = 5, critical value: 19
For Claxxon balls, Rm = 17, m ( n + m + 1) − 17 = 38, therefore W = 17
As W  19, reject H0. There is sufficient evidence at the 5% significance level that the two types of balls
do not travel the same distance.
d H
0: The balls travel a median distance of 275 metres.
H1: The balls travel a median distance of greater than 275 metres.
X ∼ B(10, 0.5), 8% significance level, one-tailed test
Three negative signs. P ( X 3 ) = 0.172 > 0.08
No reason to doubt H0. Insufficient evidence that balls travel further than 275 metres.
5
H0: Younger and older drivers take the same length of time to pass their driving test.
H1: Younger drivers take less time than older drivers.
1% significance level, one-tailed test, m = 6, n = 10, critical value: 29
Rm = 37, m (n + m + 1) − 37 = 65 therefore W = 37
As W > 29 no reason to doubt H0; there is insufficient evidence at the 1% significance level that younger
drivers take less time than older drivers to pass their driving test.
6
H0: The two doctors have the same waiting time.
H1: The two doctors do not have the same waiting time.
5% significance level, two-tailed test, m = 3, n = 7, critical value: 7
Rm = 26, m (n + m + 1) − 26 = 7 therefore, W = 7
As W  7, there is just about reason to doubt H0, so just sufficient evidence at the 5% significance level that
waiting times are different.
7
H0: The two samples are drawn from identical distributions.
H1: The two samples are not drawn from identical distributions.
2.5% significance level, two-tailed test, m = 13, n = 14, therefore normal approximation required.
Rm = 231, m (n + m + 1) − 231 = 133
therefore W = 133
µ = 1 m (n + m + 1) = 1 × 13 × (13 + 14 + 1) = 182
2
2
1
2
σ = nm ( n + m + 1) = 1 × 14 × 13 × (13 + 14 + 1) = 1274
12
12
3


133.5 − 182 
= Φ ( −2.354 ) = 1 − 0.9907 = 0.0093 < 0.0125
P (W 133 ) = Φ 

1274 


3
Therefore reject H0; there is sufficient evidence at the 2.5% significance level that the two samples are not
from identical distributions.
8
a T
he samples are no longer matched pairs, but just 12 observations from each population (though one
might consider they are not randomly drawn).
b T
he three pairs of tied values occur within each sample, e.g. 4.94 appears in Chesford’s data twice, but not
in Amerston’s. So, this means they can be ranked k and k + 1 in either ordering, and it will not affect the test.
59
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4 Non-parametric tests
c
A
1
A
13
A
2
A
14
A
3
C
15
C
4
A
16
A
5
A
17
C
6
A
18
C
7
C
19
A
8
C
20
C
9
C
21
C
10
A
22
A
11
C
23
C
12
C
24
H0: Chesford and Amerston have the same crime rate.
H1: Chesford and Amerston do not have the same crime rate.
5% significance level, two-tailed test, m = 12, n = 12, therefore normal approximation required. Using
Amerston as m, and ranking from low to high crime rate:
Rm = 130, m (n + m + 1) − 130 = 170 therefore W = 130
µ = 1 × 12 × 25 = 150
2
2
σ = 1 × 12 × 12 × 25 = 300
12
 130.5 − 150 
P (W 130 ) = Φ 
 = Φ ( −1.126 ) = 1 − 0.8698 = 0.1302 > 0.025

300 
Therefore no reason to doubt H0; the crime rates in Chesford and Amerston are the same.
9
Entries N have been omitted for clarity
1
2
M
M
M
3
4
5
6
11
3
4
10
4
5
9
5
6
8
6
7
7
7
5
9
5
6
8
6
7
7
7
8
6
6
7
7
7
8
6
6
M
9
5
5
9
5
5
M
10
4
4
M
11
3
3
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
M
W
3
M
M
Rm
60
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4
Worked solutions
Sampling distribution:
w
3
4
5
6
7
P(W = w)
2
15
2
15
4
15
4
15
3
15
Therefore, lowest possible significance level for two-tailed test would be 2 .
15
Exam-style questions
1
a H0: Median ratio of pupils to teachers is the same in 2000 and 2010.
H1: Median ratio of pupils to teachers is lower in 2010 than in 2000.
X ∼ B(9, 0.5), 10% significance level, one-tailed test, 2 negative signs P ( X 2 ) = 0.08984 < 0.1, therefore
reject H0; there is evidence to show that the median ratio of pupils to teachers is lower in 2010, so quality
is increasing.
b The Wilcoxon matched-pairs signed-rank test has a lower probability of Type II error.
2
H0: Median number of pages set for reading is 40.
H1: Median number of pages set for reading is not 40.
2% significance level, two-tailed test, n = 14, critical value 15
Difference
9
20
−8
25
3
−5
−2
−1
38
13
17
7
15
22
|Difference|
9
20
8
25
3
5
2
1
38
13
17
7
15
22
Rank
7
11
6
13
3
4
2
1
14
8
10
5
9
12
Signed rank
7
11
−6
13
3
−4
−2
−1
14
8
10
5
9
12
P = 92, Q = 13, so T = 13 < 15, therefore reject H0; the median number of pages set is not 40.
3
H0: Drinking Blue Stallion does not improve concentration.
H1: Drinking Blue Stallion does improve concentration.
5% significance level, one-tailed test, m = 6, n = 7, critical value 29
1
2
3
4
5
6
7
8
9
10
11
12
13
32
(BS)
44
(BS)
51
(Co)
58
(BS)
59
(Co)
60
(BS)
62
(Co)
67
(Co)
68
(BS)
72
(BS)
73
(Co)
74
(Co)
81
(Co)
Rm = 32, m ( n + m + 1) − Rm = 52 therefore W = 32. W > 29, so no reason to doubt H0; Blue Stallion drinkers’
reaction times are drawn from an identical distribution to the control groups’ reaction times.
4
a
H0: The median coefficient of friction is the same with both oils.
H1: The median coefficient of friction is not the same with both oils.
X ~ B (15, 0.5 ), 5% significance level, two-tailed test, 3 negative signs P ( X 3 ) = 0.01758 < 0.025, therefore
reject H0; there is a difference in median coefficient of friction between the two oils.
b C
ritical value for n = 15, 5% significance level, two-tailed test is 25. As T = 33 > 25, there is no reason to
doubt the null hypothesis. This changes the conclusion from above.
5
a The data does not appear to be symmetric.
bH0: The median amount of time for pain to be relieved is 30 minutes.
H1: The median amount of time for pain to be relieved is more than 30 minutes.
X ~ B (12, 0.5 ), 5% significance level, one-tailed test, 3 negative signs
P ( X 3 ) = 0.0730 > 0.05, therefore no reason to doubt H0; on average pain is relieved within 30 minutes.
6
H0: There is no preference for one sports kit manufacturer over the other.
H1: There is preference for one sports kit manufacturer over the other.
X ~ B (100, 0.5 ) , therefore use a normal approximation, X ~ N ( 50, 25 )
1% significance level, two-tailed test, 36 ‘negative’ signs
(
)
P(X 36) = Φ 36.5 − 50 = Φ(−2.7) = 1 − 0.9965 = 0.0035 < 0.005,
5
therefore reject H0; there is evidence of a difference in preference for the two manufacturers of the sports kit.
61
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4 Non-parametric tests
7
a H0: North African and Central American television ownership rates are drawn from identical distributions.
H1: North African and Central American television ownership rates are not drawn from
identical distributions.
5% significance level, two-tailed test, m = 5, n = 10, critical value 23
Rank
Value
Region
1
84.8
NA
2
93.5
NA
3
93.9
NA
4
99.5
CA
5
99.8
CA
6
104.8
NA
7
109.7
CA
8
110.9
NA
9
125.9
CA
10
129.4
CA
11
134.6
CA
12
152.0
CA
13
158.7
CA
14
158.9
CA
15
166.1
CA
Rm = 20, m (n + m + 1) − Rm = 60, therefore W = 20. W < 23, reject H0; there is sufficient evidence of a
difference between television ownership rates in North Africa and Central America.
b In order to perform a two-sample t-test the samples must be drawn from distributions with identical
variance, but clearly the two standard deviations are not close to being the same.
8
a H
0: Median score on first paper is the same as on the second paper.
H1: Median score on first paper is the lower than on the second paper.
5% significance level, one-tailed test, n = 10, critical value 10
Difference
10
−19
11
16
20
|Difference|
28
−4
21
8
2
10
19
11
16
20
28
4
21
8
2
Rank
4
7
5
6
8
10
2
9
3
1
Signed rank
4
−7
5
6
8
10
−2
9
3
1
P = 46, Q = 9, so T = 9 < 10 therefore reject H0; there is evidence to support the claim that the median
mark on the second paper is higher than on the first paper.
b This converts the test into a two-tailed test, and the critical value is now 8. As T > 8 there is no reason to
doubt H0, which is that the median scores on the two papers are the same. For a given significance level,
in stating that one median is lower than the other, the critical region becomes larger than just looking
for a generic difference (either higher or lower). Hence, it is not contradictory that the first test should
reject, whilst the second test finds no reason to doubt the null hypothesis.
9
H0: The two samples are drawn from identical distributions.
H1: The two samples are not drawn from identical distributions.
5% significance level, two-tailed test, m = 15, n = 20, normal approximation required
Rm = 340, m (n + m + 1) − Rm = 200, therefore W = 200.
(
)
1
1
2
2
W ~ N µ, σ : µ = 2 m (n + m + 1) = 270, σ = 12 nm (n + m + 1) = 900
200.5 − 270
P (W 200 ) = Φ
= Φ ( −2.317 ) = 1 − 0.9898 = 0.0102 < 0.025
30
(
)
Therefore, reject H0; the two samples are not drawn from identical distributions.
62
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Worked solutions
10
4
aH0: Median birth rates have not changed from 2000 to 2005.
H1: Median birth rates have decreased from 2000 to 2005.
5% significance level, one-tailed test, n = 7, critical value 3
Difference
−2.55
−0.45
0.73
−0.28
−0.17
−0.29
0.06
|Difference|
2.55
0.45
0.73
0.28
0.17
0.29
0.06
Rank
Signed rank
7
5
6
3
2
4
1
−7
−5
6
−3
−2
−4
1
P = 7, Q = 21, so T = 7 > 3, therefore no reason to doubt H0; there has been no change in median birth rate
from 2000 to 2005.
b Wilcoxon rank-sum test
c
5% significance level, one-tailed test, m = n = 7, critical value 39. As W = 50 > 39, this does not change the
conclusion that there is no reason to doubt the null hypothesis.
11
H0: The median weight of the first-born twin is the same as that of the second-born.
H1: The median weight of the first-born twin is greater than that of the second-born.
8% significance level, one-tailed test, n = 45, normal approximation required
1
T ~ N µ, σ 2 : µ = 4 n ( n + 1) = 517.5
s 2 = 1 n ( n + 1)( 2n + 1) = 7848.75
24
(
)
T = 437
 437.5 − 517.5 
 = Φ ( −0.9030 ) = 1 − 0.8167 = 0.1833 > 0.08
P (T 437 ) = Φ 
7848.75 
Therefore, no reason to doubt H0; the median weight of the two twins is the same.
12
a
H0: Winston and Jamal have the same median time.
H1: Winston has a lower median time than Jamal.
X ~ B ( 5, 0.5 ) , 5% significance level, one-tailed test, no negative signs
P(X = 0) = 0.03125 < 0.05, therefore reject H0; there is sufficient evidence that Winston has a lower median
time than Jamal.
b The paired-sample sign test can only be used if the samples are drawn under the same conditions for
each point. In this case the races being different will have different underlying conditions, so the test is
not valid.
c
Use a Wilcoxon rank-sum test.
H0: Winston and Jamal’s times are drawn from identical distributions.
H1: Winston’s times are lower than Jamal’s.
5% significance level, one-tailed test, m = n = 5, critical value 19.
Using Jamal as m Rm = 34, m ( n + m + 1) − Rm = 21, therefore W = 21.
W > 19, therefore, no reason to doubt H0; Winston and Jamal’s times are drawn from identical
distributions.
As only the second non-parametric test is valid, this suggests that there is insufficient evidence to pick
Winston ahead of Jamal. For example, Winston’s times may have come in races where the wind was
behind, whereas Jamal’s may have all been into a headwind.
If the times had come from the same races (so under the same conditions) there might have been
sufficient evidence to pick Winston ahead of Jamal, as he would have beaten him five times out of five.
13 a H0: Median black-fly damage is identical on crops treated by organic or chemical pesticides.
H1: Median black-fly damage is not identical on crops treated by organic or chemical pesticides.
5% significance level, two-tailed test, n = 9, critical value 5
63
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4 Non-parametric tests
Difference
−4.6
1.2
−7.6
−4.0
3.4
−1.1
−3.2
−5.8
−3.3
|Difference|
4.6
1.2
7.6
4.0
3.4
1.1
3.2
5.8
3.3
7
2
9
6
5
1
3
8
4
−7
2
−9
−6
5
−1
−3
−8
−4
Rank
Signed rank
P = 7, Q = 38, so T = 7 > 5, therefore no reason to doubt H0; type of pesticide makes no difference to
prevalence of black-fly damage.
b The paired-sample t-test could be used to test whether the mean damage is the same.
c
H0: Mean difference of black-fly damage between crops treated by organic or chemical pesticides is zero.
H1: Mean difference of black-fly damage between crops treated by organic or chemical pesticides is not zero.
5% significance level, two-tailed test
Estimated standard deviation of differences
s=
9  (−4.6)2 + 1.2 2 + … + (−3.3)2
2
− ( 6.4 − 3.6 ) 
8 
9

