Solutions to Suggested Problems
Section 11.5
1. We must show that f (x) ≥ 0 for a ≤ x ≤ b and
2
f (x) = x9 ≥ 0 for all x, and
x3
dx =
9
27
Z 3 2
x
0
3. f (x) =
1
x2
3
=
0
Rb
a
f (x)dx = 1. Here
1 3
(3 − 0) = 1.
27
> 0 for all x 6= 0, and
Z ∞
Z ∞
1
b 1
dx
=
lim
dx
2
b→∞ 1 x2
1 x
b
1
1
= lim − + 1 = 1.
= lim −
b→∞
x 1 b→∞
b
f (x)dx =
1
Z
5. f (x) = xe−x ≥ 0 for all x ≥ 0, and using IBP with u = x and dv = e−x dx:
Z ∞
Z b
f (x)dx = lim
b→∞ 0
0
xe−x dx
b
= lim −xe−x − e−x 0
b→∞
= lim (−be−b − e−b + 1) = 1.
b→∞
c
. Therefore,
17. We must have 1 = 01 cx2 (1 − x)dx = [c( 31 x3 − 14 x4 )]10 = 12
c = 12. We also observe that if c = 12, then f (x) ≥ 0 for x ≤ 1.
R
R 4000
R 4000
cx−2 dx = 1. Note that
Z 4000
1
1
3c
−2
−1 4000
1=
cx dx = −cx 1000 = c −
+
=
.
4000 1000
4000
1000
19. (a) We must have
1000
f (x)dx =
1
1000
4000
3 .
Therefore, c =
(b)
4000 −1 4000
x dx = −
x
P(3000 ≤ X ≤ 4000) =
3
3
3000
3000
4000
1
1
1
= −
−
= .
3
4000 3000
9
Z 4000
4000
−2
21. Using integration by parts with u = x and dv = e−x dx, we have
Z
xe−x dx = −xe−x +
Z
e−x dx = −(x + 1)e−x + c.
(a)
Z ∞
P(X ≥ 1) =
1
b
xe−x dx = lim −(x + 1)e−x 1
b→∞
2
= lim [−(b + 1)e−b + 2e−1 ] = .
b→∞
e
(b)
P(0 ≤ X ≤ 5) =
Z 5
0
5
xe−x dx = −(x + 1)e−x 0
−5
= 1 − 6e
23. First, f (x) =
1
√
2 x
> 0 for all x > 0. Second,
Z 4
f (x)dx =
1
e5 − 6
=
.
e5
Z 4
1
1
√ dx =
2 x
√
√ 4 √
x 1 = 4 − 1 = 1.
Therefore, f (x) is a probability
density
√function.
x √
Rx 1
If 1 < x < 4, then F(x) = 1 2√t dt = t 1 = x − 1. Thus,


0√
F(x) =
x−1


1
2
x≤1
1<x≤4
x > 4.
25. First, if −1 ≤ x ≤ 0, then f (x) = 1 + x ≥ 0; and if 0 < x ≤ 1 then
f (x) = 1 − x ≥ 0. Second,
Z 1
Z 0
Z 1
(1 + x)dx + (1 − x)dx
0
0
1 2
1 2 1
= x+ x
+ x− x
2 −1
2 0
1 1
=
+ = 1.
2 2
f (x)dx =
−1
−1
Therefore, f (x) is a probability density function.
If −1 < x < 0, then
Z x
1
1
1 2 x
= x + x2 +
F(x) =
(1 + t)dt = t + t
2 −1
2
2
−1
and if 0 < x < 1, then
Z 0
F(x) =
Z x
(1 + t)dt +
−1
0
x
1
t2
1
1
= + x − x2 .
(1 − t)dt = + t −
2
2 0 2
2
Thus,

