Section 3.6

Recall that (1/2) =
y–1/2 e–y dy = 
(Multivariable Calculus is
required to prove this!)
0
Perform the following change of variables in the integral:
w =  2y
y = w2 / 2
0<w<
0 < y <

2 e
– w2 / 2
dy = w dw
dw
=


0
– w2 / 2
2 e
————— dw = 1

From this, we see that
0



– w2 / 2
2 e
————— dw = 2

–
– w2 / 2
e
———
2

dw = 1
–
Let – < a <  and 0 < b < , and perform the following change of
variables in the integral:
x = a + bw
w=
dw =
<x<
<w<
We
shall come
back
to this
derivation
later.
A random
variable
having
this
p.d.f. is said
to have a normal distribution
with mean  and variance 2, that is, a N(,2) distribution.
Right now skip to the following:
A random variable Z having a N(0,1) distribution, called a standard
normal distribution, has p.d.f.
– z2 / 2
e
f(z) = ———
2
for – < z < 
We let (z) = P(Z  z), the distribution function of Z. Since f(z) is a
symmetric function, it is easy to see that (– z) = 1 – (z). Tables Va
and Vb in Appendix B of the text display a graph of f(z) and values of
(z) and (– z).
Important Theorems in the Text:
If X is N(,2), then Z = (X – ) /  is N(0,1).
Theorem 3.6-1
If X is N(,2), then V = [(X – ) / ]2 is 2(1).
Theorem 3.6-2
We shall discuss these theorems later. Right now go to Class
Exercise #1:
1. The random variable Z is N(0, 1). Find each of the following:
P(Z < 1.25) = (1.25) = 0.8944
P(Z > 0.75) = 1 – (0.75) = 0.2266
P(Z < – 1.25) = (– 1.25) = 1 – (1.25) = 0.1056
P(Z > – 0.75) = 1 – (– 0.75) = 1 – (1 – (0.75)) = (0.75) =
0.7734
P(– 1 < Z < 2) = (2) – (– 1) = (2) – (1 – (1)) = 0.8185
P(– 2 < Z < – 1) = (– 1) – (– 2) = (1 – (1)) – (1 – (2)) =
0.1359
P(Z < 6) = (6) = practically 1
a constant c such that P(Z < c) = 0.591
P(Z < c) = 0.591
(c) = 0.591
a constant c such that P(Z < c) = 0.123
P(Z < c) = 0.123
(c) = 0.123
– c = 1.16
(– c) = 0.877
c = 0.23
1 – (– c) = 0.123
c = – 1.16
a constant c such that P(Z > c) = 0.25
P(Z > c) = 0.25
1 – (c) = 0.25
c  0.67
a constant c such that P(Z > c) = 0.90
P(Z > c) = 0.90
1 – (c) = 0.90
(– c) = 0.90
– c = 1.28
c = – 1.28
1.-continued
z0.10
P(Z > z) =   1 – (z) = 
z0.10 = 1.282
z0.90
P(Z > z) =   1 – (z) = (–z) =   1 – (–z) = 1 –  
P(Z > –z) = 1 –   z1– = –z
z0.90 = – z0.10 = – 1.282
a constant c such that P(|Z| < c) = 0.99
P(– c < Z < c) = 0.99
P(Z < c) – P(Z < – c) = 0.99
(c) – (– c) = 0.99
(c) – (1 – (c)) = 0.99
(c) = 0.995
c = z0.005 = 2.576


– w2 / 2
2 e
————— dw = 2

– w2 / 2
e
———
2

–
dw = 1
–
Let – < a <  and 0 < b < , and perform the following change of
variables in the integral:
w = (x – a) / b
x = a + bw
–  < x <

– <w<
(x – a)2
– ———
2b2
e
———— dx
b2
–
dw = (1/b) dx
= 1
The function of x being integrated
can be the p.d.f. for a random
variable X which has all real
numbers as its space.
The moment generating function of X is M(t) = E(etX) =