= 3.416
Test statistic T = 6.4 − 3.6 = 2.459
3.416
9
Critical value from t-distribution with 8 degrees of freedom is 2.306
As T > 2.306, reject H0; there is sufficient evidence of a difference in the mean black-fly damage between
the two pesticides.
d Paired sample t-test as Wilcoxon signed-rank test doesn’t take into account the magnitude of the
differences. Different regions may respond better to different pesticides.
14 a H0: Ship-building times in Guangnan and Jiangzhou are drawn from identical distributions.
H1: Ship-building times in Guangnan and Jiangzhou are not drawn from identical distributions.
10% significance level, two-tailed test, m = 3, n = 4, critical value 6
Ranking: 23J, 27J, 32G, 34J, 40J, 43G, 54G
Rm = 16, m (n + m + 1) − Rm = 8
Therefore W = 8. W > 6, so no reason to doubt H0; the ship-building times are the same.
b H0: The mean ship-building times in Guangnan and Jiangzhou are equal.
H1: The mean ship-building times in Guangnan and Jiangzhou are not equal.
10% significance level, two-tailed test, degrees of freedom 3 + 4 − 2 = 5
Critical value from t-distribution is 2.015
Sample means xG = 43 and x J = 31
Estimate of shared variance
2 × 121 + 3 × 170
3 = 82.4
s2 =
5
43 − 31
Test statistic T =
= 1.731
82.4 13 + 14
As T < 2.015, there is no reason to doubt H0; the mean ship-building times are the same
for both companies.
(
c
15
)
Advantage is that it does not rely on the underlying populations being drawn from a normal
distribution. Disadvantage is that it does not use all the information (it only uses the relative sizes and
not the exact values) of the data points to test the hypothesis.
1
× 10 × (10 + 1) = 55,
55 assuming the first ten ranks were all from the sample of size m.
2
b Let ω be the maximum value at which the null hypothesis would be just rejected. Using a normal
approximation
a R
m=
(
)
1
W ∼ N µ, σ 2 : µ = m (n + m + 1) = 125,
2
64
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
4
Worked solutions
σ 2 = 1 nm (n + m + 1) = 875
12
3




+ 0.5 − 125 
ω
+ 0.5 − 125 
ω


P (W ω ) = Φ
⇒ 0.01 = Φ




875
875




3
3
= −2.326 this yields ω = 84.78
Therefore W < 84.8.
Mathematics in life and work
Use a matched-pairs Wilcoxon signed-rank test, as the nesting locations are the same throughout.
H0: Median number of eggs from earlier year to later year is unchanged.
H1: Median number of eggs from earlier year to later year has decreased.
5% significance level, one-tailed test, n = 12, critical value 17
For 2000 to 2005, T = 19 and for 2005 to 2010 T = 18. In both cases T > 17, so there is no reason to doubt H0; the
median number of eggs is unchanged.
However, for 2000 to 2010, T = 9 and T < 17, so the null hypothesis is rejected in favour of H1; the median number
of eggs has decreased.
This demonstrates that even if over shorter periods there is no evidence to show egg numbers are decreasing, in
the longer run the evidence supports this hypothesis.
Location
A
B
C
D
E
F
G
H
I
J
K
L
2000
154
239
107
167
130
245
280
68
179
294
273
249
2005
99
201
121
162
129
258
254
90
162
278
242
252
Difference
55
38
−14
5
1
−13
26
−22
17
16
31
−3
|Difference|
55
38
14
5
1
13
26
22
17
16
31
3
Rank
12
11
5
3
1
4
9
8
7
6
10
2
Negative ranks
5
4
8
2
Location
A
B
C
D
E
F
G
H
I
J
K
L
2005
99
201
121
162
129
258
254
90
162
278
242
252
2010
109
193
76
140
118
246
230
108
153
203
269
212
Difference
−10
8
45
22
11
12
24
−18
9
75
−27
40
|Difference|
10
8
45
22
11
12
24
18
9
75
27
40
Rank
3
1
11
7
4
5
8
6
2
12
9
10
Negative ranks
3
Location
6
9
A
B
C
D
E
F
G
H
I
J
K
L
2000
154
239
107
167
130
245
280
68
179
294
273
249
2010
109
193
76
140
118
246
230
108
153
203
269
212
Difference
45
46
31
27
12
−1
50
−40
26
91
4
37
|Difference|
45
46
31
27
12
1
50
40
26
91
4
37
9
10
6
5
3
1
11
8
4
12
2
7
Rank
Negative ranks
1
8
65
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
5 Probability Generating Functions
5 Probability generating functions
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in
this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Prerequisite knowledge
1
Mean, E(X ) = 2 × 0.2 + 5 × 0.5 + 7 × 0.3 = 5
Variance, Var(X ) = 22 × 0.2 + 52 × 0.5 + 72 × 0.3 – 52 = 3
iiFrom the expansion, the coefficient of t r is
P ( X = r ) = kα r
b
GX (1) =
kα
1
=1⇒α =
1−α
k +1
a G
eometric distribution requires a set of trials
that have two outcomes (success or failure,
equivalently, pass or fail his test) with these
Variances, Var(X ) = 2, Var(Y ) = 5 × 0.4 × (1 – 0.4) =
outcomes being independent from trial to trial
1.2
and with a fixed probability of success. These
−2 
2
3
1


−
2
×
−
3
(
)
(
)
(−are
2) ×stated
(−3) ×in
(−the
4) question.
1
9
9
1
1
1
assumptions
Once
3
× − x +
+ − x + …
 = 16 1 + ( −2 ) × − 4 x +
2 = 9 × 16 1 − 4 x
2
4
6
4
Sudhir passes his test, he does not retake it,
(4 − x )




which is as in the geometric distribution: once
3
−2 
1

( −2 ) × ( −3) × − 1 x 2 + (−2) × (−3)
× (−4)
1
9 
1
1
there
has+ been
= 9 ×  1 − x  = 1 + ( −2 ) × − x +
− xa success
+ … the trials end.
16
4
16
4
2
4
6
4



 2
3
1
2 1 2 2
1
2
1
b
G
t
=
t
+
×
t
+
× t3+
× t4 +…
(
)
2
3
X

3
3 3
3
3
3
3
−2 ) × ( −3 )
(
(−2) × (−3) × (−4)
1
1
+
× − x +
+ − x + …
2
4
6
4
2
3


1 
2
2
2
G (t ) = t  1 + t + t + t + …
3 
3
3
3
9
1
3 2 1 3
9
9
272
9 X3

=
+
x+
x +
x …
1+ x + x + x … =
16
2
16
16
16 32
256
256
This is an infinite geometric series with a
1
3 2 1 3
9
9
27 2
9 3
x+ x + x … =
+
x+
x +
x …
common ratio of 2 t and initial term
2
16
16
16 32
256
256
3
1
t . Therefore
3
1t
Exercise 5.1A
t
GX (t ) = 3
=
1 − 23 t 3 − 2t
1 1
3
1
1
1 GX (t ) = + t + t 2 + t 3 + t 4
1 = 1 as required.
2 5
20
10
20
GX (1) = 3 −
2
2
(
(
)
)
)
( )
( )
( )
( )
}
2
4
Expectations, E(X ) = 2, E(Y ) = 5 × 0.4 = 2
{

 1010

 1010
GHG(Ht ()t=) =   (0.7
)10)10+ +   (0.7
)9 )(90.3
)t )t
(0.7
(0.7
(0.3
 0 0 
 1 1 
 10
  8 8 2 22 2 …
 10
++
) )(0.3
) )t t+ + …
(0.7
(0.3
 2  (0.7
 2
a
i
( )
( )
( )
}
c GX (t ) =
5
()
()
( ) ( )
pt
1 − (1 − p )t
GX (t ) = e −λ + e −λ λt +
e −λ λ 2 2 e −λ λ 3 3 … e −λ λ r r
t +
t + +
t
r!
2
6

( λt )2 + ( λt )3 + …
= e −λ  1 + λt +

2!
3!