0



x + 1 x2 + 1
F(x) = 1 2 1 22
2 + x − 2x



1
x ≤ −1
−1 < x ≤ 0
0<x≤1
x > 1.
27.
√
(a) P(X ≤ 3) = F(3) = 12 3 − 2 = 12 .
(b)
P(4 ≤ X ≤ 5) = P(4 < X ≤ 5) = P(X ≤ 5) − P(X ≤ 4)
√
√
1√
3− 2
1√
= F(5) − F(4) =
5−2−
4−2 =
.
2
2
2
(c) f (x) = F 0 (x) =
√1 ,
4 x−2
2 < x < 6.
3
29.
Z 1
µ = E(X) =
0
2 3 1
2 1
2
x(2 − 2x)dx = x − x
= 1− =
3 0
3 3
Z 1
1
1
x2 f (x)dx − µ2 =
x2 (2 − 2x)dx −
9
0
0
1
2 3 1 4
1
1 2 1 1
=
x − x
− = − − = .
3
2 0 9 3 2 9 18
Z
Var(X) =
31.
3
µ = E(X) =
4
2
3 2 3 x4
3 16 16
x(2x − x )dx =
x −
=
−
= 1.
4 3
4 0 4 3
4
Z 2
2
0
Z 2
3 2 2
x2 f (x)dx − µ2 =
x (2x − x2 )dx − 1
4
0
0
2
5
3
32
1
3 1 4 x
x −
−1 =
8−
−1 = .
=
4 2
5 0
4
5
5
Z
Var(X) =
33. By integration by parts, we have
Z
2xe
−2x
1 −2x
e +c
dx = − x +
2
and
Z
2 −2x
2x e
2 −2x
dx = −x e
Z
+
2xe
−2x
1 −2x
dx = − x + x +
e + c.
2
2
Then,
Z ∞
Z b
−2x
µ = E(X) =
2xe dx = lim
2xe−2x dx
b→∞ 0
0
b
1 −2x
1 −2b 1
1
= lim − x +
e
= lim − b +
e
+
= .
b→∞
b→∞
2
2
2
2
0
4
Z ∞
b
1
x2 f (x)dx − µ2 = lim
2x2 e−2x dx −
b→∞ 0
4
0
b
1 −2x
1
e
= lim − x2 + x +
−
b→∞
2
4
0
1 −2b 1
1 1
= lim − b2 + b +
e
+
− = .
b→∞
2
2
4 4
Z
Var(X) =
1
xe−x/10 dx = −(x + 10)e−x/10 + c
37. By integration by parts, we have 10
and
Z
1 2 −x/10
x e
dx = −(x2 + 20x + 200)e−x/10 + c.
10
Then,
R
∞
b 1
µ = E(X) =
xe−x/10 dx
x f (x)dx = lim
b→∞
10
0
0
h
ib
= lim −(x + 10)e−x/10 = lim [−(b + 10)e−b/10 + 10] = 10
Z
Z
b→∞
b→∞
0
Z ∞
b 1
x2 e−x/10 dx − 100
x2 f (x)dx − µ2 = lim
b→∞
10
0
0
h
ib
= lim −(x2 + 20x + 200)e−x/10 − 100
Z
Var(X) =
b→∞
0
−b/10
2
= lim [−(b + 20b + 200)e
b→∞
and σ(X) =
+ 200] − 100 = 100,
p
Var(X) = 10.
41.
Z 1
E(X) =
Z 1
x f (x)dx =
0
0
If m is the median of X, then
1
2
=
Rm
0
3
3x dx = x4
4
3
3x2 dx
= m3 .
1
3
= .
4
0
Thus, m =
3
q
1
2.
43. By integration by parts xe−x dx = −(x + 1)e−x + c. Then
R
Z ∞
E(X) =
Z b
x f (x)dx = lim
b→∞ 0
−b
0
= lim [−(b + 1)e
b→∞
If m is the median of X, then
1
2
=
b
xe−x dx = lim −(x + 1)e−x 0
b→∞
+ 1] = 1.
R m −x
−m . Thus, m = ln(2).
0 e dx = 1 − e
5
Mean and Variance Notes
1.
x
R
R
(a) if 0 < x < 1, P(x) = 0x p(x)dx = 0x (4x − 4x3 )dx = 2x2 − x4 0 , so the
cumulative distribution function is


