Chapter 15 – Continuous random variables

and their probability distributions

Solutions to Exercise 15A
24

 3 a and c
 x3 3 ≤ x ≤ 6



1 f (x)  y
x < 3 or x > 6


0

Z ∞ Z 6
24 2.5
f (x) dx = dx
−∞ 3 x
3
2
#6
1.5
"
−12
=
x2 3
1
−12 12
= + 0.5
6 3
∞
x
Z
f (x) dx = 1
−∞ 0.5 1
f (x) ≥ 0 b
∴ is a probability density f unction
Z 0.5
Pr (X < 0.5) = 12x2 − 12x3 dx
0
Z ∞  0.5
2 f (x) dx = 1 = 4x − 3x 3 4
−∞ 0
Z 2 4 3
x2 + kx + 1 dx = 1 = −
0
8 16
8 3
=
#2
x3 kx2 −
"
+ +x =1 16 16
3 2 0 5
=
8 4k 16
+ +2=1
3 2
−11
2k =
3
−11
k=
6

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
  Z 7
ke
−y
y≥0 1
b Pr (T ≤ 7) =

f (y) =  (t − 5)(11 − t) dt

4 a
0

y<0 5 36

−1 7 2
Z
∞
= t − 16t + 55 dt
Z
f (y) dy = 1 36 5
−∞ #7
−1 t3
"
Z ∞
ke−y dy = 1 = − 8t + 55t
2

0
36 3 5
!
consider 1 125
= − 200 + 275
Z a 36 3
lim ke−y dy = 1 !!
a→∞ 343
0
− − 392 + 385
lim [−ke−y ]a0 = 1 3
a→∞ !
1 350 322
lim (−ke−a + ke0 ) = 1 = −
a→∞ 36 3 3
−k lim (e−a ) + k = 1 28
a→∞ =
108
k=1 7
= ≈ 0.259
 27
−y
e
 y≥0
b f (y) = 

c
0

y<0 Pr(5.5 < T < 7)
Pr(T < 7|T > 5.5) =
Z 2 Pr(T > 5.5)
Pr (Y ≤ 2) = e−y dy R7 1
0
5.5 36
(t − 5)(11 − t) dt
= [−e−y ]20 =R
11 1
(t − 5)(11 − t) dt
= −e−2 + e0 5.5 36

1 ≈ 0.244
=1− ≈ 0.865
e2
d
Pr(T < 7)
Pr(T < 7|T < 10) =
5 a Pr(T < 10)
y R7 1
(t − 5)(11 − t) dt
5 36
0.4 =R
10 1
5 36
(t − 5)(11 − t) dt
0.2
≈ 0.28
x
1 2 3 4 5 6 7 8 9 1011

678

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 6 a Z ∞
7 a Z ∞
f (x) dx = 1 f (t) dt = 1
−∞ −∞
17
Z ! ∞
 
1 −t
Z
k sin π(x − 7) dx = 1 ke 200 dt = 1
7 10 0
!#17
π Consider
"
−10
k cos (x − 7) =1
π 10 7
Z a   −t
lim ke 200 dt = 1
−10 10 a→∞
k cos(π) + k cos(0) = 1 0
π π " 
−t
 #a
10 10 k × lim −200e 200 =1
k+ k=1 a→∞
π π 0
π
 
−a
!
k= QED k × lim −200e 200 + 200e 0
=1
20 a→∞
 !
b i −a
π
17
π 200k × lim 1 − e 200 =1
Z !
Pr (X ≥ 16) = sin (x − 7) dx a→∞
7 20 10
!#17 200k = 1
π
"
−1
= cos (x − 7) 1
2 10 7
k=
! 200
−1 1 9π = 0.005
= cos(π) + cos
2 2 10
 
R∞ 1 −t
≈ 0.024 b Pr (T ≥ 1000) = 1000 e 200 dt
200
ii Pr ≤ X ≤ 13) Consider
Z (12 Z a  
1 200 −t
13
π π
!
sin (x − 7) dx lim e dt
a→∞ 1000 200
12 20 10
"  −t  #a
!#13
π = lim −e 200
"
−1
= cos (x − 7) a→∞
1000
2 10 12  
−a
!
π = lim −e +e
! ! −5
−1 1 5π 200
= cos + cos a→∞
2 10 2 10
1
=
!
−1 3π
= cos e5
2 5
1
∴ Pr (T ≥ 1000) = ≈ 0.007
≈ 0.155 e5

679

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 8 a y 9 a y

k 3
2
1
x
x 0 0.2 0.4 0.6 0.8 1
−1 0 1
Z 3
Z ∞ 4
b f (x) dx = 1 b Pr (0.25 < X < 0.75) = 1
3x2 dx
−∞ 4
Z 0 Z 1  3
k(1 + x) dx + k(1 − x) dx = 1 4
= x3 1
−1 0
4
2 0 #1
x2
" # "
x 1 27 1
x+ + x− = = −
−1
2 2 0 k 64 64
! ! 13
0 − −1 +
1
+ 1−
1
−0=
1 = ≈ 0.406
2 2 k 32
1 1 1
+ =
2 2 k 10 a y
k=1
0.1
!
−1 1
c Pr ≤X≤
2 2
x
−10 −5 0
Z 1
2 5 10
= −1 f (x) dx
2
1
Z 0 Z
2
= −1
1 + x + dx + 1 − x dx
2 0

#1
#0
x2 x2 2
" "
= x+ + x−
2−1 2 0
2
! !
1 1 1 1
=0− − + + − −0
2 8 2 8
3 3
= +
8 8
3
=
4

