5Continuous random variables
IA Probability
5.5 The normal distribution
Definition (Normal distribution). The normal distribution with parameters
µ, σ
2
, written N(µ, σ
2
) has pdf
f(x) =
1
√
2πσ
exp
−
(x − µ)
2
2σ
2
,
for −∞ < x < ∞.
It looks approximately like this:
µ
The standard normal is when µ = 0, σ
2
= 1, i.e. X ∼ N(0, 1).
We usually write
ϕ
(
x
) for the pdf and Φ(
x
) for the cdf of the standard normal.
This is a rather important probability distribution. This is partly due to the
central limit theorem, which says that if we have a large number of iid random
variables, then the distribution of their averages are approximately normal. Many
distributions in physics and other sciences are also approximately or exactly
normal.
We first have to show that this makes sense, i.e.
Proposition.
Z
∞
−∞
1
√
2πσ
2
e
−
1
2σ
2
(x−µ)
2
dx = 1.
Proof. Substitute z =
(x−µ)
σ
. Then
I =
Z
∞
−∞
1
√
2π
e
−
1
2
z
2
dz.
Then
I
2
=
Z
∞
−∞
1
√
2π
e
−x
2
/2
dx
Z
∞
∞
1
√
2π
e
−y
2
/2
dy
=
Z
∞
0
Z
2π
0
1
2π
e
−r
2
/2
r dr dθ
= 1.
We also have
Proposition. E[X] = µ.
Proof.
E[X] =
1
√
2πσ
Z
∞
−∞
xe
−(x−µ)
2
/2σ
2
dx
=
1
√
2πσ
Z
∞
−∞
(x − µ)e
−(x−µ)
2
/2σ
2
dx +
1
√
2πσ
Z
∞
−∞
µe
−(x−µ)
2
/2σ
2
dx.
The first term is antisymmetric about
µ
and gives 0. The second is just
µ
times
the integral we did above. So we get µ.
Also, by symmetry, the mode and median of a normal distribution are also
both µ.
Proposition. var(X) = σ
2
.
Proof.
We have
var
(
X
) =
E
[
X
2
]
−
(
E
[
X
])
2
. Substitute
Z
=
X−µ
σ
. Then
E
[
Z
] = 0,
E[Z
2
] =
1
σ
2
E[X
2
].
Then
var(Z) =
1
√
2π
Z
∞
−∞
z
2
e
−z
2
/2
dz
=
−
1
√
2π
ze
−z
2
/2
∞
−∞
+
1
√
2π
Z
∞
−∞
e
−z
2
/2
dz
= 0 + 1
= 1
So var X = σ
2
.
Example. UK adult male heights are normally distributed with mean 70” and
standard deviation 3”. In the Netherlands, these figures are 71” and 3”.
What is
P
(
Y > X
), where
X
and
Y
are the heights of randomly chosen UK
and Netherlands males, respectively?
We have
X ∼ N
(70
,
3
2
) and
Y ∼ N
(71
,
3
2
). Then (as we will show in later
lectures) Y − X ∼ N(1, 18).
P(Y > X) = P(Y − X > 0) = P
Y − X − 1
√
18
>
−1
√
18
= 1 − Φ(−1/
√
18),
since
(Y −X)−1
√
18
∼ N (0, 1), and the answer is approximately 0.5931.
Now suppose that in both countries, the Olympic male basketball teams are
selected from that portion of male whose hight is at least above 4” above the
mean (which corresponds to the 9
.
1% tallest males of the country). What is the
probability that a randomly chosen Netherlands player is taller than a randomly
chosen UK player?
For the second part, we have
P(Y > X | X ≥ 74, Y ≥ 75) =
R
75
x=74
ϕ
X
(x) dx +
R
∞
x=75
R
∞
y=x
ϕ
Y
(y)ϕ
X
(x) dy dx
R
∞
x=74
ϕ
X
(x) dx
R
∞
y=75
ϕ
Y
(y) dy
,
which is approximately 0.7558. So even though the Netherlands people are only
slightly taller, if we consider the tallest bunch, the Netherlands people will be
much taller on average.