 10
  10 1010 10
 10
++
(0.3
) )t t
(0.3
 10
 10
 
= (0.7 + 0.3t )
Using the Maclaurin series result this equals
= e −λe λt = e λt − λ = e λ(t −1) as required.
10
3
( )
(1 − αt )−1 = 1 + αt + (αt )2 + (αt )3 + …
−1
kα t (1 − α t ) = kα t + k(α t )2 + k(α t )3 + k(α t ) 4 + …
⇒ P ( X = 1) = kα
66
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
5
Worked solutions
Exercise 5.2A
1
E( X ) = G′X (1) =
3 t 3 + 1 t 10
GX(t ) = 15 + 25 t + 10
10
2
9
2
9
G′X (t ) = 5 + 10 t + t
9 + 1 = 2.3 = E X
G′X (1) = 25 + 10
( )
G′′X (t ) = 95 t + 9t 8
G′′X (1) = 95 + 9 = 10.8
Var( X ) = G′′X (1) + G′X (1) − G′X (1)


GX′′ (t ) =
2
24
2
6
b G
′X (t ) = 125 ( 3 + 2t ) and G′′X (t ) = 125 ( 3 + 2t )
6
× 25 = 1.2
Therefore E ( X ) = G′X (1) =
125
3
2 (1 − p )
p2
=
a GY (1) =
a ( 5 − b ) + b (1 + a )
=1
(5 − b )2
Substituting a = 4 − b gives
( 4 − b )(5 − b ) + b (1 + 4 − b ) = 4 = 1
5−b
(5 − b )2
⇒ b = 1 and a = 3
c Expanding
(2 + m )2 = 1 ⇒ (m + 2 )2 = 25 , as m > 0,
25
m + 2 = 5, m = 3
(
)
1 t 2 +…
(1 + 3t )(5 − t )−1 = 15 (1 + 3t ) 1 + 15 t + 25
4
4m m 2
+
t
a Expanding brackets: GZ (t ) = t −1 +
25
25
25
Therefore Z can take values {−1, 0,1}
b G
Z (1) =
)
3
2 (1 − p ) p
1 1− p
+ 2 − 2 = 2 as required
p2
p
p
p
so GY′ (1) =
2
Var( X ) = G′′X (1) + G′X (1) − G′X (1)


24
2
= 125 × 5 + 1.2 − 1.2 = 0.72
(
2 (1 − p ) p
1 − (1 − p )
1+ a
=1⇒ a = 4−b
5−b
a ( 5 − bt ) + b (1 + at )
b GY′ (t ) =
(5 − bt )2
5
3
1
a GX (1) = k ( 3 + 2 × 1) = 125k = 1 ⇒ k = 125
p
= 1 as required
p2 p
(1 − (1 − p )t )3
Var ( X ) =
= 10.8 + 2.3 − 2.32 = 7.81
=
2 (1 − p ) p
so GX′′ (1) =
2
p
(1 − (1 − p ))2
Mean is expectation, so equal to 1. Therefore
P (Y = 1) is the coefficient for t from the
16
expansion: 1 1 t + 3t so P (Y = 1) =
25
5 5
(
)
Exercise 5.3A
6( 2 + 3t ) × 25t − 25 × ( 2 + 3t )
c GZ′ (t ) =
625t 2
(2 + 3t )( 3t − 2 )
=
25t 2
Therefore the mean (expectation) is
(2 + 3)( 3 − 2 ) = 0.2
GZ′ (1) =
25
2
G′′ (t ) =
Z
(
18t × 25t 2 − 50t 9t 2 − 4
625t 4
8 = 0.32
G′′Z (1) = 25
)=
8
25t 3
1
GX +Y (t ) = GX (t ) × GY (t )
(
) 0.05
(8 + 12t )
t
= 0.005 ( 6 + 3t + t )( 8 + 12t )
(t ) = 0.005 ( 48 + 24t + 8t + 72t + 36t
GX +Y (t ) = 0.1t 2 6 + 3t + t 2 ×
2
GX +Y
4
2
4
2
4
5
+ 12t 6
)
P ( X + Y is odd ) is the sum of coefficients of tr where
r is odd. P ( X + Y is odd ) = 0.005 ( 24 + 36 ) = 0.3
2
Let Yi be the score on an individual dice, then
2
2 3
1
1
1
1
1
1
G′′Z (1) + G′X (1) − G′X (1)  = 0.32 + 0.2 − 0.04 =
GYt (t ) = t + t 2 + t 3 + t 4 + t 5 + t 6
5
6
6
6
6
6
6


1
2
= t (1 + t + t 2 + ... + t 5)
2 3
6
G′′Z (1) + G′X (1) − G′X (1)  = 0.32 + 0.2 − 0.04 =
5


using formula for geometric sum
pt
1 1−t6
t 1−t6
1
4 GX (t ) =
therefore
=
t
×
=
1 − (1 − p )t
6
1−t
6(1 − t )
p
(1
(1
p
)
t
)
pt
(1
p
)
×
−
−
−
×
−
−
p
G′X (t ) =
=
(1 − (1 − p)t )2
(1 − (1 − p)t )2
Standard deviation is
(
) (
)
67
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
5 Probability Generating Functions
So for X = Y1 + Y 2 + Y 3
(
) 
t 1 − t 6
GX (t ) = 
 6(1 − t )
as required.
0.5(0.5 + 0.5t )5 × (1 − 0.5t )−1
3
t3 1−t6 
= 216
 1 − t 

= 0.015625 + 0.0859375t + 0.19921875t 2 + …
3
Therefore
P ( A + B < 3 ) = P ( A + B = 0 ) + P ( A + B = 1)
P ( X 16 ) is the sum of the coefficients of t16, t17
and t18 Begin by noting
1−t 6 = 1+t +t2 +t 3 +t 4 +t 5
1−t
(
3
(
therefore P ( X 16 ) =
3
GX1 (t ) = 0.75t
−2
GX (t ) = 0.75t
= 0.015625 + 0.0859375 + 0.19921875 = 0.301
6
a T
his is a case of the sum of two geometric
distributions, each with a PGF of
0.25t
G X i (t ) =
1 − 0.75t
So
)
6 + 3 +1
5
=
216
108
5
+ 0.25t and X = X 1 + X 2 + … + X 25
−2
+ P( A + B = 2 )
(3 s.f.)
)
1−t 
so 
= … + 6t 13 + 3t 14 + t 15
 1 − t 
6
5 25
+ 0.25t 
0.25t 
G X1 + X 2 (t ) = G X1 (t ) × G X 2 (t ) = 
1
−
 0.75t 
b GX′ (t ) =
To find expectation
4
E(X ) = GX′ (1) = 25 × (−1.5 + 1.25) × (0.75 + 0.25)24
= −6.25
c
0.125(1 − 0.75)2 + 1.5(1 − 0.75) × 0.0625
=8
(1 − 0.75) 4
GX (t ) = (0.25t )2(1 − 0.75t )−2
= (0.25t )2(1 + 1.5t + 1.6875t 2 + …)
P ( X < 5 ) = 0.252 (1 + 1.5 + 1.6875 ) = 0.262 to
a G′X (t ) = 0.3 + 0.74t + 0.6t 2 + 0.16t 3
3 s.f.
E(X ) = G′X (1) = 0.3 + 0.74 + 0.6 + 0.16 = 1.8
G′′X (t ) = 0.74 + 1.2t + 0.48t 2
Var(X ) = G′′X (1) + G′X (1) − G′X (1) 


0.125t (1 − 0.75t )2 + 1.5(1 − 0.75t ) × 0.0625t 2
(1 − 0.75t ) 4
E(X ) = GX′ (1) =
GX′ (t ) = 25 × (−1.5t −3 + 1.25t 4) × (0.75t −2 + 0.25t 5 )24
Exam-style questions
2
2
= (0.74 + 1.2 + 0.48) + 1.8 − 1.8 = 0.98
1
 n
n  n
n −1
a G X (t ) =   (1 − p ) +   (1 − p ) pt
0
1 
 n
n −2
+   (1 − p ) p 2t 2
2
GY (t ) = 0.3 + 0.5t + 0.2t 2 as [ GY (t )] 2 = GX (t )
b i ii P (Y = 1) = 0.5
iii 2E (Y ) = E ( X ) ⇒ E (Y ) = 0.9 and
2Var (Y ) = Var ( X ) ⇒ Var (Y ) = 0.49
5
The number of heads Alberta gets, A, is a
geometric sequence, but if she succeeds (gets a
tail) on her first throw there are no heads. This can
be thought of as a geometric distribution with the
0.5
values starting at zero. Hence GA (t ) =
1 − 0.5t
The number of heads Bruno gets, B, is simple
binomial, so GB (t ) = ( 0.5 + 0.5t )5
GA + B (t ) = GA (t ) × GB (t ) =
0.5 ( 0.5 + 0.5t )
1 − 0.5t
5
= 0.5(0.03125 + 0.15625t + 0.3125t + …)
= 0.015625 + 0.078125t + 0.15625t + …
2
(
)n
GX′′(t ) = n (n − 1)p 2((1 − p) + pt )n − 2
Therefore
E(X ) = GX′ (1) = np ((1 − p) + p)n −1 = np and
Var(X ) = GX′′(1) + GX′ (1) − GX′ (1) 