680

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 R ∞ 1000
b Pr (−1 ≤ X < 1) b Pr (X ≥ 2000) = dx
2000 x2
Z 1
Consider
= f (x) dx Z a " #a
−1 1000 −1000
lim dx = lim
Z 0 a→∞ 2000 x2 a→∞ x
1 2000
= (10 + x) dx −1000 1000
!
−1 100 =lim +
Z 1
a→∞ a 2000
1
+ (10 − x) dx 1
=
0 100 2
#0 1
x2
"
1 ∴ Pr (X ≥ 2000) =
= 10x + 2
100 2 −1
#1
x2
"
1 Z ∞
+ 10x − 12 a Pr (X ≥ 1.5) = f (x) dx
100 2 0
1.5
!! Z 2
1 1
!
= 0 − −10 + 1
100 2 = 2 1 − 2 dx
1.5 x
!
1 "
1
!#2
+ 10 − −0 = 2 x+
2 x 1.5
19
= = 0.19
! !!
1 3 2
100 =2 2+ + +
2 2 3
2
∞
=
Z
11 a f (x) dx = 1 3
−∞
Z ∞
k b Pr (X ≤ 1.8 | X ≥ 1.5)
2
dx = 1
1000 x Pr (1.5 ≤ X ≤ 1.8)
=
Consider Pr (X ≥ 1.5)
Z a
k 1
!
dx = 1
R 1.8
lim 2 1 − 2 dx
a→∞ 1000 x2 1.5 x
" #a =
−k 2
lim =1
a→∞ x 3
1000
" !#1.8
3 1
!
−k k
lim + =1 = 2 x+
a→∞ a 1000 2 x 1.5
! !!
k 9 5 3 2
=1 =3 + − +
1000 5 9 2 3
k = 1000 17
=
30

681

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z ∞
13 a Pr (X ≥ 8) = f (x) dx 14 a Pr (X ≤ 0.5 | = 0.45)
8 Z 0.5
Z ∞
1 −x f (x) dx
= e 5 dx −∞
8 5 Z 0 Z 0.5
Consider
Z a " #a 0.2 dx + 0.2 + 1.2x dx
1 −x −x −1 1
lim e 5 dx = lim −e 5
a→∞ 8 5 a→∞ #1
x 3x2 2
"
8 1
−a −8
! = + +
= lim −e 5 + e5 5 5 5 0
a→∞ !
1 1 3
−8 = + +
=e 5 5 10 20
−8 9
∴ Pr (X ≥ 8) = e 5 ≈ 0.202 = = 0.45
20
Pr (X ≥ 12) b Pr (X > 0.5 | X > 0.1)
b Pr (X ≥ 12 | X ≥ 8) =
Pr (X ≥ 8)
Pr (X > 0.5)
R ∞ 1 −x =
e 5 dx Pr (X > 0.1)
12 5
= 1 − (a)
−8 = R∞
e5 f (x) dx
Consider 0.1
Z a " #a 0.55
−x
lim
1 −x
e 5 dx lim −e 5 =R
1 1 6
a→∞ 12 5 a→∞
+ x dx
−8
= −8
12 0.1 5 5
e5 e5 11
8 −a −12
! = R1
= e5 lim −e 5 + e 5 4 0.1 1 + 6x dx
a→∞
11
8 −12 = 
=
1
e5 × e 5
4 x + 3x2
−4 0.1
=e 5
11
−4
= !!
Pr (x ≥ 12 | X ≥ 8) = 1 3
∴ e5 ≈ 0.449 4 (1 + 3) − +
10 100
11
=
387
25
275
= ≈ 0.711
387

682

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 15 a y ii Pr
Z ∞(X ≥ 1)
e−x dx
1 1
Consider
x Z a
0 lim e−x dx = lim [−e x ]a1
a→∞ 1 a→∞

Z 0.5
= lim (−e−a + e−1 )
a→∞
b i Pr (X < 0.5) = e−x dx 1
0 =
e
= [−e−x ]0.5
0 1
= −e−0.5 + e0 ∴ Pr (X ≥ 1) = = e−1
e
1 iii
=1− √ Pr (X ≥ 1)
e Pr (X ≥ 1 | X > 0.5) =
1 Pr (X > 0.5)
= 1 − e− 2 1
= e
1
√
e
1
= √
e
1
= e− 2

683

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to Exercise 15B
1 a f (x) = 2x, 0 < x < 1 consider
Z a
Z ∞ 1
E(X) = x f (x) dx lim dx = lim [ln |x|]a1
a→∞ 1 x a→∞
−∞
Z 1 = lim (ln a − ln 1)
= 2x2 dx a→∞
0 = lim ln a
" #1 a→∞
2 ∴ E(X) does not exist
= x3
3 0
2
= 2 a 1
3
1 b 2.097
b f (x) =
√ , 0<x<1
2 x
Z ∞ c 1.132
E(X) = x f (x) dx
−∞ d 0.4444
Z 1 √
x
= dx ∞
2
Z
0
" #1 3 a µ= x f (x) dx
1 3 −∞
= x2 Z 1
3 0 = 2x4 − x2 + x dx
1 0
= " #1
3 2 5 1 3 1 2
= x − x + x
5 3 2 0
c f (x) = 6x − 6x2 , 0 < x < 1
2 1 1
Z ∞ = − +
E(X) = x f (x) dx 5 3 2
−∞ 12 − 10 + 15
Z 1 =
30
= 6x2 − 6x3 dx 17
0 = ≈ 0.567
" #1 30
3
= 2x − x4 3 Z µ
2 0 b Pr (X ≤ µ) = f (x) dx
3 1 −∞
=2− = 17
2 2
Z
30
= 2x3 − x + 1 dx
1 0
d f (x) =
, x≥1 # 17
x2 "
1 4 1 2 30
= x − x +x
Z ∞
E(X) = x f (x) dx 2 2 0
−∞ !4 !2
1 1 17 1 17 17
= +
Z
1 −
= dx 2 30 2 30 30
0 x
≈ 0.458