x<0
0
P(x) = 2x2 − x4 0 ≤ x ≤ 1


1
x>1
(b) if m is the median, then
1
2
Rm
=
0
m
p(x)dx = 2x2 − x4 0 = 2m2 − m4 .
√
So we have m4 − 2m2 + 21 = 0, which has solutions m2 = 1 ± 22 , so m =
q
q
√
√
2
1 ± 2 . The only one of these which is in [0, 1] is m = 1 − 22 , so the
q
√
median is m = 1 − 22 .
(c)
Z ∞
µ = E(X) =
Z 1
xp(x)dx =
−∞
0
4
4
(4x −4x )dx = x3 − x4
3
5
2
4
1
4 4
8
= − = .
3 5 15
0
(d)
Z ∞
Var(X) =
2
2
x p(x)dx − µ =
Z 1
−∞
0
2 6 1 64
64
4
= x − x
(4x − 4x )dx −
−
225
3 0 225
3
5
2
64
11
= 1− −
=
3 225 225
p
σ(X) = Var(X) =
√
11
.
15
2.
x m
R
R dx
m
a
(a) If m is the median, then 12 = am p(x)dx = am b−a
= b−a
− b−a
.
= b−a
a
Solving this for m we have
1
a
b−a
b+a
m = (b − a) ·
+
=
+a =
2 b−a
2
2
6
This is not surprising because it is halfway between a and b.
The mean is
b
Z b
x
x2
a2
(b + a)(b − a) b + a
b2
µ=
dx =
−
=
=
.
=
2(b − a) a 2(b − a) 2(b − a)
2(b − a)
2
a b−a
Again, this is not surprising.
(b)First find the variance
Var(X) =
Z b 2
x
x3
dx − µ =
b−a
3(b − a)
a
b3 − a3
2
b
a2 + 2ab + b2
−
4
a
a2 + 2ab + b2 a2 + ab + b2 a2 + 2ab + b2
=
−
3(b − a)
4
3
4
2
2
2
(a − b)
a − 2ab + b
=
=
12
12
=
−
So,
p
b−a
Var(x) = √ .
12
Note we must take b − a rather than a − b in the square root.
σ=
(c) We have:
P(µ − σ ≤ X ≤ µ + σ) =
Z µ+σ
Z µ+σ
p(x)dx =
µ−σ
µ−σ
1
dx
b−a
1
1
µ+σ
=
[x]µ−σ =
· [µ + σ − (µ − σ)]
b−a
b−a
1
1
b−a
1
=
· (2σ) =
· √ =√ .
b−a
b−a
3
3
3. (a) Find m such that
Rm
2
1
2
1
−
16
1
16
m
f (x)dx =
=
=
=
=
1
2
So,
8 m
dx = − 3
4
x 2
2 x
1
1
−
3
m
8
1
m3
1
(16) 3
Z m
24
7
(b)
Z ∞
µ=
Z ∞
24
b 24
dx
=
lim
dx
3
b→∞ 2 x3
2 x
12 b
12 12
= lim − 2 = lim − 2 +
b→∞
x 2 b→∞
b
4
= 3.
x f (x)dx =
2
Z
(c)
σ =
2
Z ∞
2
x f (x)dx − µ
2
2
Thus σ =
Z ∞
24
Z b
24
dx − 9 = lim
dx − 9
b→∞ 2 x2
x2
24 b
24
= lim −
− 9 = lim − + 12 − 9
b→∞
b→∞
x 2
b
= 3.
=
2
√
3.
5. (a)
Z 6
Z 6
t
√
dt
3
3 2 t −2
Make the substitution u = t − 2, du = dt, and use t = u + 2. This will require
the change of bounds of integration t = 3 gives u = 1 and t = 6 gives u = 4.
So
µ=
Z 4
u+2
t f (t)dt =
Z 4 1/2
u
Z 4
du =
du +
u−1/2 du
2
2u1/2
1
1
1 3/2 4 h 1/2 i4 1 3/2
u
+ 2u
=
4 − 13/2 + 2 41/2 − 11/2
=
3
3
1
1
8 1
13
=
− +4−2 = .
3 3
3
µ =
1
8
(b)
σ2 =
=
=
=
=
Hence, σ =
p
t2
√
dt − µ2
3
3 2 t −2
Z 4
Z 4 2
2
(u + 2)
u + 4u + 4
2
du − µ =
du − µ2
1/2
1/2
2u
2u
1
1
Z 4 3/2
Z 4
Z 4
u
du +
2u1/2 du +
2u−1/2 du − µ2
2
1
1
1
"
#4 "
#4
i4
h
u5/2
4u3/2
+
+ 4u1/2 − µ2
5
3
1
1
1
2
4
13
34
1
(32 − 1) + (8 − 1) + 4(2 − 1) −
= .
5
3
3
35
Z 6
t 2 f (t)dt − µ2 =
Z 6
34/35 Note: the same substitution was used here as in (a).
6. (a)
Z 2
Z 1
µ =
2
Z 2
xp(x)dx =
x dx +
0
2
x3
1 3 1
2
x 0+ x −
=
3
3 1
8
1
1
+ 4 − − 1 + = 1.
=
3
3
3
0
(2x − x2 )dx
1
(b) First we must find σ
σ2 =
Z 2
Z 1
x2 p(x)dx − µ2 =
x3 dx +
0
0
3
1
4
2x
x
1 4 1
x 0+
−
− µ2
=
4
3
4 0
1
16
2 1
=
+
−4−
−
−1
4
3
3 4
7
1
=
−1 = .