2
2
 n
 n
n−3
+   (1 − p ) p 3t 3 + … +   p nt n = (1 − p ) + pt
3
 n
b GX′ (t ) = np ((1 − p) + pt )n −1 and
0.5(0.5 + 0.5t )5
(1 − 0.5t )−1 = 1 + 0.5t + 0.25t 2 + …
2
2
= n(n − 1)p 2 + np − n 2p 2 = np(1 − p)
(
)
2
1
a GX (1) = k 1( 3 + 4 ) + (1 + 1) = 11k = 1 ⇒ k = 11
b The coefficient of the t2 term is P(X = 2).
4
Therefore P(X = 2) =
11
68
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
5
Worked solutions
( )
()
−1
2
1
1

t
t
t
a 
a
(2 + 8t + 12t 2) and GX′′(t ) = (8 + 24t )
= t 1 + +
+ …
c GX (t ) = t 1 −
11
11
b
b
b
b 
b

Therefore E(X ) = GX′ (1) = 2 and
a
a
a
2
3
2
t +
t +…
=
t+
32
10
a +1
Var(X ) = GX′′(1) + GX′ (1) − GX′ (1)  =
+2−4=
(a + 1)2
(a + 1)3
11
11 

d p = a and 1 − p = 1
3 GY (t ) = pt + (1 − p)pt 2 + (1 − p)2pt 3 + (1 − p)3pt 4 + …
a +1
a +1
2
3
2 −2
2
= pt (1 + (1 − p )t + (1 − p ) t 2 + (1 − p ) t 3 + …)
2
t
6 a G′X (t ) = 4te
and G′′X (t ) = 4(1 + 4t 2)e2t − 2
This is an infinite geometric series with common
so E(X ) = G′X (1) = 4
ratio (1 − p )t and first term pt.
2
pt
Therefore GY (t ) =
and Var(X ) = G′′X (1) + G′X (1) − G′X (1) 


1 − (1 − p )t
c G
X′ (t ) =
= 20 + 4 − 4 2 = 8
To find expectation and variance
GY′ (t ) =
(
)
p × 1 − (1 − p )t − pt × − (1 − p )
(1 − (1 − p )t )
E(Y ) = G′Y (1) =
G′′Y (t ) =
GY′′ (1) =
4
p
(1 − (1 − p ))2
2 (1 − p ) p
(1 − (1 − p )t )
(
2 (1 − p ) p
1 − (1 − p )
Var (Y ) =
)
p
=
(1 − (1 − p )t )
p
1
= 2=p
p
=
4
2
)
1
= 66c = 1 ⇒ c =
66
(
)
(
(
1
2
c GQ′′ (t ) = 66 12 (1 + t ) + 36
)
)
so, Var (Q ) = GQ′′(1) + GQ′ (1) − GQ′ (1)


(
2
( )
)
2
5
a
GX (1) =
b G
X′ (t ) =
× et −1
= e(t −1)(2t + 3)
(
1
(1 + t )4 + 2 ( 2 + 3t )2
4356
2
a
=1⇒ b = a +1
b −1
a ( b − t ) + at
(b − t )2
a ( b − 1) + a a 2 + a a + 1
=
= a
E( X ) = G′X (1) =
a2
(b − 1)2
)(
(
)
)
−1

1
t t 2 …
3 + t 2 1 + +
+ 
5
5 25


(
)
Coefficient is: 1 1 + 3 × 1 = 28
25 125
5
2t (5 − t ) + (3 + t 2)
(5 − t )2
so E(X ) = G X′ (1) =
H ′Z (t ) =
3
4
(
)
2
4t 3 + t 2 ( 5 − t ) + 2( 5 − t )( 3 + t )
so E( Z ) = H Z′ (1) =
2
2
(5 − t )
4
256 + 128 = 3 = 2 × E X
( )
256
2
as required
8
)
(
1
t
3 + t2 1−
5
5
 3 + t2
2
d H Z (t ) = [ G X (t )] ⇒ H Z (t ) = 
 5 − t 
= 0.724
d GP +Q (t ) = GP (t ) × GQ (t )
=
c GX′ (t ) =
2
1
46
46
= 66 12(1 + 1) + 36 + 33 − 33
=
2 +t − 3
2 −2
a GX (1) = 3 + a = 1 ⇒ a = 1
5 −1
GX (t ) =
b GQ′ (t ) = 1 4 (1 + t )3 + 12 ( 2 + 3t )
66
Therefore,
1
46
3
E (Q ) = GQ′ (1) =
4 (1 + 1) + 12 ( 2 + 3 ) =
66
33
= e2t
2 (1 − p )
p2
a GQ (1) = c (1 + 1) + 2 ( 2 + 3 )
b GX +Y (t ) = G X (t ) × GY (t ) = e2t
b The coefficient of t 2 represents P ( X = 2 ).
2 (1 − p ) p
1 1− p
+ 2− 2 = 2
p2
p
p
p
(
2
7
so
3
3
2
(
)
a G
D1 (t ) = k t + 2t 2 + 3t 3 + 4t 4 + 5t 5 + 6t 6 ,
When t = 1, 21k = 1
therefore k = 1
21
1
1 + 4t + 9t 2 + 16t 3 + 25t 4 + 36t 5 ,
b GD′ 1 (t ) =
21
′ (1) = 13
GD
1
3
1
4 + 18t + 48t 2 + 100t 3 + 180t 4 ,
and GD′′1 (t ) =
21
′′ (1) = 50
GD
1
3
(
)
(
)
69
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
5 Probability Generating Functions
′ (1) = 13 and
E( X ) = GD
1
3
11
a E
( X ) = 0 × (1 − p ) + 1 × p = p and
Var ( X ) = [02 × (1 − p ) + 12 × p] − p 2 = p (1 − p )
2
′ (1)  = 20
Var ( X ) = GD′′1 (1) + GD′ 1 (1) − GD
9
 1 
c Let D be the sum of the scores on the three
dice, then
3
1
GD (t ) =
t + 2t 2 + 3t 3 + 4t 4 + 5t 5 + 6t 6
9261
(
)
P ( D 16 ) is given by the coefficients of the
t16, t17 and t18 terms.
1
26
P ( D 16 ) =
432 + 450 ) + 540 + 216 ) =
9261 ((
147
a GV (t ) = a t 2 + 2 t −2
b
b
a+2
GV (1) =
=1⇒b = a +2
b
GV′ (t ) = 2a t − 4 t −3
b
b
E(V ) = GV′ (1) = 2a − 4 = − 23 ⇒ b = 6 − 3a
b
Solving simultaneously yields a = 1 and b = 3
b GX (t ) = P ( X = 0 ) × t 0 + P ( X = 1) × t 1 = (1 − p ) + pt
−λ 2
−λ 3
c GX (t ) = e −λ + e −λ λt + e λ t 2 + e λ t 3
2
6
+…+

( λt )2 + ( λt )3 + …
= e −λ  1 + λt +

2!
3!


Using the Maclaurin series result
9
P(V = 0) is the coefficient of the constant
term. Using the binomial expansion, this
3
6  3
coefficient is   1 2 = 160
729
 3 3 3
1
10 a GY (1) = k (1 + a ) = 1 ⇒ k =
1+ a
( )( )
Substituting k = 1 gives
1+a
2
= λ2 + λ − λ2 = λ
as required.
( (
−λ 1− (1− p )+ pt
)) = e λp(t −1)
f
KZ′ (t ) = λ pe λp(t −1) and KZ′′ (t ) = ( λ p ) e λp(t −1)
KZ′ (1) = λ p and KZ′′ (1) = ( λ p ) so
2
2
E ( Z ) = KZ′ (1) = λ p and
Var ( Z ) = ( λ p ) + λ p − (λ p)2 = λ p
2
2
(
H′Y (1) = λ and H′′Y (1) = λ 2 so
e KZ (t ) = HY ( G X (t )) = e
as required
2
= 6 ak + 3 ak − ( 3ak ) = 9 ak (1 − ak ) = 2
−λ 1 − t
= e λt − λ = e ( ) as required.
Var ( X ) = H′′Y (1) + H′Y (1) − H′Y (1) 


G′Y (t ) = 3 akt 2 and G′′Y (t ) = 6 akt
Var ( X ) = GY′′ (1) + GY′ (1) − GY′ (1) 


=e e
E ( X ) = H′Y (1) = λ and
6
−λ λt
−λ 1− t
−λ 1 − t
d H′Y (t ) = λ e ( ) and H′′Y (t ) = λ 2e ( )
b L
et V be the sum of six independent
observations, then
6
1
2
HV (t ) = [ GV (t )] =  t 2 + t −2 
3 
 3
e −λ λ r r
t
r!
)
g Po ( λ p )
Mathematics in life and work
9a
9a
a
=
=2
1−
1+ a
1+ a
(1 + a )2
L
et X be the score of the voter:
GX (t ) = 0.4t + 0.25 + 0.35t −1
1
2 1
2a 2 − 5a + 2 = 0 ⇒ a = or 2 and k = or
2
3 3
I f sample is random, then scores for each voter
will be independent, so let Y be the total score of
three voters, hence
(
b H Z (t ) = k10 1 + at 3
)
10
P ( Z 3 ) is given by the sum of coefficients
of the constant term, t, t 2 and t 3. Using a
binomial expansion
(
)
H Z (t ) = k10 1 + 10at 3 + …
1
2
For a = , k = : P ( Z 3 ) = 0.104
2
3
For a = 2, k = 1 : P ( Z 3 ) = 0.000356
3
(
GY (t ) = 0.4t + 0.25 + 0.35t −1
)
3
I n the sample of three, these provisos imply that
Y 2 (2 yes, 1 non-vote or 3 yes votes).
Therefore, P (Y 2 ) is the sum of the coefficients
of the t2 and t3 terms.
P (Y 2 ) = 0.4 3 + 3 × 0.4 2 × 0.25 = 0.184
70
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
Summary Review
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in
this publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Warm-up Questions
1
2σ z
< 0.2
n
σ = 0.17 and for a 99% confidence interval z = 2.576
Width of the confidence interval is
2 × 0.17 × 2.576
< 0.2
n
n > 4.3792…
n > 19.17…
nMIN = 20
2
H0: μ = 17
H1: μ ≠ 17
This is a two-tailed test with 2.5% in each tail ⇒
z = ±1.96
x = 17.8 + 22.4 + 16.3 + 23.1 + 11.4 = 18.2
5
x − µ 18.2 − 17
=
= 1.12
σ
2.4
n
5
−1.96 < 1.12 < 1.96 ⇒ not in the critical region.
Accept H0: μ = 17. Accept the lecturer’s claim.
3
C ~ N(91, 3.2 2) and S ~ N(72, 2.6 2)
X = C1 + ... + C6 + S1 + ... + S6 + 550
E(X ) = 6 × 91 + 6 × 72 + 550 = 1528
Var(X ) = 6 × 3.22 + 6 × 2.62 + 02 = 102
X ~ N (1528, 102 )