684

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 ! Z ∞
1 1 1 1
4 f (x) = + cos x, −π ≤ x ≤ π 6 a E = f (x) dx
2π 2π X −∞ x
Z ∞
E(X) =
Z 1
x f (x) dx
−∞ = 12x − 12x2 dx
Z π 0
x x cos x
= + dx = [6x2 − 4x3 ]10
−π 2π 2π
using the CAS calculator = 0 =6−4=2
Alternatively, notice that the function Z ∞
is symmetrical about the y-axis, so the b E(e ) = x
e x f (x) dx
−∞
mean must be 0. Z 1
= 12x2 e x (1 − x) dx
Z ∞ 0
5 f (y) dy = 1 using the CAS calculator = 1.858
−∞
Z B
Ay dy = 1
0 7 a Pr
Z (X ≤ 1)
#B 1  1
e dx = −e
−x −x
"
A 2
y =1 0 0
2 0
2 = −e1 + e−0
AB
=1 1
2 =1−
≈ 0.632
e
1 AB2 = 2

Z ∞
Z 2
µ= y f (y) dy b Pr (1 ≤ X ≤ 2) = e−x dx
−∞ 1

= [−e−x ]21
Z B
2= Ay dy2
0 = −e−2 + e−1
" #B
A 3 e−1
2= y = ≈ 0.233
3 0 e2
AB = 63 m
Z
2 1
c e−x dx =

2 0 2
⇒B=3

1  −x m 1
−e 0 =
1 ⇒ A(3)2 = 2
Sub in 2
1
2 −e−m + e0 =
A= 2
9
1
1 − e−m =
2
1
e−m =
2
em = 2
m = ln 2 ≈ 0.693

685

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z ∞ Z m
1 −x 1
8 a f (x) dx = 1 10 e 4 dx =
−∞ 0 4 2
Z 1 " #m
−x 1
k dx = 1 −e 4 =
0 0
2
[kx]10 = 1 −m 1
−e 4 + e0 =
k=1 2
−m 1
m
e 4 =
2
Z
1
b 1 dx = m
0 2 e4 = 2
1 m
[m]10 = = ln 2
0 4
1 m = 4 ln 2 ≈ 2.773 minutes
m=
2
Z ∞
9 fZ(x) = 5(x − 1) , 0 ≤ x ≤ 1 4
11 a µ = x f (x) dx
m
1 −∞
5(x − 1)4 dx = Z 1 Z 2
2
0
= x dx +
2
2x − x2 dx
1 0 1
0 =
[(x − 1)5 ]m #1
#2
2 x3
"
x3
"
1 = + x − 2
(m − 1)5 − (−1)5 = 30
3 1
2 ! !
1 1 8 1
(m − 1)5 = − = + 4− − 1−
2 3 3 3
!1 1 10
1 5 = +4−
m−1= 3 3
2
=1
! 15
1 m
m=1−
Z
≈ 0. 1294 1
2 b f (x) dx =
0 2
if m ≤ 1,
Z m
1
x dx =
0 2
" 2 #m
x 1
=
2 0 2
m2 1
=
2 2
m =1
2

m=1

686

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 12 f (x) = 30x4 − 30x5 , 0 < x < 1 π π
!
13 f (x) = sin (x − 7) , 7 ≤ x ≤ 17
Z ∞ 20 Z10m
a µ= x f (x) dx 1
f (x) dx =
−∞ −∞ 2
Z 1
π π
Z m !
= 30x5 − 30x6 dx 1
sin (x − 7) dx =
0 7 20 10 2
" #1 !#m
30 π
"
= 5x − x7 6 −1 1
7 cos (x − 7) =
0 2 10 7
2
30 π
!
=5− − cos (m − 7) + cos 0 = 1
7 10
5
µ = ≈ 0.714 π
!
7 cos (m − 7) = 0
Z m 10
1 π π
b f (x) dx = (m − 7) =
−∞ 2 10 2
Z m
1 m−7=5
30x4 − 30x5 dx =
0 2
" #m m = 12
1
6x5 − 5x6 =
0
2 Z ∞

s − 5m =
1
6m5 6 14 a µ = x f (x) dx
2 −∞
using the CAS calculator Z 0
x
Z 1
x 6x2
m ≈ 0.736 = dx + + dx
−1 5 0 5 5
µ = 0.714 < m QED " 2 #0
x
" 2
x 2x3
#1
= + +
10 −1 10 5 0
1 1 2
=0− + + −0
10 10 5
2
=
5
Z m
1
b f (x) dx =
−∞ 2
if m ≤ 0,
Z m
1 1
dx =
−1 5 2
" #m
x 1
=
5 −1 2
m 1 1
+ =
5 5 2
m>0

687

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z 0 Z m
1 1 6 1
∴ dx + + x dx =
−1 5 0 5 5 2
#m
x 3x2
"
1 1
+ + =
5 5 5 0 2
m 3m2 3
+ − =0
5 5 10
3
3m2 + m − = 0
2
√
−1 ± 1 + 18
m=
6
since m > 0,
√
−1 + 19
m=
6

d
15 a (kxe−kx ) = ke−kx − k2 xe−kx
dx
Z Z
−1
kxe dx =
−kx
−k2 xe−kx
k
+ ke−kx − ke−kx dx
Z
−1
= ke−kx
k
− k2 xe−kx dx
Z
+ e−kx dx

−1 −1 −kx
= (kxe−kx ) + e
k k
−1 −kx
= −xe−kx − e
k
(kx + 1) −kx
=− e
k
Z ∞
b µ= x f (x) dx
−∞
Z ∞
x −x
= e λ dx
0 λ

688

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Consider c
y
!
1
a
1 − λ x
Z !
lim x e dx
a→∞ 1 λ 2
" #a
1 1
a = lim xe− λ x − λe− λ x 1.5
a→∞
0
! 1
a
= lim −e −λ
(a + λ) + e (0 + λ)
0
0.5
a→∞

x
!
a
= λ + lim −(a + λ) − e− λ 0 1 2 3
a→∞

µ=λ 1
d y = e−x is dilated by factor from
λ
the x-axis and by factor λ from
the y-axis

