STAT 430/510 Lecture 12
STAT 430/510 Probability
Lecture 12: Central Limit Theorem and
Exponential Distribution
Pengyuan (Penelope) Wang
June 15, 2011
STAT 430/510 Lecture 12
Review
Discussed Uniform Distribution and Normal Distribution
Normal Approximation to Binomial Distribution
STAT 430/510 Lecture 12
A little thing about Normal Distribution
For N(µ, SD = σ),
68% probability is within ±σ around µ.
95% probability is within ±2σ around µ.
99.7% probability is within ±3σ around µ.
So if a value is outside 3σ around µ, it is very rare to happen!
STAT 430/510 Lecture 12
Central Limit Theorem: a Generalization of Normal Approximation
to Binomial Distribution
Xi , i = 1, ..., n independently and identically follow a
distribution with expected value µ and standard deviation
σ. If n is large, then X̄ = n1 (X1 + X2 + ... + Xn ) follows
approximately a normal distribution with mean µ, and
standard deviation √σn .
STAT 430/510 Lecture 12
Example
There are 100 workers in a factory. Assume that their ages
independently and identically follows a distribution with
expected value 40 years old and standard deviation 10
years old. What is the probability that their mean age
higher than 45 years old?
Let X represent their mean age. According to the Central
Limit Theorem, X approximately follows a normal
distribution
with mean 40 and standard deviation
√
10/ 100 = 1.
P(X > 45) = P(Z >
45−40
1 )
= P(Z > 5) = 0.00.
STAT 430/510 Lecture 12
Example
There are 100 male workers and 100 female workers in a
factory. Assume that the ages of males independently and
identically follows a distribution with expected value 40
years old and standard deviation 10 years old. Assume
that the ages of females independently and identically
follows a distribution with expected value 39 years old and
standard deviation 20 years old. What is the probability
that male’s mean age is at least 2 years higher than
female’s mean age?
STAT 430/510 Lecture 12
Example
There are 100 male workers and 100 female workers in a
factory. Assume that the ages of males independently and
identically follows a distribution with expected value 40
years old and standard deviation 10 years old. Assume
that the ages of females independently and identically
follows a distribution with expected value 39 years old and
standard deviation 20 years old. What is the probability
that male’s mean age is at least 2 years higher than
female’s mean age?
The male’s mean age (X ) approximately follows normal
distribution (40, 1), and the female’s mean age (Y ) follows
normal distribution (39, 2).
Thus
√ X − Y approximately follows normal distribution
(1, 5).
√
√
So P(X − Y > 2) ≈ P(Z > (2 − 1)/ 5) = 1 − Φ(1/ 5).
STAT 430/510 Lecture 12
Example
The approximation to the Binomial distribution is just one
application of the central limit theorem.
An insurance company believes that people can be divided
into two classes: those who are accident prone and those
who are not. The company’s statistics show that an
accident-prone person will have an accident at some time
within a fixed 1-year period with probability 0.4, whereas
this probability decreases to 0.2 for a person who is not
accident prone. We assume that 30 percent of the
population is accident prone. If we draw 100 policyholders,
what is the probability that more than 30 of them have
accidents within a year of purchasing a policy?
STAT 430/510 Lecture 12
Example
The probability to have accident for each person is
0.4*0.3+0.2*0.7=0.26 .
Let X be the proportion of the people with accidents, then
√ X −0.26
approximately follows standard Normal
0.26∗0.74/100
distribution.
P(X > 0.3) = P( √
1 − Φ(0.91).
X −0.26
0.26∗0.74/100
>√
0.3−0.26
)
0.26∗0.74/100
=
STAT 430/510 Lecture 12
Exponential Random Variable
A continuous random variable X is said to have a
exponential distribution with parameter λ if the pdf of X
is
λe−λx , if x ≥ 0
f (x) =
0, if x < 0
STAT 430/510 Lecture 12
cdf of Exponential r.v.
For exponential r.v. X with parameter λ, the cdf is
1 − e−λx , x ≥ 0
F (x) =
0, x < 0
STAT 430/510 Lecture 12
Expected Value and Variance
X is exponential random variable with parameter λ.
E[X ] =
1
λ
Var (X ) =
1
λ2
STAT 430/510 Lecture 12
Example
Suppose that the length of a phone call in minutes is an
exponential random variable with parameter λ = 0.1. If
someone arrives immediately ahead of you at a public
telephone booth, find the probability that you will have to
wait
(a) More than 10 minutes?
(b) Between 10 and 20 minutes?
STAT 430/510 Lecture 12
Example: Solution
Let X denote the length of the call made by the person in
the booth.
P(X > 10) = 1 − F (10) = e−1 = 0.368
P(10 < X < 20) = F (20) − F (10) = e−1 − e−2 = 0.233
STAT 430/510 Lecture 12
Memoryless Property
The exponential distribution is memoryless!
It means: if
P(T ≥ s + t|T ≥ t) = P(T ≥ s)
for all s, t ≥ 0.
In fact, the exponential distribution is the only memoryless
continuous distribution.
STAT 430/510 Lecture 12
Example
Suppose that the length of a phone call in minutes is an
exponential random variable with parameter λ = 0.1.
Someone arrives immediately ahead of you at a public
telephone booth. Now you have already waited for 10
minutes, from now on what is the probability that you need
to wait at least for another 10 minutes?
STAT 430/510 Lecture 12
Example
Suppose that the length of a phone call in minutes is an
exponential random variable with parameter λ = 0.1.
Someone arrives immediately ahead of you at a public
telephone booth. Now you have already waited for 10
minutes, from now on what is the probability that you need
to wait at least for another 10 minutes?
From the memoryless property of exponential distribution,
P(X > 10 + 10|X > 10) = P(X > 10) = e−10∗0.1 = e−1
STAT 430/510 Lecture 12
Example
Suppose that the number of miles that a car can run before
its battery wears out is exponentially distributed with an
average value of 10,000 miles. If a person has used the
battery for some time and now he desires to take another
5000-mile trip,
then what is the probability that he or she will be able to
complete the trip without having to replace the car battery?
What can be said when the distribution is not exponential?
STAT 430/510 Lecture 12
Example: Solution
X is the lifetime of the battery
t is the number of miles that the battery had been in use
prior to the start of the trip.
From the memoryless property of exponential distribution,
P(X > t+5000|X > t) = P(X > 5000) = e−5000/10000 = 0.606
If the distribution F of X is not exponential distribution,
P(X > t+5000|X > t) =
P(X > t + 5000)
1 − F (t + 5000)
=
P(X > t)
1 − F (t)