# COMP 245 Statistics Solutions 4 ```COMP 245 Statistics
Solutions 4 - Continuous Random Variables
−x
Z
1. Begin with F(−x) + F(x) =
Z
x
f (u)du +
u=−∞
v=−∞
Z x
for the first part leads to F(−x) + F(x) = −
Z x
Z ∞
f (v)dv =
f (v)dv = 1.
u=−∞
2.
f (v)dv. Taking the change of variable v = −u
Z x
Z ∞
f (−v)dv +
f (v)dv =
f (v)dv +
v=∞
v=−∞
v=x
v=−∞
(a) Since a particle is equally likely to hit anywhere on the plate, for 0 &lt; r &lt; 1 the probability
πr2
= r2 . Hence
that it will strike inside a circle of radius r is
π12


0, x ≤ 0


 2
x , 0&lt;x&lt;1
F(x) = 


 1, x ≥ 1.
(b) P(r &lt; X &lt; s) = P(X &lt; s) − P(X &lt; r) = s2 − r2 .
(c) From 2a the cdf of X is given by F(x) = x2 for 0 ≤ x ≤ 1. So the pdf, f (x) = F0 (x) = 2x
for 0 ≤ x ≤ 1, and 0 everywhere else.
(d) The expected distance from the origin is
∞
Z
E(X) =
1
Z
x f (x)dx =
−∞
0
2x3
x.2xdx =
3
&quot;
#1
0
2
= .
3
3. A random variable X ∼ Exp(λ) has density f (x) = λe−λx for x ≥ 0. So
Z ∞
Z ∞
Z ∞
h
i
−λx
−λx ∞
E(X) =
x f (x)dx =
xλe dx = −xe
+
e−λx dx
0
−∞
0
&quot; −λx #0∞ −e
1
1
= [0 − 0] −
= 0− −
= .
λ 0
λ
λ
E X
2
Z
∞
=
∞
Z
x f (x)dx =
x λe
2
−∞
2
−λx
∞
Z
h
i
2 −λx ∞
dx = −x e
+
0
0
2xe−λx dx
0
2
2
E(X) = 2
λ
λ
2
1
1
⇒ Var(X) = E X2 − {E (X)}2 = 2 − 2 = 2 .
λ
λ
λ
= [0 − 0] +
(
4. X has range [0, 1] and pdf fX (x) =
1, 0 &lt; x &lt; 1
0, otherwise.
Thus the cdf for Y = eX is given by
Z
FY (y) = PY (Y ≤ y) = P(e ≤ y) = P(X ≤ log(y)) =
log(y)
X
−∞
log(y)
Z
fX (x)dx =
0
dx = log(y),
for 0 &lt; log(y) &lt; 1; that is, for 1 &lt; y &lt; e.
For the pdf, differentiating FY (y) wrt y gives
1
y,
(
fY (y) =
0,
1&lt;y&lt;e
otherwise.
x−&micro;
, so Y = g(X). First we note that g is clearly a continuous, monotonically
σ
increasing function of x. Therefore we have
5. Let g(x) =
0
fY (y) = fX {g−1 (y)}|g−1 (y)|.
0
Well g−1 (y) = σy + &micro;, so g−1 (y) = σ. Since X ∼ N(&micro;, σ2 ),
)
(
(x − &micro;)2
1
fX (x) = √ exp −
2σ2
σ 2π
and hence
)
(
( 2)
(σy + &micro; − &micro;)2
y
1
1
fY (y) = fX (σy + &micro;)σ = √ exp −
σ = √ exp −
=⇒ Y ∼ N(0, 1).
2
2
2σ
σ 2π
2π
6.
(a) ∀a , 0, FY (y) = PY (Y ≤ y) = P(aX + b ≤ y) = P(aX ≤ y − b).
If a &gt; 0,
!
!
y−b
y−b
FY (y) = P X ≤
= FX
,
a
a
whereas if a &lt; 0,
!
!
y−b
y−b
FY (y) = P X ≥
= 1 − FX
.
a
a
(b) The pdf of a continuous random variable is the derivative of the cdf, fY (y) = F0y (y).
So if a &gt; 0,
!
!
!
y−b
y−b
d
1 0 y−b
1
fY (y) =
FX
= FX
= fX
,
dy
a
a
a
a
a
whereas if a &lt; 0,
(
!)
!
!
y−b
y−b
d
1 0 y−b
1
fY (y) =
1 − FX
= − FX
= − fX
,
dy
a
a
a
a
a
So either way, we have
!
y−b
1
fY (y) =
fX
.
|a|
a
7.
(a) z = 1.16 or z = -1.16.
(b) z = 1.09.
(c) z = -1.35 or z = -1.69.
8.
(a) 0.3849
(c) 0.6636
(e) 0.8997
(b) 0.2517
(d) 0.1828
```