689

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to Exercise 15C
Z ∞ Z b
3
1 E(X) = x f (x) dx b f (x) dx =
−∞ −∞ 4
Z 1 Z b
3
= 2x2 dx 3x2 dx =
0 0 4
#1
3
"
2 3 [x3 ]b0 =
= x 4
3 0
3
2 b3 =
= 4
3
! 31
4 3
(E(X))2 = b= ≈ 0.909
9 4
Z ∞
E(X 2 ) = x2 f (x) dx c
−∞
Z 1 the interquartile range = b − a
= 2x3 dx ! 13 ! 13
0 3 1
#1 = −
1
" 4 4
= x4
2 0 ≈ 0.279
1
=
2
σ2 = E(X 2 ) − (E(X))2
4 a y
9 8 1
= − = 0.5
18 18 18
√ x
1 2 0
σ= √ =
3 2 6
Z a
1
b f (x) dx =
2 (384, 416) −∞ 4
Z a
1
0.5e x dx = , since a < 0.
Z a −∞ 4
1
3 a x f (x) dx = consider
−∞ 4
Z a
1
3x2 dx =
0 4
1
[x3 ]a0 =
4
1
a3 =
4
! 13
1
a= ≈ 0.630
4

690

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Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z a Z ∞
1 x 1
lim e dx = 5 a f (x) dx = 1
k→−∞ k 2 4 −∞
1 Z 9
k
lim [e x ]ak = dx = 1
k→−∞ 2 1 x
1
lim (ea − ek ) = [k ln x]91 = 1
k→−∞ 2
1 k ln 9 − k ln 1 = 1
ea =
2 k ln 9 = 1
1
a = ln 1
2 k=
ln 9
a = − ln 2 Z ∞
∞
µ=
Z
1 −x 1 b x f (x) dx
e dx = , since b > 0 −∞
b 2 4 Z 9
Z ∞ 1 8 4
1 dx = = ≈ 3.641
e−x dx = 1 ln 9 Z ln 9 ln 3
b 2 ∞
consider E(X 2 ) = x2 f (x) dx
Z h −∞
1 9
e−x dx =
Z
lim x
h→∞ b 2 = dx
1 ln 9
1
lim [−e−x ]hb =
" 2
#9
x
h→∞ 2 =
1 4 ln 3 1
lim (−e−h + e−b ) = 81 1
h→∞ 2 = −
1 4 ln 3 4 ln 3
e =
−b
2 20
1 =
−b = ln ln 3
2 σ2 = E(X 2 ) − µ2
b = ln 2 20 16
= −
the interquartile range = b − a ln 3 (ln 3)2
= ln 2 − (− ln 2) =
20 ln 3 − 16
≈ 4.948
(ln 3)2
= 2 ln 2 ≈ 1.386

691

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z a Z ∞
1
6 a f (x) dx = b µ= x f (x) dx
−∞ 4 −∞
Z a Z 1
1
2 − 2x dx = = 2x − 2x2 dx
0 4 0
1 #1
[2x − x2 ]a0 =
"
2
4 = x − x32
3 0
1
2a − a2 = 2
4 =1−
1 3
a2 − 2a + =0 1
4
√ =
3
2± 4−1
a=
Z ∞
2 E(X 2 ) = x2 f (x) dx
√ −∞
3
a=1− , 1
Z
2 = 2x2 − 2x3 dx
0
since 0 ≤ a ≤ 1 " #1
2 1
b
= x3 − x4
Z
3
f (x) dx = 3 2 0
−∞ 4
2 1
b
=
Z
3 −
2 − 2x dx = 3 2
0 4
1
" #b =
3 6
2x − x2 =
0
4 σ = E(X 2 ) − µ2
2

3 1 1
2b − b2 = = −
4 6 9
3 1
b2 − 2b + = 0 =
√ 4 18
2± 4−3
b=
2 Z a
1
b=1±
1 7 f (x) dx =
2 −∞ 4
Z a
1 2 1
b = , since 0 ≤ b ≤ 1 2xe−x dx =
2 0 4
#a
=b−a
"
1
√ e−x2 =
0
4
1 3
= + −1 2 1
2 2 −e−a + e0 =
√ 4
3 1
= −
2 2
≈ 0.366

692

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 2 3 Z b
3
e−a = f (x) dx =
4 −∞ 4
3
−a = ln
2 b
Z
x 3
4 dx =
r 0 2 4
4
a=+ b
Z
ln ≈ 0.5364,
3 2x dx = 3
0
since a > 0 " #b
Z b
3 x2 = 3
f (x) dx = 0
−∞ 4
Z b b =3 2
−x23 √
2xe dx =
0 4 b = 3, since 0 ≤ b ≤ 2
"
2
#b
3 the interquartile range = b − a
e−x = √
0
4 = 3−1
2 3 ≈ 0.732
e−b + e0 =
4 Z ∞
−b2 1
e = b µ= x f (x) dx
4 −∞
1 Z 2 2
x
−b2 = ln = dx
4
√ 0 2
b = + ln 4 ≈ 1.1774, " 3 #2
x
since b > 0 =
6 0
the interquartile range = b − a 8 4
= −=
≈ 0.641 6 Z 3∞
E(X 2 ) = x2 f (x) dx
−∞
Z a
1 2 3
f (x) dx =
Z
8 a x
−∞ 4 = dx
Z a 0 3
x 1
dx =
" 4 #2
x
0 2 4 =
Z a 8 0
2x dx = 1 16
0 =
=2
" #a 8
x2 = 1 σ2 = E(X 2 ) − µ2
0 16 2
=2− =
a =12
9 9
a = 1, since 0 ≤ a ≤ 2

