6주차

Introduction to Probability and Statistics
6th Week (4/12)

Special Probability Distributions (1)

Chebyshev’s Inequality
Chebyshev’s inequality guarantees that in any
data sample or probability distribution, "nearly all"
values are close to the mean

The precise statement being that no more than
1/k2 of the distribution’s values can be more than
k standard deviations away from the mean.

The inequality has great utility because it can be
applied to completely arbitrary distributions
(unknown except for mean and variance), for
example it can be used to prove the weak law of
Pafnuty Lvovich large numbers.
Chebyshev (1821 –
1894)

Law of Large Numbers

The law of large numbers
(LLN) is a theorem that
describes the result of
performing the same
experiment a large number of
times.

According to the law, the
average of the results obtained
from a large number of trials
should be close to the
expected value, and will tend to
become closer as more trials
are performed.

Law of Large Numbers

Why is it important?

Other Measures of Central Tendency

Percentiles: A Practical Example

Skewness, Kurtosis, and Moment

Discrete Probability Distribution

What kinds of PD do we have to know to
solve real-world problems?

Discrete Uniform Distribution

• Consider a case with rolling a fair dice

• Each random variable has same probability → Uniform distribution

Discrete Uniform Distribution

• Probability density
function :
• Expectation:

• Variance :

• Example

Suppose that we have a box containing 45 numbered balls. In this case, we
randomly select a ball and its number is X:
(1) Probability distribution for X
(2) Expectation and Variance for X
(3) P(X>40)

• Solution
(1)

(2)

(3)

(Discrete) Binomial Distribution

Bernoulli experiment: Only two kinds of results are possible

p = 0.85
q = 1- p = 0.15


Binomial Distribution

Some Properties of the Binomial Distribution


(1)

(2)
(3)
(4)


μ = n/2 을 중심으로 좌우대칭 :
대칭이항분포 (symmetric binomial distribution)

Tail in right

Tail in left

Example
Two factories, S and L, produce smart phones and their failure ratios are 5%. If you
buy 7 phones from S and 13 phones from L, what is the probability to have at least
one failed phone? And what is the probability that you have one failed phone?
Assume that the failure rates are independent.

Solution X (from S) and Y (from L) :
X ∼ B (7, 0.05) , Y ∼ B (13, 0.05) , X, Y : Independent

X + Y ∼ B (20, 0.05)

Only one phone is failed

At least one phone is failed

Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment
provided
1. The experiment is performed a fixed number of
times. Each repetition of the experiment is called a
trial.
2. The trials are independent. This means the
outcome of one trial will not affect the outcome of the
other trials.
3. For each trial, there are two mutually exclusive
outcomes, success or failure.
4. The probability of success is fixed for each trial of
the experiment.

Notation Used in the
Binomial Probability Distribution
• There are n independent trials of the experiment
• Let p denote the probability of success so that
1 – p is the probability of failure.
• Let x denote the number of successes in n
independent trials of the experiment. So, 0 < x < n.

EXAMPLE Identifying Binomial Experiments
Which of the following are binomial experiments?
(a) A player rolls a pair of fair die 10 times. The number
X of 7’s rolled is recorded.
(b) The 11 largest airlines had an on-time percentage of
84.7% in November, 2001 according to the Air Travel
Consumer Report. In order to assess reasons for
delays, an official with the FAA randomly selects flights
until she finds 10 that were not on time. The number of
flights X that need to be selected is recorded.
(c ) In a class of 30 students, 55% are female. The
instructor randomly selects 4 students. The number X
of females selected is recorded.

EXAMPLE Constructing a Binomial Probability
Distribution
According to the Air Travel Consumer Report,
the 11 largest air carriers had an on-time
percentage of 84.7% in November, 2001.
Suppose that 4 flights are randomly selected
from November, 2001 and the number of on-time
flights X is recorded. Construct a probability
distribution for the random variable X using a
tree diagram.

(Discrete) Multinomial Distribution

(Discrete) Geometric Distribution

Repeat Bernoulli experiments until the first success. => Number of Trial is X

Slot Machine:

How many should I try if I
get the jackpot?


Repeat Bernoulli experiments until the first success. => Number of Trial is X

: 성공 : 실패

(Discrete) Negative Binomial Distribution
Repeat Bernoulli experiments until the rth success.

Crane Game:

How many should I try if I
want to get three dolls?

Repeat Bernoulli experiments until the rth success.

(Discrete) Hypergeometric Distribution

It is similar to the binomial distribution. But the difference is the method of
sampling

Binomial experiment: Sampling with replacement
Hypergeometric experiment: Sampling without replacement

Normal shooting Russian roulette

Each trial has same probability Each trial may have different probability


A box contains N balls, where r balls are white (r<N)
Suppose that we randomly select n balls from the box, what is the number
of white balls (X)?

Assumption: Sampling without replacement
n 개의
items
N개
추출
의
items
개
개
개
개

Total 50 chips are in a box. Among those, 4 are out of order (failed
chips). If you select 5 chips:
(1) Probability distribution for the failed chip in these selected chips
(2) Probability to have one or two failed chips for this case
(3) Mathematical expectation and variance

(1) Random variable: X

(2)

(3) N = 50, r = 4, n = 5

Multivariate Hypergeometric Distribution

개
개
개
개
개
개

 X1 , X2 , X3 : Joint Probability Function

In a box, there are 3 red balls, 2 blue balls, and 5 yellow balls. You
select 4 balls.
(1) Joint probability function for X, Y, and Z
(2) Probability to select 1 red ball, 1 blue ball, and 2 yellow balls.

