Chapter 5B: Normal and Exponential Distributions  
MATH/STAT 394: Probability I  
Arman Jahangiri  
University of Washington Department of Mathematics  
Summer 2026  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
1
 
Learning Goals  
By the end of this lecture, students should be able to:  
▶ define the standard Normal distribution;  
▶ standardize a Normal random variable;  
▶ compute Normal probabilities using the standard Normal CDF;  
▶ use the 68–95–99.7 rule;  
▶ define the Exponential distribution;  
▶ compute Exponential probabilities;  
▶ understand the memoryless property of the Exponential distribution.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
2
Normal Distribution  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
3
The Normal Distribution N(µ, σ2)  
Definition  
A continuous random variable X has a Normal dis-  
tribution with mean µ and variance σ2 if  
X ∼ N(µ, σ2),  
and its density is  
ꢀ
ꢁ
1
(x − µ)2  
√
f (x) =  
exp  
−
,
x ∈ R.  
2σ2  
2πσ2  
The parameter µ controls the center, while σ2 controls the spread. The special case where µ = 0, σ = 1 is called  
Standard Normal.  
Preview  
This is the most fundamental distribution in probability, with so many important fetures. Central Limit  
Theorem (CLT) will later shows one of its key aspects.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
4
PDF Verification of Normal  
fX (x) ≥ 0 is straightforward, so it suffices to show  
Claim  
Z
∞
√
2
−x  
e
/2 dx = 2π.  
−∞  
Let  
Z
∞
2
−x  
I =  
e
/2 dx.  
−∞  
Then  
ꢀ
ꢁ ꢀ  
ꢁ
Z
Z
Z
∞
∞
2
2
2
2
−x  
−y  
+y  
I2  
=
e
/2 dx  
e
/2 dy  
=
e−(x  
)/2 dx dy.  
2
−∞  
−∞  
R
Using polar coordinates,  
Therefore,  
x = r cos θ,  
y = r sin θ,  
dx dy = r dr dθ.  
Z
Z
2π  
∞
2
I2  
=
e−r /2r dr dθ.  
0
0
Let u = r2/2, so du = r dr. Then  
Z
Z
Z
2π  
∞
2π  
I2  
=
e−u du dθ =  
1 dθ = 2π.  
0
0
0
Thus,  
√
Chapter 5B: Normal and Exponential Distributions  
I = 2π.  
Arman Jahangiri  
Summer 2026  
5
Mean and Variance of N(µ, σ2)  
Proposition  
Suppose  
X ∼ N(µ, σ2),  
with density  
Then,  
ꢀ
ꢁ
1
(x − µ)2  
√
f (x) =  
exp  
−
.
2σ2  
2πσ2  
E(X) = µ,  
Var(X) = σ2.  
Proof: By definition,  
Z
∞
E(X) = −∞ xf (x) dx.  
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Chapter 5B: Normal and Exponential Distributions  
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Let  
z = x − µ.  
Then  
x = z + µ,  
dx = dz.  
Therefore,  
Z
∞
2
2
1
/(2σ  
e−z  
) dz.  
√
E(X) = −∞(z + µ)  
2πσ2  
Splitting the integral,  
Z
Z
|
∞
∞
2
2
2
2
1
1
/(2σ  
/(2σ  
e−z  
) dz +µ  
e−z  
) dz .  
√
√
2πσ2  
E(X) = −∞ z  
2πσ2  
−∞  
|
{z  
}
{z  
}
=0  
=1  
Hence,  
E(X) = µ.  
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Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
7
Variance of N(µ, σ2)  
Recall:  
ꢂ
ꢃ
Var(X) = E (X − µ)2  
.
Using LOTUS,  
Z
∞
Var(X) =  
(x − µ)2f (x) dx.  
−∞  
Let  
z = x − µ.  
Then  
Z
∞
2
2
1
Var(X) = −∞ z2  
e−z  
) dz.  
/(2σ  
√
2πσ2  
Now substitute  
Then  
z
u =  
,
z = σu,  
dz = σ du.  
σ
Z
∞
2
1
Var(X) = σ2 −∞ u2  
e−u /2 du.  
√
2π  
The remaining integral equals 1, since it is the variance of the standard Normal distribution N(0, 1).  
Therefore,  
Var(X) = σ2.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
8
Standard Normal CDF  
The standard Normal CDF is denoted by  
It is defined by  
Φ(z).  
Z
z
2
1
e−t /2 dt.  
√
Φ(z) = P(Z ≤ z) =  
2π  
−∞  
Important  
There is no elementary closed-form expression for Φ(z).  
So Normal probabilities are usually computed using tables, calculators, or software. See ”normal-table.pdf” in  
Canvas for the Standard Normal table.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026  
9
Symmetry of the Standard Normal  
The standard Normal PDF is symmetric about 0, i.e., φ(z) = φ(−z). Therefore, P(Z ≤ −z) = P(Z ≥ z). In  
terms of the CDF, Φ(−z) = 1 − Φ(z). This is why some Normal PDF (not CDF) tables skip the negative values  
and only provide ϕ(z) for z ≥ 0.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 10  
Converting N(0, 1) to N(µ, σ2)  
Proposition  
If  
Z ∼ N(0, 1),  
then  
X = µ + σZ  
has the Normal distribution with mean µ and variance σ2, i.e.,  
X ∼ N(µ, σ2).  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 11  
Converting N(µ, σ2) to N(0, 1)  
Proposition  
If  
X ∼ N(µ, σ2),  
then  
X − µ  
Z =  
∼ N(0, 1).  