693

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z ∞ Z ∞
9 a f (x) dx = 1 E(X ) =
2
x2 f (x) dx
−∞ −∞
Z 10 Z 10
1
kx(100 − x2 ) dx = 1 = (100x3 − x5 ) dx
0 0 2500
Z 10 " #10
1 1 6
k 100x − x3 dx = 1 = 4
25x − x
0 2500 6 0
#10
x4
" !
1 1000000
k 50x − 2
=1 = 250000 −
4 0 2500 6
k(5000 − 2500) = 1
!
1 15000 − 10000
=
1 25 6
k=
2500 5000
=
= 0.0004 150
100
Z ∞ =
b µ= x f (x) dx 3
−∞ σ2 = E(X 2 ) − µ2
Z 10
100x2 − x4 100 256
= dx = −
0 2500 3 9
#10 44
1 100 3 x5 =
"
= x − 9
2500 3 5 0 √
2 11
1 100000 100000
! σ= ≈ 2.21
= − 3
2500 3 5
!
1 2000 Z ∞
= 10 a f (x) dx = 1
25 15 −∞
80
Z a
= k a2 − x2 dx = 1
15 −a
16 #a
x3
"
=
3 k a x− 2
=1
3 −a
a3 a3
! !!
k a − 3
− −a +
3
=1
3 3
4 3
a k=1
3
3
k=
4a3

694

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z ∞ √
b µ= x f (x) dx ∴ a=2 5
−∞
a
3x 3x3
Z
= − 3 dx 11 a y
−a 4a 4a
" 2 #a
3x 3x4 3k
= −
8a 16a3 −a
3a2 3a4 3a2 3a4
! !
= − − −
8a 16a3 8a 16a3 x
=0 1 2 3 4 5 6
Z ∞
E(X 2 ) = x2 f (x) dx b
−∞
Z ∞
Z a f (x) dx = 1
3x2 3x4
= − 3 dx −∞
−a 4a 4a 3 6
Z Z
" 3 #a k(3 − x) dx + k(3 − x) dx = 1
x 3x5
= − 0 3
4a 20a3 −a #3 " 2 #6 !
x2
"
x
a3 3a5
! k 3x − + − 3x = 1
=2 − 2 0 2 3
4a 20a3 ! ! !!
 9  36 9
a2 3a2
! k 9− −0 + − 18 − −9 =1
=2 − 2 2 2
4 20
k×9=1
5a2 3a2
= − 1
10 10 k=
2
9
a
=
5
σ2 = E(X 2 ) − µ2
a2
=
−0
5
a2
=
5
but σ = 2
∴ σ2 = 4
a2
=4
5
a2 = 20
√
a = ± 20
√
sa = ±2 5
note: the way the question is
stated implies that a is positive,

695

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 c Z ∞
σ2 = E(X 2 ) − µ2
µ= x f (x) dx 27
−∞ = −9
Z 3 Z 6 2
x x 9
= (3 − x) dx + (x − 3) dx = = 4.5
0 9 3 9 2
Z 3 Z 6 2
x x2 x x
= − dx + − dx
0 3 9 3 9 3
" 2 3 6
x3 x3 x2
# " #
x
= − + −
6 27 0 27 6 3
! ! !
9 27 216 36
= − −0 + −
6 27 27 6
!
27 9
− −
27 6
1 1
= +2+
2 2
= 3 QED
Z ∞
E(X ) =
2
x2 f (x) dx
−∞
3 2
x3
Z
x
= − dx
0 3 9
6
x3 x2
Z
+ − dx
3 9 3
" 3 3 #6
x4
" 4
x3
#
x x
= − + −
9 36 0 36 9 3
! !
27 81
= − −0
9 36
64 216
! !
81 27
+ − − −
36 9 36 9
9 9
=3− + 36 − 24 − + 3
4 4
9
= 18 −
2
27
=
2

696

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to Exercise 15D
1 E(X) = 4 b E(V) = E(100 − 1.5X)
C = 300X + 100 = 100 − 1.5E(X)
E(C) = E(300X + 100) 3 17
= 100 −
×
2 24
= 300E(X) + 100
4800 51
= 1300 = −
48 48
1583
= = $98.94
Z ∞ 16
2 a E(X) = x f (x) dx Var (V) = Var (100 − 1.5X)
−∞
= 2.25Var (X)
Z 1
3x3
= + x2 dx
0 2 ≈ 0.1086
" 4 #1
3x x3 p
sd (V) = Var (V) = $0.33
= +
8 3 0
3 1 17
= + = ∞
Z
8 3 24 3 a E(3X) = 3x f (x) dx
Z ∞ −∞
E(X ) =
2
x2 f (x) dx Z 1
9x3
−∞ = dx
Z 1 −1 2
3x4
= + x3 dx " 4 #1
2 9x
0 =
" 5 #1 8 −1
3x x4
= + 9 9
10 4 0 − =0
=
8 8
3 1 11
= + =
Z ∞
10 4 20 E(9X ) =
2
9x2 f (x) dx
−∞
Var (X) = E(X 2 ) − E(X)2 1
27x4
Z
11 17 2 = dx
= − −1 2
20 242 #1
27x5
"
≈ 0.048 =
10 −1
27 27 27
+= = = 5.4
10 10 5
Var (3X) = E(9X 2 ) − E(3X)2
27
=
5

697

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 b E(3 − X) = 3 − E(X) c 1, 5.4
1
= 3 − E(3X) d Let g(x) be the function required
3
we know g(x) = ax2 and since
=3−0=3
−1 ≤ x ≤ 1, −3≤ 3x ≤ 3
Var(3 − X) = Var (X) 2
ax −3 ≤ 3x ≤ 3