P ( X > 1550 ) = P  Z > 1550 − 1528 


102
= P ( Z > 2.178 )
= 1 − 0.9853 = 0.0147
= 1 − P ( Z 2.178 )
A Level Questions
1
∑ x = 5 and
x = 5
N
∑ x 2 = 11
s x2 =
for N observations
1 11 − 52 
N − 1 
N 
71
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
∑ y = 10
∑ y 2 = 160 for 10 observations
and
2

s y2 = 1 160 − 10  = 150
9
10  9
y = 10 = 1
10
So, the pooled estimate is:
+ 9 150
( N − 1) × N 1− 1 11 − 25
N
9
2
sp =
N + 10 − 2
(
) ( )
11 − 25 + 150
N
N +8
s p2 =
Given that s p2 = 12
12 ( N + 8) = 11 − 25 + 150
N
12N + 96 = 161 − 25
N
2
12N − 65N + 25 = 0
(12N − 5)(N − 5) = 0
N= 5
12
We know that N must be an integer, so N = 5.
2
H0: median = 400 ml
H1: median < 400 ml
The deviations from the median are: −10, −3, −15, 10, −8, −30, 30, −3, −12, −9, −25, 42, −4, −28, −19, 4.
There are 4 positives and 12 negatives.
Under H0, X ~ B (16, 0.5 )
0.0384 < 0.05 ⇒ this result is in the critical region. Reject H0 and accept H1.
The customers’ complaints are justified.
3


F( x ) = 


Y=
New limits are: 1 → 1, 4 → 16
For 1 y 16 , G ( y ) = 1
63
N =5
or
P ( X 12 ) = 0.0384
⇒
X2



G( y) = 



1 
63 
1 x 4
x>4
X= Y
((
y
)
3
)
 3 
− 1 = 1  y 2 − 1
63 

y <1
0
3
y2
1
Differentiating:
 1 1
2

g ( y ) =  42 y
 0

x <1
0
1 3
( x − 1)
63
1

− 1

1 y 16
y > 16
1 y 16
otherwise
72
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
At the median, G(y) = 0.5 and y = m
3
1 m 2 − 1 = 0.5

63 

i
3
m 2 = 32.5
m = 10.18
16
ii E (Y ) = ∫ y g ( y ) d y
1
=
4
1 16 23
y dy
42 ∫1
16
 5
= 1 y 2 
105  
1
= 1 [1024 − 1] = 9.74
105
i G X (1) = 1 ⇒ 2 + a = 1 ⇒ a = 3
5
ii For P(X = 2), we need the coefficient of t2.
)(
)
2 + 3t 3 = 2 + 3t 3 7 − 2t −1 = 1 2 + 3t 3 1 − 2 t −1
(
)
7 − 2t
7
7
3

2 + 3t = 1 2 + 3t 3 1 + −1 − 2 t + (−1)(−2 ) − 2 t
 ( ) 7
7 − 2t 7
2!
7

The term in t2 is:
(
)
(
(
( )
)
( ) = 27 × 494 t
1 × 2 × (−1)(−2 ) − 2 t
7
2!
7
iii G X′ (t ) =
2
2
( ) + …

2
= 8 t2
343
So P ( X = 2 ) = 8
343
(7 − 2t )9t 2 − (2 + 3t 3 )( −2 )
(7 − 2t )2
5 × 9 − 5 × ( −2 ) 11
E ( X ) = G X′ (1) =
=
25
5
5
Let d = score before eating fruit – score after eating fruit
H0: µd = 0
there is no difference between the two sets of results
H1: µd < 0
there is an increase in the results
This is a one-tailed test with p = 0.05, v = 13, so the critical value is −1.771
Calculating the differences and squares:
Student
1
2
Before
15
10
After
16
Differences
−1
1
Squared
∑ d = −12
xd =
and
3
4
5
6
7
8
9
10
11
12
13
14
7
12
18
16
15
13
10
5
19
20
14
15
7
11
14
19
15
12
15
3
−4
−2
−1
1
3
−2
11
7
18
19
19
18
−1
−2
1
1
−5
−3
9
16
4
1
1
9
4
1
4
1
1
25
9
∑ d 2 = 86
−12
6
=−
14
7

(−12)2  = 1 × 530 = 530
s d2 = 1 86 −
13 
14  13
7
91
73
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
The test statistic is:
− 67
= −1.329
530
91
14
−1.329 > −1.771 ⇒ not in the critical region. Accept H0
There is insufficient evidence to claim that eating fruit improves mathematical ability.
The claim is not justified.
6
H0: gender and preferred brand are independent
H1: gender and preferred brand are not independent
Table of expected frequencies:
A
B
C
Male
28.57
37.71
13.71
Female
21.43
28.29
10.29
X2 = ∑
=
(O − E )2
E
( 32 − 28.57 )2 + ( 36 − 37.71)2 + (12 − 13.71)2 + (18 − 21.43)2 + ( 30 − 28.29)2 + (12 − 10.29)2
28.57
37.71
13.71
21.43
28.29
10.29
= 0.4118 + 0.07754 + 0.2133 + 0.5490 + 0.1034 + 0.2842
= 1.639
For a 5% test with v = 1 × 2 = 2, the critical value is 5.991.
1.639 < 5.991 ⇒ not in the critical region. Accept H0.
Gender and brand are independent. There is no difference in the preferences between males and females.
If the sample is n times larger, then χ2 will also be n times larger. For χ2 to be in the critical region it must be
greater than 5.991.
1.639n > 5.991
n > 3.655
Since n is an integer, nMIN = 4.
x
x
⌠ x
 x2 
1 2
x2 4
7
 6 dx =  12  = 12 − 12 = 12 x − 4
 2
⌡2
 0
x <2
 1
2
F( x ) = 
2x4
x −4
12

x>4
 1
(
(
)
Y = X3 ⇒ X = 3 Y
New limits are: 2 → 8, 4 → 64
For 8 y 64
G(y ) = 1
12
((
3
)
y
)
2
)
 2

− 4 = 1  y 3 − 4
12 

So, the CDF is:
 0


 1  23
y − 4
G( y) = 
12 


 1

y <8
8 y 64
y > 64
74
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
Differentiating:
 1 −1
3

g ( y ) =  18 y
 0

8 y 64
otherwise
64
E (Y ) = ∫ y g ( y ) d y =
8
=
64
1 64 23
1  53 
y dy =
y
∫
18 8
30  
8
1  5
4 − 2 5  = 33.1
30 
8
i
H0: median difference = 0 There is no change in the amount of litter in the street
H1: median difference < 0 There has been a reduction in the amount of litter in the street
Let the difference be after minus before.
Calculating the differences and ranks gives:
Site
A
B
C
D
E
F
G
H
I
J
Before poster
campaign
85
146
137
120
79
95
153
144
108
127
After poster
campaign
78
120
110
128
61
65
121
131
88
104
Difference
−7
−26
−27
8
−18
−30
−32
−13
−20
−23
Rank
−1
−7
−8
2
−4
−9
−10
−3
−5
−6
Sum of the positive ranks:
P =2
Absolute sum of the negative ranks:
Q = 53
Therefore T = 2.
For a one-tail test at the 1% level, T  5 to reject H0.
Since T = 2  5 we can reject H0 and accept H1.
There has been a reduction in the amount of litter in the street.
iiThe test tells us if there has been a significant change, but it does not establish cause and effect. In this
case, the reduced amount of litter may not be as a result of the poster campaign.
9
λ=
(0 × 7 ) + (1 × 20 ) + (2 × 39) + ( 3 × 16 ) + ( 4 × 14 ) + (5 × 2 ) + (6 × 1) + (7 × 1)
100
λ = 225 = 2.25
100
H0: data can be modelled by Po(2.25)
H1: data cannot be modelled by Po(2.25)
e −2.25 × 2.25r
, which gives:
r!
10.540, 23.715, 26.679, 20.009, 11.255, 5.065, 1.899, 0.6105, 0.2275
The expected values are calculated using 100 ×
The last three expected values are too small as they must be greater than 5, so the final four categories are
combined to get an observed value of 4 and an expected value of 7.802.
(O − E )2
= 1.189 + 0.5820 + 5.690 + 0.803 + 0.6695 + 1.853 = 10.8
E
At the 2.5% level with v = 4, the critical value is 11.14
10.8 < 11.14
X2 = ∑
⇒ Accept H0. The data can be modelled by Po(2.25).
10 i
GY (t ) = k (5t − at 4)
GY (1) = 1
E (Y ) = 2
⇒
GY′ (t ) = k (5 − 4at 3)
⇒ 1 = k (5 − a )
1
⇒
2 = k (5 − 4a )
2
75
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
2 ÷ 1
2 = 5 − 4a
5−a
10 − 2a = 5 − 4a
⇒
a = −5
2
Substituting in 1
⇒ k= 2
1=k 5+ 5
2
15
( )
ii
(
)
GY (t ) = 2 5t + 5 t 4 = 1 (2t + t 4)
2
3
15
1
3
′
GY (t ) = (2 + 4t )
3
GY″ (t ) = 1 (12t 2) = 4t 2
3
Var (Y ) = GY″ (1) + GY′ (1) − [ G Y′ (1)]
2
G ′ (1) = E(Y ) = 2
Y
GY″ (1) = 4 × 12 = 4
Var (Y ) = 4 + 2 − 2 2 = 2
iii
()
3
HZ (t ) = 1 (2t + t 4)3
3
3
= 1 ( 2t ) + 3(2t )2(t 4) + 3 ( 2t ) (t 4)2 + (t 4)3 
27
= 1 8t 3 + 12t 6 + 6t 9 + t 12 
27
P(Z  6) is the sum of the coefficients of t with powers  6.
8 12 20
P ( Z 6) =
+
=
27 27 27
11 i
x = 2478 = 45.05 s x2 = 343.75 = 6.25
55
55
y = 3981 = 56.87 s y2 = 857.5 = 12.25
70
70
For a 90% confidence interval, we need p = 0.95
+ 12.25
( 45.05 − 56.87 ) ± 1.645 × 6.25
55
70
⇒
z = 1.645
−12.7 µ x − µ y −10.9
ii
H 0: µ x − µ y = 0
H 1: µ x − µ y ≠ 0
The test statistic is
45.05 − 56.87 = −22.0
6.25 + 12.25
55
70
For a two-tail test at the 10% significance level, z = ±1.645
−22.0 < −1.645 ⇒ it is in the critical region. Reject H0.
μx is not the same as μy.
12 For 1  x  3,
F ( x ) = ∫ 1 dx = x + c
2
2
When x = 1, F ( x ) = 0 ⇒
1 +c = 0
2
⇒ c = −1
2
76
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
F ( x ) = x − 1 = 1 (x − 1)
2 2 2