σ
Standardization converts a general Normal random variable into a standard Normal.  
Use  
To compute probabilities involving X, convert them into probabilities involving Z, and then use the Normal  
probability table.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 12  
Example: Standardizing  
Suppose  
Find  
X ∼ N(70, 102).  
P(X < 85).  
Standardize:  
ꢀ
ꢁ
X − 70  
85 − 70  
P(X < 85) = P  
<
.
10  
10  
Thus  
P(X < 85) = P(Z < 1.5) = Φ(1.5).  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 13  
The 68–95–99.7 Rule  
If  
X ∼ N(µ, σ2),  
then approximately:  
P(|X − µ| < σ) ≈ 0.68,  
P(|X − µ| < 2σ) ≈ 0.95,  
P(|X − µ| < 3σ) ≈ 0.997.  
Interpretation  
Most Normal observations lie within a few standard deviations of the mean.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 14  
Exponential Distribution  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 15  
Exponential Distribution  
The Exponential distribution models waiting times.  
Examples:  
▶ waiting time until a phone call;  
▶ waiting time until a radioactive particle decays;  
▶ waiting time until a customer arrives;  
▶ lifetime of certain components.  
It is the continuous analog of waiting for the first success.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 16  
Definition of Exponential Distribution  
Definition  
A random variable X has the Exponential distribution with rate λ > 0 if its PDF is  
f (x) = λe−λx  
,
x > 0.  
We write  
X ∼ Exp(λ).  
For x ≤ 0,  
f (x) = 0.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 17  
Exponential CDF and Survival Function  
If  
X ∼ exp(λ),  
then for x > 0,  
F(x) = P(X ≤ x) = 1 − e−λx  
P(X > x) = 1 − F(x) = e−λx  
.
.
Therefore,  
Survival function  
P(X > x) = e−λx  
.
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 18  
Mean and Variance of Exponential  
If  
X ∼ Exp(λ),  
then  
and  
1
E(X) =  
,
λ
1
λ2  
Var(X) =  
.
Interpretation  
A larger rate λ means events occur more frequently, so waiting times become shorter on average.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 19  
Proof of the Mean  
The density of an exponential random variable is  
f (x) = λe−λx  
,
x ≥ 0.  
Thus  
Z
∞
E(X) =  
xλe−λx dx.  
0
Use integration by parts.  
Let  
u = x,  
dv = λe−λx dx.  
Then  
du = dx,  
v = −e−λx  
.
Therefore,  
Z
ꢄ
h
i∞  
∞
E(X) = −xe−λx  
+
e
−λx dx.  
0
0
The boundary term equals 0, so  
ꢅ∞  
Z
∞
1
E(X) =  
e
−λx dx =  
−
e−λx  
.
λ
0
0
Hence  
1
E(X) =  
.
λ
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 20  
Proof of the Variance  
We first compute  
Using the density,  
E(X2).  
Z
∞
E(X2) =  
x2λe−λx dx.  
0
Applying integration by parts twice gives  
2
λ2  
E(X2) =  
.
Now use  
Var(X) = E(X2) − (E(X))2.  
Therefore,  
ꢀ
ꢁ
2
2
λ2  
1
Var(X) =  
−
.
λ
Hence  
1
λ2  
Var(X) =  
.
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 21  
Memoryless Property of Exp(λ)  
Memoryless Property  
If  
X ∼ exp(λ),  
then for s, t ≥ 0,  
P(X > s + t | X > s) = P(X > t).  
Interpretation:  
Given that we have already waited s units of time, the additional waiting time behaves like a fresh Exponential  
random variable.  
Remark: Exponential distribution is the only memoryless r.v. among all the continuous r.v.s. Also, the only  
memoryless discrete r.v. is the Geometric r.v. that we studied in Chapter 3.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 22  
Proof of Memorylessness  
Using conditional probability,  
P(X > s + t)  
P(X > s)  
P(X > s + t | X > s) =  
.
For an Exponential random variable,  
Thus,  
P(X > x) = e−λx  
.
P(X > s + t)  
P(X > s)  
e−λ(s+t)  
e−λs  
=
= e−λt  
.
But  
e−λt = P(X > t).  
Therefore,  
P(X > s + t | X > s) = P(X > t).  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 23  
Geometric vs Exponential  
Both distributions model a waiting time until the first event.  
Distribution Time type  
Function Formula  
p
1−p  
,
ln(1−p)x  
Geom(p)  
discrete  
PMF  
P(X = x) =  
e
,
x ∈ Z+  
exp(λ)  
continuous PDF  
f (x) = λe−λx  
x ∈ R+  
Geometric PMF: probability at integer times  
Arman Jahangiri Chapter 5B: Normal and Exponential Distributions  
Exponential PDF: density over continuous time  
Summer 2026 24  
Geometric
is the discrete-time analogue of Exp
onential.  
Summary  
▶ The Normal distribution is bell-shaped and central in statistics.  
▶ Standard Normal:  
Z ∼ N(0, 1).  
▶ Standardization:  
X − µ  
∼ N(0, 1).  
σ
▶ Exponential random variables model waiting times.  
▶ If X ∼ exp(λ), then  
P(X > x) = e−λx  
.
▶ The Exponential distribution is memoryless.  
Next lecture  
Joint distributions and transformations.  
Arman Jahangiri  
Chapter 5B: Normal and Exponential Distributions  
Summer 2026 25