∴ g(x) = 

1 0 x < −3 or x > 3

= Var (3X)
9 Z ∞
3 g(x) dx = 1
= = 0.6 −∞
5 Z 3
ax2 dx = 1
−3
#3
x3
"
a =1
3 −3

9a − (−9a) = 1
1
a=
18
 2
 x
−3 ≤ x ≤ 3


g(x) = 

∴

18
x < −3 or x > 3

0




 (x − 1)2
−2≤ x≤4


e h(x) = 


 18
0

 otherwise

698

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to Exercise 15E
Z x
1 a F(x) = f (t) dt c
Pr (X ≥ 2)
0 Pr (X ≥ 2 | X < 3) =
Z x
1 Pr (X < 3)
= dt e−4
0 5 =
" #x
t F(3)
= e−4
5 0 = ≈ 0.0183
x 1 − e−9

 if 0 < x ≤ 5
5



F(x) = 




0 if x ≤ 0 3 a F(6) = 1

0 x ≥ 5

k(6)2 = 1

b Pr (X ≤ 3) = F(3) 1
k=
36
3
= b
5 ! !
1 1
Pr ≤ X ≤ 1 = Pr (X ≤ 1) − Pr X <
2 2
2 a y !
1
= F(1) − F
2
x
!
1 1 1
0 = (1) −
36 36 4
1
=
b Pr (X ≥ 2) = 1 − Pr (X < 2) 48
= 1 − F(2)
= 1 − (1 − e−4 )
= e−4

699

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to Technology-free questions
√
Z 2 2
1 a k x dx = 1 E(X) = , so:
3
1 Z 1
" # √2 x(a + bx2 ) dx =
2
1
k x2 =1 0 3
2 1
Z 1
2
1 1
! x(ax + bx3 ) dx =
k ×2− ×1 =1 0 3
2 2 " #1
1 2 1 4 2
1 ax + bx =
k=1 2 4 0
3
2
1 1 2
k=2 a+ b=
2
2 4 3
!
Z 1.1 1 1 4
b Pr (1 < X < 1.1) = 2x dx 2 × (2) − (1): − b = −1
2 3 3
1
 1.1 1 1
= x2 b=
1
6 3
= 1.21 − 1 b=2
1
= 0.21 Substituting in (1) gives a = .
3
Z 1.2
c Pr (1 < X < 1.2) = 2x dx 1
1 3 The graph of sin x, 0 ≤ x ≤ π, has
 1.2 2
x-intercepts at (0,0) !and (π, 0) and a
= x2 π 1
1 maximum at , .
= 1.44 − 1 2 2
Also, the graph is symmetrical about the
= 0.44 π
live x = , so the area under the curve
2
π 1
from 0 to is .
Z 1 2 2
(a + bx2 ) dx = 1
!
2 Rm1 1
0 Alternatively, solve 0 sin x dx = .
" #1 2 2
1
ax + bx3 =1 Note that the symmetry also implies
3 0
π
!
1
a+ b=1
1 that the mean is .
3 2

1 1
4 a Pr (1 < x < 3) = (3 − 1) − (1 − 1)
4 4
1
=
2

700

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 0.5
b Pr (X > 2 | 1 < X < 3)
Z
b P(X < 0.5) = (12x2 − 12x3 ) dx
Pr (X > 2 ∩ | < X < 3) 0
= 0.5
Pr (1 < X < 3)

= 4x − 3x 3 4
Pr (2 < X < 3) 0
=
Pr (1 < X < 3) =4×
1
−3×
1
Pr(X < 3) − Pr(X < 2) 8 16
= ! 5
1 =
16
2
1 1
(3 − 1) − (2 − 1) Z 1
= 4 4 6 a k (x2 − x3 ) dx = 1
1 0
2 1
"
1
#1
1 k x3 − x4 =1
= (2 − 1) 3 4 0
2 !
1 1 1
= k − =1
2 3 4
c 1
Pr(X > 4 ∩ X > 2) k=1
Pr(X > 4 | X > 2) = 12
P(X > 2)
k = 12
Pr(X > 4) (Note that this agrees with the
=
Pr(X > 2) function given in Qn. 5 above.)
1
= 4
3 b f (x) = 12x2 − 12x3
4
1 f 0 (x) = 24x − 36x2
=
3 = 12x(2 − 3x)
2
5 a = 0 if x = 0 or x =
3
Since the graph of y = f (x) is that
y
2 , 16 of a negative cabic with x inter-
2
3 9 cepts (0, 0) and (1, 0), x = must
3
correspond to a local maximum. So
2
the mode is .
3
x
0 1 1
2

701

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 π m  πx 
2
Z
8 Pr(X < m) =
! Z
2 3 cos dx
Pr X < = (12x2 − 12x3 ) dx 0 4 4
3 0   πx m
 2 = sin
4 3 4 0
= 4x − 3x3
 πm 
0
= sin
8 16 4
=4× −3×  πm 
27 81 sin = 0.5
4
=
16 πm π
27 =
4 6
2
1
! m=
 ! Pr X < 3
1  2 3
c Pr X <  X < = !
3 3 2 4
x(x + 2)
Z
Pr X < 9 a E(X) = dx
3 16
0

1
!  1 1
Z 4
4 3
Pr X < = 4x − 3x
3
= x2 + 2x dx
3 0 16 0
1 1  x3 4
= , so = +x 2
9 16 3 0
1 7
=
 !
1 2 3
Pr X <  X < = 9 = 3
3  3 16 16 a
(x + 2)
Z
5
27 b dx =
0 16 32
Z a
5
0.2
(x + 2) dx =
Z
7 a Pr(X < 0.2) = 3x2 dx 0 2
0  x2 a 5
+ 2x =
  0.2
= x3 2 0 2
0
a2 5
= 0.008 + 2a =
2 2
Z 0.2 a + 4a − 5 = 0
2
3x2 dx
(a + 5)(a − 1) = 0
b Pr(X < 0.2|X < 0.3) = Z0 0.3
3x2 dx a = −5 or a = 1
0
∴a=1
0.008
=
0.027
8
=
27

702

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z 1
x3 1  1
10 a c(1 − x ) dx = c x −2 b loge m =
−1 3 −1 4
1
=
4c m = e4
3
4c 3
For PDF =1 c loge m =
3 4
3
3 m = e4
∴c=
4 3 1
IQR = e 4 − e 4 ≈ 0.833
b 0

Z 1  −n(1 − x)n 1 13 For a continuous variable X,

11 n(1 − x) n−1
dx = P(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95
0 n 0
Here, µ = 330 and σ = 5,
=1 µ + 2a = 330 + 10 = 340
µ − 2σ = 330 − 10 = 320
Z m m
1 
so (320, 340) is the required
12 a dx = loge (x)
0 x 1 (approximate) interval for 95%
= loge m of cans.
For median
1 14 Since the variance
√ is 4, the standared
loge m = deviation is 4 = 2.
2
1 µ + 2σ = 250 + 4 = 254
m = e2
For the interquartile range: µ − 2σ = 250 − 4 = 246
so (2 × 6, 254) is an (approximate) 95%
interval.