0
 1
F ( x ) =  ( x − 1)
2

1

x <1
1 x 3
x>3
G(y) = P(Y  y)
Y = X3
(
G( y ) = P X 3 y
⇒
)
1


 1
= P X y 3  = F y 3 


 

0

  1 
G( y) =  1  y 3 − 1

 2

1

y <1
1 y 27
y > 27
 1 
For 1  y  27, G ( y ) = 1  y 3 − 1
2

1 −2
1
⇒ g(y ) = 6 y 3 = 2
6y 3
 1

2
g (y ) =  6 y 3

 0
1 y 27
otherwise
g( y)
0.2
1
6
27,
0
10
27
27 1
1
1
E (Y ) = ∫ y g (y) d y = ∫
27
6
20
25
y
1
y 3 dy
27
P(median Y  mean) = |P(Y < 10) – 0.5| = |G(10) – 0.5|
1  13 
10 − 1 − 0.5 = 0.0772 (3 s.f.)
2 

13
∑x = 2623, ∑x 2 = 1 376 081
x = 2623 = 524.6
5
15
4


 4
=  3 × 1 y 3  = 3  y 3  = 3 ( 81 − 1) = 10
24  
24
 4 6 1
1
=
5
1
54
s2 =
1
26232 
1 376 081 −
= 13.8
4 
5 
77
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
⇒
2.5% in each tail ⇒
95% confidence interval
There are 4 degrees of freedom
Therefore, the confidence interval is: 524.6 ± 2.776 13.8 = 524.6 ± 4.6118
5
[520, 529] to 3 s.f.
Let the first sample be sample A and the second sample be sample B.
H0: μA = μB
H1: μA ≠ μB
For sample B: ∑ x = 5216,
⇒
t4, 0.975 = 2.776
p = 0.975
(from tables)
= 521.6
∑x 2 = 2 720 780, x = 5216
10
2

s 2 B = 1  2720780 − 5216  = 12.71
10 
9
For the combined sample:
s 2p =
4 × 13.8 + 9 × 12.71
= 13.05
13
T = 524.6 − 521.6 = 1.516
13.05 1 + 1
5 10
(
)
10% significance level and 2-tail test ⇒ p = 0.95
There are 13 degrees of freedom
⇒
⇒
t13, 0.95 = 1.771
(from tables)
not in the critical region ⇒ accept H0.
1.5164 < 1.771
There is no significant evidence of a difference in the population means before and after the adjustments.
14 i
∞
∫0
Ae −λt dt = 1
∞
 − A e −λt  = 1
 λ
0
[0] −  − Aλ  = 1
A=λ
ii
A =1
λ
1
∫0 λe
−λt
dt ≈
16
100
1
 −e −λt  ≈ 16

0 100
 −e −λ  − [ −1] ≈ 16


100
16
−λ
e ≈1−
100
( )
λ ≈ −ln (1 − 16 ) = 0.174
100
−λ ≈ ln 1 − 16
100
For median:
T
∫0.174e
−0.174t
dt = 0.5
0
78
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
T
 −e −0.174t  = 0.5
0
[– e– 0.174T ] – [–1] = 0.5
e– 0.174T = 0.5
– 0.174T = ln 0.5
T = 3.98 years (3 s.f.)
15 G(y) = P(Y  y)
Y = X3
(
G( y ) = P X 3 y
⇒
)
1


 1
= P X y 3  = F y 3 


 
For 1  x  4, F ( x ) =
2
1
∫ 15x dx = 15 x
When x = 1, F ( x ) = 0 ⇒
2
F ( x ) = x − 1 = 1 (x 2 − 1)
15 15 15
i
+c
1
+ c=0
15

0

 2


G ( y ) =  1  y 3 − 1
15 



1

2
⇒ c=−
1
15
y <1
1 y 64
y > 64
Let m be the median value of Y.
G(m) = 0.5
1  m 23 − 1 = 0.5

15 
2
m 3 − 1 = 7.5
m 3 = 8.5
2
ii
m = 24.8
(3 s.f.)
−1 
−1

For 1 y  64, g ( y ) = 1  2 y 3  = 2 y 3
15  3
45

64
E (Y ) = ∫ y g ( y ) d y =
1
64
2 64 23
y dy
45 ∫1
64
5


 5
= 2  3 y 3  = 2  y 3  = 2 (1024 − 1)
45  5 
75  
75
1
1
= 27.28 = 27.3 (3 s.f.)
16 H0: μO – μI = 0
H1: μO – μI ≠ 0
Outdoor times – Indoor times: 0
.1, 2.1, –0.1, 0.2, 2.4, 0.5, 2.8, –2.6
∑x = 5.4, ∑x 2 = 25.08,
s=
x = 5.4 = 0.675
8
1  25.08 − 5.4 2  = 1.750
8 
7 
79
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
For the combined sample: t = 0.675 = 1.091
1.750
8
5% significance level and 2-tail test ⇒ p = 0.975
There are 7 degrees of freedom ⇒
1.091 < 2.365
There is no significant evidence that there is a difference between the indoor and outdoor swimming times.
t7, 0.975 = 2.365
⇒ not in the critical region
⇒
(from tables)
accept H0.
222.8
17 x = 10 = 22.28
4.12 = 0.6766
9
s=
⇒
2.5% in each tail ⇒
95% confidence interval
There are 9 degrees of freedom
Therefore, the confidence interval is:
⇒
t9, 0.975 = 2.262
p = 0.975
(from tables)
2
22.28 ± 2.262 0.6766 = 22.28 ± 0.4840
10
[21.8, 22.8] to 3 s.f.
3
3
18 E 2 x < 3 = 80⌠
 32 dx = 80  − 3  = 80  [ −1] −  − 3  = 80 × 0.5 = 40

⌡2 x
 x 2
 2 
4
4
 32 dx = 80  − 3  = 80   − 3  − [ −1] = 80 × 0.25 = 20
E 3x < 4 = 80⌠
  4 

⌡3 x
 x  3
5
5
⌠ 3 dx = 80  − 3  = 80   − 3  −  − 3  = 80 × 0.15 = 12
E 4x <5 = 80
  5   4 
⌡4 x 2
 x  4
6
6
 32 dx = 80  − 3  = 80   − 3  −  − 3  = 80 × 0.1 = 8
E 5x <6 = 80⌠
 x 5
  6   5 
⌡5 x
H0 : f ( x ) = 32
x
H1 : f ( x ) = 32 does not fit the data.
x
10% significance level ⇒ p = 0.9
There are 3 degrees of freedom
X2 =
fits the data.
⇒
χχ3232,, 00.9.9 == 66..251
251 (from tables)
( 36 − 40)2 + (29 − 20)2 + (9 − 12 )2 + (6 − 8)2
40
20
12
8
= 0.4 + 4.05 + 0.75 + 0.5 = 5.7
⇒
5.7 < 6.251
f ( x ) = 32
x
19
x = 42.5 = 5.3125
8
s = 15.519 = 1.4890
7
T = 5.3125 − 4.5 = 1.5434
1.4890
8
accept H0
fits the data
80
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
H0: μ = 4.5
H1: μ > 4.5
5% significance level and 1-tail test
There are 7 degrees of freedom
1.5434 < 1.895
There is not significant evidence that μ is greater than 4.5
95% confidence interval
There are 7 degrees of freedom
Therefore, the confidence interval is:
⇒
⇒
p = 0.95
t7, 0.95 = 1.895 (from tables)
⇒ not in the critical region
⇒
⇒
2.5% in each tail
⇒
accept H0.
⇒ p = 0.975
t7, 0.975 = 2.365
(from tables)
2
5.3125 ± 2.365 1.489 = 5.3125 ± 1.2450
8
[4.07, 6.56]
to 3 s.f.
20 For 1  x  3,
F ( x ) = ∫ 1 dx = x + c
2
2
When x = 1,
1
1
+c =0 ⇒ c =−
2
2
F(x ) = 0 ⇒
F ( x ) = x − 1 = 1 (x − 1)
2 2 2

0
 1
F ( x ) =  ( x − 1)
2

1

i
1 x 3
x>3
G(y) = P(Y  y)
Y = X3
1


 1
G( y ) = P X 3 y = P  X y 3  = F y 3 


 
(
⇒

0

  1 
G ( y ) =  1  y 3 − 1
2

 

1

 1

2
g (y ) =  6 y 3

 0
y <1
1 y 27
y > 27
ii
⇒
−2
g (y ) = 1 y 3 = 1 2
6
6y 3
1 y 27
otherwise
27
27 1
1
1
E (Y ) = ∫ y g ( y ) d y =∫
6
27
)
1 1 
For 1  y  27, G(y ) = 2  y 3 − 1


x <1
1
y 3 dy
27
4


 4
=  3 × 1 y 3  = 3  y 3  = 3 ( 81 − 1) = 10
24  
24
 4 6 1
1
( )
27
27 1
1
1
E Y 2 = ∫ y 2 g ( y ) d y =∫
6
4
y 3 dy
81
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
27
27
7


 7
= 3 × 1 y 3 = 1  y 3
7
6
14

1
 1
1
= ( 2187 − 1) = 156.14
14
Var(Y) = E(Y 2) – E 2(Y) = 156.14 – 102 = 56.1 (3 s.f.)
23.2 + 27.8 = 0.96298
50
60
21 s =
x − y = 25.4 − 23.6 = 1.8
Z=
1.8 = 1.8692
0.96298
Using the normal tables in reverse: z = 1.8692 ⇒ P(Z  z) = 0.9692
Two-tail test at α % significance level ⇒ α % in each tail.
2
α = (1 − 0.9692) × 100
2
α = 3.08
2
α  6.16%
()
22 Total number of goals scored = (0 × 12) + (1 × 16) + (2 × 31) + (3 × 25) + (4 × 13) + (5 × 3) = 220
Therefore, the average number of goals scored/match is 220 = 2.2 ⇒ λ = 2.2
100
H0: Total number of goals scored can be modelled by Po(2.2)
H1: Total number of goals scored cannot be modelled by Po(2.2)
The expected numbers of goals are:
 2.2 0 × e −2.2 
E 0 = 100 × 
 = 11.080
0!