6
6
9
Z 2
1
(2x2 − x3 )dx − µ2
So we have σ =
√1 .
6
Then
P(µ − σ ≤ X ≤ µ + σ) =
Z µ+σ
µ−σ
=
Z 1
p(x)dx =
x2
2
Z µ+σ
xdx +
µ−σ
(2 − x)dx
1
1
µ+σ
x2
+ 2x −
2 1
µ−σ
(µ + σ)2
1
1 (µ − σ)2
−
+ 2(µ + σ) −
−2+
2
2
2
2
"
2 #
1
1
1
1 2
1− √
= −1 + 2 1 + √ −
− 1+ √
2
6
6
6
√
2 6−1
=
6
=
(c) This is 1 − P(µ − 3σ < X < µ + 3σ), and
P(µ − 3σ < X < µ + 3σ) =
Z µ+3σ
p(x)dx
µ−3σ
But we notice that µ − 3σ = 1 − √36 < 0 and µ + 3σ = 1 + √36 > 2, we know
that
Z
2
P(µ − 3σ < X < µ + 3σ) =
p(x)dx = 1.
0
Hence the probability that X is more than 3σ away from µ is 1 − 1 = 0.
8. (a)
µ =
2
2
t2
t
t3
−
dt =
−
00
100 20000
200 60, 000 0
Z 200 t
0
= 200 −
(200)3
800 200
= 200 −
=
60, 000
6
3
(b) We are looking for P(T > µ), and since P(T > µ) = 1 − P(T ≤ µ) we
10
can use
Z µ
1
t
P(T ≤ µ) =
−
dt
100 20000
0
µ
t
t2
=
−
100 40000 0
µ
µ2
200
2002
−
=
−
100 40000 300 9 · 40000
5
=
.
9
=
So we have P(T > µ) = 1 − 59 = 94 .
(c)
σ
2
t3
=
−
dt − µ2
100 20, 000
0
3
200 t
200 2
t4
=
−
−
300 80, 000 0
3
Z 200 2
t
2
2004
2002
2 1 1
2
2
=
(200) −
−
= 200 ·
− −
3
80, 000
9
3 2 9
20, 000
=
9
√
p
Hence, we have σ = 20, 000/9 = 1003 2 .
10. First we need the mean,
Z ∞ 2
Z ∞
x −x/8
µ=
e
dx =
8z2 e−z dz
0
64
0
Above I used the substitution z = x/8, 8dz = dx. Using IBP with u =
z2 , du = 2zdz, dv = e−z dz, v = −e−z , we get
Z
2 −z
2 −z
z e dz = −z e
Z
+2
ze−z dz
using IBP again with u = z, du = dz, dv = e−z dz, v = −e−z . Then we have
Z
Z
2 −z
2 −z
−z
−z
z e dz = −z e + 2 −ze + e dz
= −z2 e−z − 2ze−z + −2e−z
11
We then have
b
µ = lim 8 · −z2 e−z − 2ze−z + −2e−z 0
b→∞
h
i
= lim 8 · −b2 e−b − 2be−b + −2e−b − (0 + 0 − 2e0 )
b→∞
= 16.
Now we must find σ.
σ2 =
Using IBP with u =
Z
Z ∞ 3
x −x/8
e
dx − µ2
64
0
3x2
x3
−x/8 dx, v = −8e−x/8
64 , du = 64 dx, dv = e
x3 −x/8
8
3
e
dx = − x3 e−x/8 +
64
64
8
Z
we have
x2 e−x/8 dx
So,
Z
1 3 −x/8 b 3 ∞ 2 −x/8
σ = limb→∞ − x e
+
x e
dx − µ2
8
8
0
0
h
i 3
= lim b3 e−b/8 − 0 + · 64µ − µ2
b→∞
8
2
= 0 + 24(16) − 16 = 128.
√
√
So we have σ = 128 = 8 2.
2
item 11. (a)
Z ∞
µ=
0
12t
dt
(t + 4)5/2
Make the suggested substitution u = t + 4 and du = dt. We will then use
that t = u − 4 and note that the bounds of integration are changed (t = 0
12
gives u = 4, the top bound is unchanged).
µ =
Z ∞
12(u − 4)
=
=
=
Z b
u−4
du
u5/2 Z b
u−3/2 du − 4
u−5/2 du
12 · lim
b→∞
4
4
h
ib 8 h
ib −1/2
−3/2
12 · lim −2u
− u
b→∞
3
4
4
2
8
8
2
−
−12 · lim √ − −
b→∞
3b3/2 3 · 8
b 2
2
−12(− ) = 8.
3
4
=
du = 12 · lim
u5/2
b→∞ 4
Z b
(b) Making the same substitution as in (a), we have,
σ
2
∞ 12(u − 4)2
12t 2
2
dt − µ =
du − 64
=
u5/2
4
0 (t + 4)5/2
Z ∞ 2
u − 8u + 16
=
du − 64
u5/2 Z
Z4
Z
Z ∞
Z
∞
=
u−1/2 du +
4
= ∞
∞
4
∞
8u−3/2 du + 16
u−5/2 du
4
because the first of these integrals diverges (goes to infinity). Therefore, the
variance of the shelf life for this problem does not exist.
13