703

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to multiple-choice questions
Z 1.3
1 B The second graph partly lies below 3 2
4 A Pr (X ≤ 1.3) = (x − 1) dx
the x-axis. Since f (x) ≥ 0 for all 1 4
x, this could not requirement a " #1.3
1 3 3
probability density function. = x − x
4 4 1
2 D An antiderivative of 4x is 2x2 . ≈ 0.0743
If the domain is of the Z 2
3
Z 0 < x < a, then
from
a a
5 E E(X) = x × (x2 − 1) dx

1 4
4x dx = 2x2 Z 2 !
0 0 3 3 3
= x − x dx
= 2a2 1 4 4
#2
=1
"
3 4 3 2
= x − x
1 16 8 1
⇒ a = √ (since a > 0),
2 27
so option D fits. =
16
(Note that the above shows Z 2
3
options A → C are not posible. 6 B E(x ) =
2
x2 × (x2 − 1) dx
Z √2 1 4
2
For option E : 1 4x dx Z 2
3 4 3 2
!
√
2
= x − x dx
1 4 4
  √2 " #2
2 3 5 1 3
= 2x 2
1 = x − x
√
2 20 4 1
! !
4 1 29
=2 −2 =
2 2 10
=3 var (X) = E(X)2 − [E(X)]2
!2
29 27
Z L
1
"
1
#k = −
3 D sin x dx = − cos x 10 16
0 2 2 0 67
1 1 =
= − cos k + 1280
2 2
=1
if cos k = −1
k=π

704

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 m
x3
Z
1 9
7 C dx = Z 20
x
0 4 2 C E(X) = x× (400 − x2 ) dx
" 4 #m
x 1 0 40 000
= 32
16 0 2 =
3
m4 1
= (using a CAS calculator)
16 2 Then the expected consulations time
m4 = 8 32
√4 for three patients is 3 × = 32 min.
3
m= 8
10 A Let s be the minimum score for
≈ 1.6818
an ‘A’.
8 E The graph of y = f (x) is that of a Then Pr (X ≥ S ) = 0.10 or
positive cubic which touches the Z s Pr (X < S )!= 0.909.
equivalently
x-axis at (2, 0). The end points π πx
Hence sin dx = 0.90
are (1, 0) and (3, 3). Click for 0 100 50
!# s
a local maxium turning point. πx
"
1
3 − cos = 0.90
f (x) = (x − 1)(x − 2)2 2 50 0
2
πs
!
3 1 1
= (x3 − 5x2 + 8x − 6) − cos + = 0.90
2 2 50 2
3 πs
!
f 0 (x) = (3x2 − 10x + 8) cos = −0.80
2 50
3 πs
= (x − 2)(3x − 4) = cos−1 (−0.80)
2 50
4 50
= 0 if x = 2 or x . s= cos−1 (−0.80)
3 π
4
Then x = must correspond to a ≈ 39.8
3
local maximum turning point. so the minimum score required is
4
!
3 1
!
2
!2 closest to 40.
f = −
3 2 3 3
2
=
9
But f (3) = 3, so the mode is 3
!
4
NOT !
3

705

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Solutions to extended-response questions
 !
 a x
 100 1 − 100 if 100 ≤ x ≤ 1000


1 f (x) = 



0

otherwise
!#1000
x2
"
R 1000 a 81a
a 100
f (x) dx = x− =−
100 200 100
2
81a 2
For f to be a probability density function − = 1 and hence a = −
2 81
!#1000
2 x2 x3
"
R 1000
b E(x) = 100 x f (x) dx = − − = 700 hours
8100 2 300 100

c The cumulative probability density function

Rx 1
= 100 f (t) dt = (x2 − 200x + 10 000)
810 000
Median = 736.4 hours





 0 if x < 0
2 F(x) = 

2x − x2 if 0 ≤ x ≤ 1



if x > 1


1


a Pr (X > 0.5) = 1 − Pr (X ≤ 0.5)

= 1 − F(0.5)
1
=
4

b Pr (X < a) = 0.8. This can be written as

F(a) = 0.8 which implies,
2a − a2 = 0.8
√ √
5− 5 5+ 5
Solving the quadratic for a gives a = or a =
√ 5 5
5− 5
But 0 < a < 1. Therefore a = .
5
c The corresponding probability density function is f (x) = 2 − 2x

706

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z 1
E(X) = x f (x) dx
0
Z 1
= 2x − 2x2 dx
0
#1
2x3
"
= x − 2
3 0
1
=
3
√ 1
!
√
Z
E X = x f (x) dx
0
Z 1 1 3
= 2x 2 − 2x 2 dx
0
" #1
4 3 4 5
= x2 − x2
3 5 0
8
=
15
π π
 !