 2.21 × e −2.2 
E1 = 100 × 
 = 24.377
1!

 2.2 2 × e −2.2 
E 2 = 100 × 
 = 26.814
2!

 2.2 3 × e −2.2 
E 3 = 100 × 
 = 19.664
3!

 2.2 4 × e −2.2 
E 4 = 100 × 
 = 10.815
4!

 2.2 5 × e −2.2 
E 5 = 100 × 
 = 4.7587
5!

E6+ = 100 – (E0 + E1 + E2 + E3 + E4 + E5) = 100 – 97.509 = 2.491
E5 < 5 and E6+ < 5
O5+ = 3 (from the table in the question)
⇒
combine
E5+ = 4.7587 + 2.491 = 7.2497
X2 =
(12 − 11.080 )2 + (16 − 24.377 )2 + ( 31 − 26.814 )2 + (25 − 19.664 )2 + (13 − 10.815)2 + ( 3 − 7.2497 )2
X2=
7.99
5% significance level
There are 4 degrees of freedom
7.99 < 9.488
Total number of goals scored can be modelled by Po(2.2).
11.080
⇒
24.377
⇒
26.814
19.644
10.815
7.2497
p = 0.95
⇒
χ 42,
0.95
not in the critical region
= 9.488
⇒
(from tables)
accept H0
82
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
23 H0: μ = 5.2
H1: μ > 5.2
∑x = 61, ∑x 2 = 384,
x = 61 = 6.1
10
1  384 − 612  = 1.1499
10 
9 
sx =
6.1 − 5.2
= 2.4751
1.1499
10
t=
5% significance level and 1-tail test
There are 9 degrees of freedom
2.4751 > 1.833 ⇒ in the critical region ⇒ reject H0. There is significant evidence that the new type of
tree produces a greater mass of fruit on average.
H0: μy = μx
H1: μy > μx
∑y = 70, ∑y 2 = 500.6,
sy =
⇒
⇒
p = 0.95
t9, 0.95 = 1.833 (from tables)
y = 70 = 7
10
1  500.6 − 702  = 1.0853
10 
9 
Estimate of the common variance:
2
2
s = 1.1499 + 1.0853 = 0.25
10
T = 7.1 − 6 = 1.8
0.25
5% significance level and 1-tail test
There are 18 degrees of freedom
1.8 > 1.734 ⇒ in the critical region ⇒ reject H0. There is significant evidence that the mean mass of
fruit produced by gardener Q's trees is greater than the mean mass of fruit produced by gardener P's trees.
⇒ p = 0.95
⇒ t18, 0.95 = 1.734
(from tables)
24 H0: coffee preferences are independent of company
H1: coffee preferences are not independent of company
Observed
Latte
Ground
Total
60
52
32
144
Company B
35
40
31
106
Total
95
92
63
250
Expected
Cappuccino
Company A
Cappuccino
Latte
Ground
Total
Company A
54.72
52.992
36.288
144
Company B
40.28
39.008
26.712
106
Total
95
92
63
250
X2 =
(60 − 54.72 )2 + (52 − 52.992 )2 + ( 32 − 36.288)2 + ( 35 − 40.28)2 + ( 40 − 39.008)2 + ( 31 − 26.712 )2
54.72
52.992
36.288
40.28
39.008
26.712
= 0.5095 + 0.0186 + 0.5067 + 0.6921 + 0.0252 + 0.6883 = 2.4404
5% significance level
v=2×1=2 ⇒
⇒ p = 0.95
χ 22, 0.95
= 5.991 (from tables)
83
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
⇒
⇒ accept H0
2.4404 < 5.991
Preferences are independent of company.
For the larger sample, the value of v is the same ⇒
not in the critical region
⇒ p = 0.99
⇒
1% significance level
To be in the critical region, we require:
2.44N > 9.21
N > 3.774
N must be an integer ⇒
6
∫0 kx
25 i
2
χ 22, 0.99
v=2
= 9.21 (from tables)
Nmin = 4
dx = 1
k  x 3 6 = 1
3  0
72k = 1
k= 1
72
 1 2

x 0x 6
f ( x ) =  72
 0
otherwise
3
3
3
4
5
5
E 2 x < 3 = 3∫ x 2 dx =  x 3  = ( 27 − 8 ) = 19
2
2
E 3x < 4 = 3∫ x 2 dx =  x 3  = ( 64 − 27 ) = 37
3
4
E 4 x < 5 = 3∫ x 2 dx =  x 3  = (125 − 64 ) = 61
4
4
ii
⇒
⇒
⇒
a = 19
b = 37
c = 61
H0: f(x) fits the data
H1: f(x) does not fit the data
X2 =
( 4 − 8)2 + (15 − 19)2 + ( 31 − 37 )2 + (59 − 61)2 + (107 − 91)2
8
19
37
61
91
= 2 + 0.842 10 + 0.97297 + 0.065573 + 2.8132
= 6.6938
10% significance level
v=4
⇒
χ 42, 0.9
= 7.779
6.6938 < 7.779 ⇒
⇒ p = 0.9
(from tables)
accept H0
f(x) fits the data
26 H0: area and preference are independent
H1: area and preference are not independent
Observed
Area 1
Area 2
Area 3
Local bus service
Total
73
36
30
139
Road surfaces
47
44
20
111
Total
120
80
50
250
Expected
Area 1
Area 2
Area 3
Total
Local bus service
66.72
44.48
27.8
139
Road surfaces
53.28
35.52
22.2
111
Total
120
80
50
250
84
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
X2 =
(73 − 66.72 )2 + ( 36 − 44.48)2 + ( 30 − 27.8)2 + ( 47 − 53.28)2 + ( 44 − 35.52 )2 + (20 − 22.2 )2
66.72
44.48
27.8
53.28
35.52
= 5.3646
5% significance level ⇒ p = 0.95
v=2×1=2
5.3646 < 5.991
Area and preference are independent. There is no association between them.
χ 22,
⇒
⇒
0.95
22.2
= 5.991 (from tables)
not in the critical region ⇒
accept H0
27 H0: μ = 1.2
H1: μ > 1.2
Assume the masses are normally distributed.
= 1.211
∑x = 12.11, ∑x 2 = 14.6745, x = 12.11
10
s=
T =
1
12.112 
14.6745 −
= 0.032128
9 
10 
1.211 − 1.2
= 1.0827
0.032128
10
10% significance level and 1-tail test ⇒ p = 0.9
There are 9 degrees of freedom ⇒
1.0827 < 1.383
There is no significant evidence that the mean mass of the greengrocer’s cabbages is greater than 1.2 kg.
t9, 0.9 = 1.383
⇒ not in the critical region ⇒
(from tables)
accept H0.
28 H0: μ = 7.5
H1: μ < 7.5
x = 70.4 = 7.04
10
s=
T =
8.48
= 0.970 68
9
7.04 − 7.5
= −1.4986
0.970 68
10
The tables are based on the upper tail, so we need to use the positive value of t.
10% significance level and 1-tailed test ⇒ p = 0.9
There are 9 degrees of freedom
1.4986 > 1.383 ⇒ in the critical region ⇒
mean is less than 7.5.
⇒
t9, 0.9 = 1.383
(from tables)
reject H0. There is significant evidence that the population
29 For A:
s=
x = 57.4 = 8.2
7
∑x = 57.4, ∑x 2 = 481.1,
1  481.1 − 57.42  = 1.3178
7 
6 
95% confidence interval
⇒
2.5% in each tail ⇒
⇒ t6, 0.975 = 2.447
There are 6 degrees of freedom
Therefore, the confidence interval is:
p = 0.975
(from tables)
2
8.2 ± 2.447 1.3178 = 8.2 ± 1.2188
7
[6.98, 9.42]
to 3 s.f.
85
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
Assume that for B, the population is also normally distributed and has the same variance as for A.
H0: μA = μB
H1: μA > μB
For B:
∑x = 37, ∑x 2 = 278.74,
s=
x = 37 = 7.4
5
1  278.74 − 37 2  = 1.1113
5 
4 
For the combined sample:
s=
T =
6 × 1.31782 + 4 × 1.11132 = 1.536 = 1.2394
10
8.2 − 7.4
0.8
=
= 1.1024
0.725 69
1
1
1.2394 × 7 + 5
5% significance level and one-tailed test
There are 10 degrees of freedom
1.1024 < 1.812
μA is not greater than μB.
⇒ p = 0.95
⇒ t10, 0.95 = 1.812
⇒ not in the critical region ⇒
(from tables)
accept H0.
Extension Questions
1
i
G X′ (t ) = λ e λ(t −1)
G ″X (t ) = λ 2e λ(t −1)
⇒
G X′ (1) = λ
⇒
G X″ (1) = λ 2
E ( X ) = λ
2
2
Var ( X ) = λ + λ − λ = λ
So, E ( X ) = Var ( X )
ii
Poisson distribution
2
i
For 0 x π
,
2
( )
x
I = ∫ x cos x 2 dx
Using the substitution u = x2
0
I = ∫ 1 cosu du
2

x
I =  1 sin x 2  = 1 sin x 2
2
0 2
( )
( )
π , 1 sin x 2 = 1
2 2
2
When x =
So the CDF is:

0

1
 sin x 2
 2
F(x ) =  1
1
 4 + x 8π


1

( )
( )
x <0
π
2
0x π
π
< x3
2
2
x>3
π
2
86
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
ii
(
)
 π