4 x개
개 y개
5개
z개
2개
3개

(1) Joint probability function:

(2)

(Discrete) Poisson Distribution
- Describe an event that rarely happens.
- All events in a specific period are mutually independent.
- The probability to occur is proportional to the length of the period.
- The probability to occur twice is zero if the period is short.

It is often used as a model for the number of events (such as the number of
telephone calls at a business, number of customers in waiting lines, number
of defects in a given surface area, airplane arrivals, or the number of
accidents at an intersection) in a specific time period.

If z > 0

Satisfy the PF
condition

 Probability function :

.

Ex.1. On an average Friday, a waitress gets no tip from 5 customers. Find the
probability that she will get no tip from 7 customers this Friday.

The waitress averages 5 customers that leave no tip on Fridays: λ = 5.
Random Variable : The number of customers that leave her no tip this Friday.
We are interested in P(X = 7).

Ex. 2 During a typical football game, a coach can expect 3.2 injuries. Find the
probability that the team will have at most 1 injury in this game.

A coach can expect 3.2 injuries : λ = 3.2.
Random Variable : The number of injuries the team has in this game.
We are interested in

.

Ex. 3. A small life insurance company has determined that on the average it receives 6
death claims per day. Find the probability that the company receives at least seven
death claims on a randomly selected day.
P(x ≥ 7) = 1 - P(x ≤ 6) = 0.393697

Ex. 4. The number of traffic accidents that occurs on a particular stretch of road
during a month follows a Poisson distribution with a mean of 9.4. Find the probability
that less than two accidents will occur on this stretch of road during a randomly
selected month.

P(x < 2) = P(x = 0) + P(x = 1) = 0.000860

Characteristics of Poisson Distribution
E(X) increases with parameter µ (or λ).
The graph becomes broadened with increasing the parameter µ (or λ).


Probability mass function Cumulative distribution function


Comparison of the Poisson
distribution (black dots) and the
binomial distribution with n=10 (red
line), n=20 (blue line), n=1000 (green
line). All distributions have a mean of
5. The horizontal axis shows the
number of events k. Notice that as n
gets larger, the Poisson distribution
becomes an increasingly better
approximation for the binomial
distribution with the same mean

Discrete Probability Distributions:
Summary

• Uniform Distribution
• Binomial Distributions
• Multinomial Distributions
• Geometric Distributions
• Negative Binomial Distributions
• Hypergeometric Distributions
• Poisson Distribution

Continuous Probability Distributions

What kinds of PD do we have to know to
solve real-world problems?

(Continuous) Uniform Distribution

(Continuous) Uniform Distributions
In a Period [a, b], f(x) is constant.

 f(x)

 E(x):

If X ∼ U(0, 1) and Y = a + (b - a) X,
(1)Distribution function for Y
(2)Probability function for Y
(3)Expectation and Variance for Y
(4) Centered value for Y

(1)

Since y = a + (b - a) x so 0 ≤ y ≤ b,

(Continuous) Uniform Distributions

(Continuous) Exponential Distribution

▶ Analysis of survival rate
▶ Period between first and second earthquakes
▶ Waiting time for events of Poisson distribution

For any positive α

From a survey, the frequency of traffic accidents X is given by
-3x
f(x) = 3e (0 ≤ x)

(1)Probability to observe the second accident after one month of the first
accident?
(2)Probability to observe the second accident within 2 months
(3)Suppose that a month is 30 days, what is the average day of the
accident?

(1)

(2)

(3) μ=1/3, accordingly 10 days.

• Survival function :

• Hazard rate, Failure rate:

A patient was told that he can survive average of 100 days. Suppose that the
probability function is given by

(1) What is the probability that he dies within 150 days.
(2) What is the probability that he survives 200 days

λ=0.01 이므로 분포함수와 생존함수 :
-x/100 -x/100
F(x)=1-e , S(x)=e

(1) 이 환자가 150 일 이내에 사망할 확률 :
-1.5
P(X < 150) = F(150) = 1-e = 1-0.2231 = 0.7769
(2) 이 환자가 200 일 이상 생존할 확률
-2.0
P(X ≥ 200) = S(200) = e = 0.1353

⊙ Relation with Poisson Process

(1) If an event occurs according to Poisson process with the ratio λ , the waiting
time between neighboring events (T) follows exponential distribution with the
exponent of λ.

(Continuous) Gamma Distribution


α : shape parameter, α > 0
β : scale parameter, β > 0

α=1 Γ (1, β) = E(1/β)

⊙ Relation with Exponential Distribution

Exponential distribution is a special gamma distribution with α = 1.

IF X1 , X2 , … , Xn have independent exponential distribution with the same
exponent 1/β, the sum of these random variables S= X1 + X2 + … +Xn results in
a gamma distribution, Γ(n, β).

If the time to observe an traffic accident (X) in a region have the following
probability distribution
-3x
f(x) = 3e , 0<x<∞

Estimate the probability to observe the first two accidents between the first
and second months. Assume that the all accidents are independent.
X1 : Time for the first accident
X2 : Time between the first and second accidents

Xi ∼ Exp(1/3) , I = 1, 2

S = X1 + X2 : Time for two accidents

S ∼ Γ(2, 1/3)
Probability function for S :

Answer:

(Continuous) Chi Square Distribution
A special gamma distribution α = r/2, β = 2

 PD

 E(X)

 Var(X)


A random variable X follows a Chi Square Distribution with a degree of
freedom of 5, Calculate the critical value to satisfy P(X < x0 )=0.95

Since P(X < x0 )=0.95, P(X > x0 )=0.05.

From the table, find the point with d.f.=5 and α=0.05

Why do we have to be bothered?

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