 20 cos 10 (x − 6) if 1 ≤ x ≤ 11


3 a f (x) = 


if x < 1 or x > 11

0

For the median:
Let
R q1 q1 be the first quartile
R q3and q3 be the third quartile. Then
1
f (x) dx = 0.25 and 1 f (x) dx = 0.75
π
! !
R q1 1
From the earlier calculation, 1 f (x) dx = sin (a − 6) + 1 .
2 10
π
!
1
For q1 : sin (q1 − 6) + 1 = 0.25
2 10
π
!
sin (q1 − 6) = −0.5
10
π π
(q1 − 6) = −
10 6
−5
q1 − 6 =
3
13
q1 =
3

707

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 π
!
1
For q3 : sin (q3 − 6) + 1 = 0.75
2 10
π
!
sin (q3 − 6) = 0.5
10
π π
(q3 − 6) =
10 6
5
q3 − 6 =
3
23
q3 =
3
Interqurtile range = q3 − q1
23 13 10
= − =
3 3 3
b The graph of y = f (x) is symmetrical about the line x = 6, so the mean is 6,
Alternatively:
π π
Z 11 Z 11 !
E(X) = x f (x) dx = x cos (x − 6) dx
1 20 1 10
Since the integrand is not a ‘standard’ function fn integration, use a CAS calculator
for its evaluation. This gives E(X) = 6.
For the variance, first find E(XZ2 ).
π π
Z 11 11 !
E(X ) =
2
x f (x) dx =
2 2
x cos (x − 6) dx
1 20 1 10
Using the integration command of a CAS calculates gives E(X 2 ) ≈ 40.7358 Hence
var (X) = E(X 2 ) − [E(X)]2
≈ 40.7358 − 36
= 4.736 correct to 3 decimal places.
Z a
f (x) dx = 0.5
1
π a
π
Z !
cos (x − 6) dx = 0.5
20 1 10
! #a
π π
"
10
× sin (x − 6) dx = 0.5
20 π 10 1
π π
! !
sin (a − 6) − sin − = 1
10 2
π
!
sin (a − 6) = 0
10
a=6

708

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 4 a Pr
Z (4 ≤ Y ≤ 5)
5 Z 5
2
f (y) dy (not f (x) dx) = y − 1 dy
4 25 4
" #5
2 1 2
= y −y
25 2 4
7
=
25
Z 6
b E(Y) = y f (y) dy
1
Z 6
2
= y(y − 1) dy
25 1
" #6
2 1 3 1 2
= y − y
25 3 2 1
!
2 1 13
= × 54 + =
25 6 3
Also E(X) = 8 × 0.6 = 4.80
!
13
The expected money received = $ 4.80 + 4 × = $22.13
3
Z 4
5 E(X − c) = 2
(x2 − 2cx + c2 ) f (x) dx
2
Z 4
1
= x2 − 2cx + c2 )(x − 2) dx
2 2
1 4 3
Z
=
(x − (2c + 2)2 + (c2 4c)x − 2c2 ) dx
2 2
1
= (3c2 − 20c + 34)
3
2
If E(X − c)2 =
3
implies 3c2 − 20c + 34 = 2
3c2 − 20c + 32 = 0
(3c − 8)(c − 4) = 0
8
Therefore c = or c = 4
3

k(x − 1)(3 − x)
 if 1 ≤ x ≤ 3
6 f (x) = 

0 if x < 1 or x > 3


709

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z 3
a f (x) dx = 1
1
Z 3
Therefore k −x2 + 4x − 3 dx = 1
1
3
Hence k =
4
Z 3
3
b E(X) = x f (x) dx
4 1
Z 3
3
= x(−x2 + 4x − 3) dx
4 1

=2
var (X) = E(X 2 ) − [E(X)]2
3 3
Z
= x f (x) dx − 4
4 1
21
= −4
5
1
= = 0.2
5
3R3 5
c Pr (X > 2.5) = 2.5
−x2 + 4x − 3 dx =
4 32
(Note: the integral, in this question have been evaluated using the ‘Integral’
command of a CAS calculator.)
 k
 12(x − 1)3 if 0 ≤ x ≤ 4



7 f (x) = 


if x < 0 or x > 4

0


Z 4 Z 4
k 1
a f (x) dx = 1 ⇒ dx = 1
0 (x + 1)
12 3
0
" #k
k 1
− =1
12 2(x + 1)2 0
!
k 1 1
− + =1
12 50 2
k
=1
25
k = 25

710

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 Z 4
25 1
b E(X + 1) = dy
0 (x + 1)
12 2
" #4
25 1
= −
12 (x + 1) 0
!
25 1
= − +1
12 5
5
=
3
E(X + 1) = E(X) + 1
E(X) = E(X + 1) − 1
5 2
= −1=
3 3

d P(X ≤ c) = c
Z c
25 1
dx = c
0 (x + 1)
12 3
" #c
25 1
− =c
24 (x + 1)2 0
!
25 1
− −1 =c
24 (c + 1)2
−13 2
Using the ‘solve’ command of a CAS calculator gives c = or c = 0 or c = .
6 3
2
But c > 0, so c = .
3





 kx if if 0 ≤ x < 2
8 f (x) =  k(4 − x) if 2 ≤ x < 4




if x < 0 or x > 4


0


1
a The area of the triangle = × 4 × 2k y
2
= 4k
1 (2,2k)
Therefore k =
4

0 4
x

711

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.
 b Since the graph of y = f (x) is symmetrical about x = 2, E(X) = 2.
var (X) = E(X 2 ) − [E(X)]2
1 2 3 1 4 2
Z Z
= x dx + x (4 − x) dx − 4
4 0 4 2
14
= −4
3
2
=
3

c Pr (|X − µ| < 1) = Pr (|X − 2| < 1)

= Pr (1 < X < 3)
1 2 1 3
Z Z
= x dx + (4 − x) dx
4 1 4 2
3
=
4
d Pr (X > a) = 0.6 or equivalently Pr (X ≤ a) = 0.4.
Since the graph of y = f (x) is symmetrical about x = 2, Pr (X ≤ 2) = 0.5.
Hence 0 < a < 2.
Z a
1 4
x dx = 0.4 =
0 4 10
" #a
1 2 2
x =
8 0 5
1 2 2
a =
8 5
16
a2 =
5
√
4 4 5
a= √ = (a > 0).
5 5

712

Cambridge Senior Maths AC/VCE ISBN 978-1-107-56747-4 © Evans et al. 2016 Cambridge University Press
Mathematical Methods 3&4 Photocopying is restricted under law and this material must not be transferred to another party.