P
< X < π  = P X < π − P X <
 6


π
6 
()
1
1  1
π 
= + π
− sin
8π   2
6 
4
()
1
1
1
= +
−
= 0.354
8 
4
4
3
Let the difference be after minus before.
H0: Median difference = 0 There is no change in the number of flowers.
H1: Median difference > 0 There is an increase in the number of flowers.
Calculating the ranks and signed ranks, we get:
Plant
A
B
C
D
E
F
G
H
I
J
K
L
Number of flowers
before spraying
3
7
1
5
2
8
4
4
5
9
1
6
Number of flowers 1
week after spraying
5
8
5
5
2
2
7
4
0
20
9
15
Difference
2
1
4
0
0
−6
3
0
−5
11
8
9
Rank
2
1
4
−6
3
−5
9
7
8
Notice that three plants have a difference of zero, so we ignore them and reduce n by 3.
P = 34 and Q = 11
⇒
T = 11
This is a one-tail test at the 5% level with n = 9
T>8
There has been no significant change in the number of flowers.
4
i
⇒
P(X = 1) = (k – 5) × 1! = k – 5
P(X = 2) = (k – 5) × 2! = 2(k – 5)
∴Gx(t) = (k – 5) + (k – 5)t + 2(k – 5)t2
Gx(t) = (k – 5)(1 + t + 2t2)
Gx(1) = 1 ⇒
1 = (k – 5)(1 + 1 + 2) ⇒
(
)
1 =k −5
4
⇒ k = 21
4
G X (t ) = 14 1 + t + 2t 2
G X (t ) = 14 + 14 t + 12 t 2
T  8 to reject H0
P(X = 0) = (k – 5) × 0! = k – 5
⇒
Accept H0
ii
G'X (t ) = 14 + t
µ = G'X (1) = 54
G″X(t) = 1
Var ( X ) = G '' X (1) + G ' X (1) − ( G ' X (1))
2
( ) = 1611
= 1+ 5 − 5
4
4
( ) + 161
11 2 5
16 = 5 4
2
2
⇒
1
Var( X ) = 25 µ 2 + 16
87
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
5
i
π
I = ∫ ekxsin x dx
0
By parts: I = −ekxcosx + k ∫ekxcosx dx
(
By parts again: I = −ekxcos x + k ekxsin x − k ∫ekxsin x dx
Notice the integral is equal to I.
I = – ekx cos x + kekx sin x – k2 I
(1 + k2)I = kekx sin x – ekx cos x
 kekxsin x − ekxcos x 
I=

1 + k2

0
)
π
 ekπ   −1 
I= 2
− 2
 k + 1   k + 1 
I=
ekπ + 1
k2 + 1
The integral must sum to 1 (total probability). Therefore:
ekπ + 1 = k2 + 1
ekπ = k2
ii
ekπ + 1
=1
k2 + 1
y
y = ekπ
4
y = k2
2
0
–1
1
2
k
The only solution is when k < 0.
6
H0: hair colour and eye colour are independent
H1: hair colour and eye colour are not independent
The table of expected values is:
Hair colour
Eye colour
Blue
Green
Brown
Total
Blonde
4
7.25
13.75
25
Brown
3.84
6.96
13.2
24
Black
5.92
10.73
20.35
37
Red
2.24
4.06
7.7
14
Total
16
29
55
100
We need all expected values to be greater than 5 to apply the c 2 test, so merge blue eye and green eye
columns to get the following table of observed and (expected) values.
88
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
Hair colour
Eye colour
∑
Blue or
green
Brown
Total
Blonde
21 (11.25)
4 (13.75)
25
Brown
10 (10.8)
14 (13.2)
24
Black
7 (16.65)
30 (20.35)
37
Red
7 (6.3)
7 (7.7)
14
Total
25
55
100
(Ok − E k )2 = (21 − 11.25)2 + (10 − 10.8)2 + (7 − 16.65)2 + (7 − 6.3)2 + ( 4 − 13.75)2
Ek
+
11.25
∑
10.8
16.65
6.3
13.75
(14 − 13.2 ) + ( 30 − 20.35 ) + (7 − 7.7 )
13.2
20.35
7.7
2
2
2
(Ok − E k )2 = 25.781…
Ek
2
There are 3 degrees of freedom. So the critical value of χ 3 at the 0.1% level is 16.27.
Since 25.782 > 16.27, there is sufficient evidence to reject H0. Therefore, you can conclude that hair colour
and eye colour are not independent.
7
i
The sample space for the difference between the two dice is:
Difference
Dice 1
1
Dice 2
2
3
4
5
1
0
1
2
3
4
5
2
1
0
1
2
3
4
3
2
1
0
1
2
3
4
3
2
1
0
1
2
5
4
3
2
1
0
1
6
5
4
3
2
1
0
Therefore:
x
0
1
2
3
4
5
P(X = x)
1
6
5
18
2
9
1
6
1
9
1
18
Therefore, the PGF is:
5 t + 2t 2 + 1t 3 + 1t 4 + 1 t 5
G X (t ) = 16 + 18
9
6
9
18
5
4
3
4
5
G'X (t ) = 18 + 9 t + 6 t 2 + 9 t 3 + 18 t 4
35
E( X ) = G X (1) = 18
20 4
2
G''X (t ) = 94 + t + 12
9 t + 18 t
20 35
G''X (1) = 94 + 1 + 12
9 + 18 = 9
35 35
Var( X ) = 35
9 + 18 − 18
ii
6
( ) = 665
324
aE(X) = Var(X)
35 = 665 ⇒
a 18
324
2
a = 19
18
89
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
Summary REVIEW
8
Let the difference be after minus before.
H0: Difference = 0
There is no change in the test scores.
H1: Difference > 0
Test scores have increased.
Calculating the ranks and signed ranks, we get:
A
B
C
D
E
F
G
H
I
J
Test scores before tuition
20
15
17
18
8
15
19
24
6
23
Test scores after tuition
22
16
14
18
18
17
25
20
18
18
Difference
2
1
−3
0
10
2
6
−4
12
−5
2.5
1
−4
8
2.5
7
−5
9
−6
Rank
Notice that 1 person has a difference of zero, so we ignore this and reduce n by 1. Notice also that two of the
differences are equal, so the ranks (2 and 3) are averaged.
P = 30 and Q = 15
This is a 1-tail test at the 2.5% level with n = 9 ⇒
T>5
There has been no significant increase in test scores.
9
i
⇒
ii
T  5 to reject H0
Accept H0
π
When x = 2
∴k =
⇒ T = 15
⇒
k
−π
π2 π2
e =1
4
4e 2
π2
Let y = (kx2ex) sin x
Using the product rule for the expression within the brackets and for the overall expression:
dy
= kx 2e x cos x + (kx 2e x + 2kx e x )sin x = kx 2e xcos x + kx 2e xsin x + 2kxe xsin x
dx
(
)
(kx 2e x + 2kx e x )sin x = kx 2e xcos x + kx 2e xsin x + 2kxe xsin x
= kx e x(xcos x + xsin x + 2sin x)
−π
dy 4e 2 x
= 2 xe (xcos x + xsin x + 2sin x)
dx
π

−π
 4e 2 x e x(xcos x + xsin x + 2sin x)
Therefore the pdf is f ( x ) =  π2
 0

10 Since a, b, c forms a geometric progression: GX(t) = a + art + ar2t2
GX(1) = 1
⇒
1 = a + ar + ar2
GX′(t) = ar + 2ar2t
⇒
π
2
otherwise
⇒ 1 = a(1 + r + r2)
G′X (1) = E ( X ) = ar + 2ar 2 = a(r + 2r 2)
Simultaneous equations:
1 = a(1 + r + r2)
1
24
2
19 = a(r + 2r ) 2
2 ÷ 1
a(r + 2r 2)
24
19 = a(1 + r + r 2)
24(1 + r + r2) = 19(r + 2r2)
24 + 24r + 24r2 = 19r + 38r2
0 = 14r2 – 5r – 24
0 = (7r + 8)(2r – 3)
r = − 78 or r = 23
0x 90
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886
WORKED SOLUTIONS
3
Since the geometric progression is increasing ⇒ r = 2
1
= 4
Therefore a =
1 + 23 + 94 19
4 + 6 t + 9 t2
So G X (t ) = 19
19 19
G X ′ (t ) = 6 + 18 t
19 19
G′′X(t ) = 18
19
24 24
Var( X ) = 18
19 + 19 − 19
⇒
G′′X(1) = 18
19
( ) = 222
361
2
11 Let the difference be new minus original.
H0: Difference = 0
There is no change in the median number of customers per hour.
H1: Difference ≠ 0
There is a change in the median number of customers per hour.
Calculating the ranks and signed ranks, we get:
A
B
C
D
E
I
J
K
L
M
N
O
251 700 632 348 372
366
571
336
515
324
198
337
380 837 632 485 395
237
258
465
714
523
69
337
−129 −313 129
199
199 −129
Original location
(median number
224 108
of customers
per hour)
613
New location
(median number
of customers
per hour)
361 202
484
Difference
137
94
−129 129 137
9
2
Rank
−5
5
F
0
G
H
137
23
9
1
9
−5
−13
Notice that zero ranks have been ignored and tied ranks have been averaged.
P = 63 and Q = 28
⇒
5
11.5 11.5
−5
T = 28
This is a 2-tail test at the 2% level with n = 13 ⇒ T  12 to reject H0
T > 12
There has been no significant change in the median number of customers per hour. The market research
was correct.
12 i
⇒ accept H0
GX(t) = q + pt
iiGY(t) = [GX(t)]n = (q + pt)n This represents the binomial distribution.
iii GY′(t) = np(q + pt)n – 1
⇒ GY′(1) = np(q + p)n – 1
We know that q + p = 1 ⇒ GY′ (1) = E ( X ) = np
0
iv
GY″(t) = (n – 1) np2(q + pt)n – 2 ⇒ GY″(1) = (n – 1) np2(q + p)n – 2
We know that q + p = 1 ⇒ GY″(1) = (n – 1) np2
Var(X) = (n – 1)np2 + np – (np)2
= n2p2 – np2 + np – n2p2
= np – np2
= np(1 – p)
We know that q + p = 1
Var(X) = npq
⇒
q=1–p
91
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Further Probability & Statistics 9780008271886