(x – a)2
– ———
2b2
etx e
———— dx
b2
e
————
b2
=
–
(x – a)2 – 2b2tx
– ——————
2b2
dx
=
–

(x – a)2 – 2b2tx
– ——————
2b2
exp{
}
——————————
b2
Let us consider the exponent
dx
(x – a)2 – 2b2tx
– —————— .
2b2
–
(x – a)2 – 2b2tx
– —————— =
2b2
x2 – 2ax + a2 – 2b2tx
– ————————— =
2b2
x2 – 2(a + b2t)x + (a + b2t)2 – 2ab2t – b4t2
– —————————————————
2b2
=
[x – (a + b2t)]2 – 2ab2t – b4t2
– ———————————— . Therefore, M(t) =
2b2

(x – a)2 – 2b2tx
exp{ – ——————
}
2b2
—————————— dx =
b2
–

[x – (a+b2t)]2
2
2
b2t2
bt
exp{ – ——————
}
at + ——
2b2
exp{at + ——}
——————————
dx = e
2
2
b2
–
M(t) =
e
b2t2
at + ——
2
M (t) = (a + b2t) e
M (t) = (a + b2t)2 e
for – < t < 
b2t2
at + ——
2
b2t2
at + ——
2
+
b2 e
b2t2
at + ——
2
E(X) = M (0) =
E(X2) = M (0) =
a
a2 + b2
Var(X) = a2 + b2 – a2 = b2
Since X has mean  = a and variance 2 = b2 , we can write the p.d.f of X
as
(x – )2
– ———
22
e
f(x) = ———— for – < x < 
2
A random variable having this p.d.f. is said to have a normal distribution
with mean  and variance 2, that is, a N(,2) distribution.
A random variable Z having a N(0,1) distribution, called a standard
normal distribution, has p.d.f.
– z2 / 2
e
f(z) = ———
2
for – < z < 
We let (z) = P(Z  z), the distribution function of Z. Since f(z) is a
symmetric function, it is easy to see that (– z) = 1 – (z). Tables Va
and Vb in Appendix B of the text display a graph of f(z) and values of
(z) and (– z).
Important Theorems in the Text:
If X is N(,2), then Z = (X – ) /  is N(0,1).
Theorem 3.6-1
If X is N(,2), then V = [(X – ) / ]2 is 2(1).
Theorem 3.6-2
2. The random variable X is N(10, 9). Use Theorem 3.6-1 to find each
of the following:
6 – 10
X – 10
12 – 10
P(6 < X < 12) = P( ——— < ——— < ———— ) =
3
3
3
P(– 1.33 < Z < 0.67) = (0.67) – (– 1.33) =
(0.67) – (1 – (1.33)) = 0.7486 – (1 – 0.9082) = 0.6568
X – 10
25 – 10
P(X > 25) = P( ——— > ———— ) = P(Z > 5) =
3
3
1 – (5) = practically 0
2.-continued
a constant c such that P(|X – 10| < c) = 0.95
P(|X – 10| < c) = 0.95
P(|Z| < c/3) = 0.95
X – 10
c
P( ——— < — ) = 0.95
3
3
(c/3) – (– c/3) = 0.95
(c/3) – (1 – (c/3)) = 0.95
c/3 = z0.025 = 1.960
c = 5.880
(c/3) = 0.975
3. The random variable X is N(–7, 100). Find each of the following:
X+7
0+7
P(X > 0) = P( ——— > —— ) = P(Z > 0.7) = 1 – (0.7) =
10
10
0.2420
a constant c such that P(X > c) = 0.98
X + 7 c +7
P( —— > —— ) = 0.98
P(X > c) = 0.98
10
10
P(Z > (c+7) / 10) = 0.98
((c+7) /10) = 0.02
c = – 27.54
1 – ((c+7) /10) = 0.98
(c+7) /10 = z0.98 = – z0.02 = – 2.054
3.-continued
X2 + 14X + 49
the distribution for the random variable Q = ——————
100
X2 + 14X + 49
From Theorem 3.6-2, we know that Q = —————— =
100
2
X+7
—— must have a 2(1) distribution.
10