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Chapter 5
Basic Probability Distributions
:: Sunu Wibirama ::
http://te.ugm.ac.id/~wibirama/notes/
CONTENTS
• 5.1. Random variables
• 5.2. The probability distribution for a discrete random
variable
• 5.3. Numerical characteristics of a discrete random
variable
• 5.4. The binomial probability distribution
• 5.5. The Poisson distribution
• 5.6 Continuous random variables: distribution
function and density function
• 5.7 Numerical characteristics of a continuous
random variable
• 5.8. The normal distribution
Ch. 2 & 3
Ch. 5
Ch. 4
Ch. 2-4 : we used observed sample (what did actually happen)
Ch. 5 : combine ch.2-4 by presenting possible outcome along
with the relative frequency we expect
5.1 Random Variables
Definition 5.1
• A random variable is a variable that
assumes numerical values associated
with events of an experiment.
• Example: x = number of girls among
14 babies. x is random variable
because its values depend on
chances
Table 5.0.
Classification of random variables:
discrete
continuous
Definition 5.2
• A discrete random variable is one that can
assume only a countable number of values.
• A continuous random variable can assume
any value in one or more intervals on a line.
5.2 The probability distribution for a discrete
random variable
Definition 5.3
• The probability distribution for a discrete
random variable x is a table, graph, or formula
that gives the probability of observing each
value of x. We shall denote the probability of x
by the symbol p(x).
Table of Probability Distribution
for Random Variable x
x
x1
x2
...
xn
p
p1
p2
...
pn
Example
A balanced coin is tossed twice and the
number x of heads is observed. Find the
probability distribution for x.
Solution
• Let Hk and Tk denote the observation of a head and a tail,
respectively, on the k-th toss, for k = 1, 2. The four simple
events and the associated values of x are shown in Table 5.1.
SIMPLE EVENT
DESCRIPTION
PROBABILITY
NUMBER OF HEADS
E1
H1 H 2
0.25
2
E2
H1 T 2
0.25
1
E3
T1H2
0.25
1
E4
T1T 2
0.25
0
Probability distribution for x
• P(x = 0) = p(0) = P(E4) = 0.25.
• P(x = 1) = p(1) = P(E2) + P(E3) = 0.25 + 0.25 = 0.5
• P(x = 2) = p(2) = P(E1) = 0.25
Probability distribution for x, the number of heads in two tosses of a coin
x
0
1
2
P(x)
0.25
0.50
0.25
Properties of the probability distribution for a
discrete random variable x
0  p ( x)  1
 p( x)  1
all x
Table 5.2.
x
0
1
2
P(x)
0.25
0.50
0.25
5.3 Numerical characteristics of a discrete random
variable
• Mean or expected value
• Variance and standard deviation
5.3.1 Mean or expected value
Definition 5.4
• Let x be a discrete random variable with probability distribution p(x).
Then the mean or expected value of x is :
μ = E(x)=  xp(x)
all x
Definition 5.5
• Let x be a discrete random variable with probability distribution p(x) and
let g(x) be a function of x . Then the mean or expected value of g(x) is :
E[g(x)] =  g(x)p(x)
all x
Example 5.6
• Refer to the two-coin tossing experiment and the probability
distribution for the random variable x Demonstrate that the
formula for E(x) gives the mean of the probability distribution
for the discrete random variable x.
Solution :
If we were to repeat the two-coin tossing experiment a large
number of times – say 400,000 times, we would expect to
observe x = 0 heads approximately 100,000 times, x = 1 head
approximately 200,000 times and x = 2 heads approximately
100,000 times. Calculating the mean of these 400,000 values
of x, we obtain
Example cont’d
Calculating the mean of these 400,000 values of x, we obtain
x 100,000( 0 )+ 200,000( 1 )+100, 000( 2 )

μ
=
n
400,000
1
1
1
= ( 0 )+ ( 1 )+ ( 2 )=  p(x)x
4
2
4
all x
Thus, the mean of x is 1
5.3.2 Variance and standard deviation
Definition 5.6
• Let x be a discrete random variable with probability distribution p(x). Then
the variance of x is
σ = E[( x - μ) ]
2
2
• The standard deviation of x is the positive square root of the variance of x:
σ= σ
2
Example 5.7
• Refer to the two-coin tossing experiment and
the probability distribution for x. Find the
variance and standard deviation of x.
Solution
• In Example 5.6 we found the mean of x is 1. Then
2
σ = E[( x-μ) ] =  ( x-μ)2 p(x)
2
2
x=0
1
1
1 1
 ( 0  1 )2   +( 1  1 )2   +( 2  1 )2   =
4
2
4 2
and
1
σ= σ =
 0.707
2
2
5.4 The binomial probability distribution
• Example 5.8 Suppose that 80% of the jobs submitted to a
data-processing center are of a statistical nature. Then
selecting a random sample of 10 submitted jobs would be
analogous to tossing an unbalanced coin 10 times, with the
probability of observing a head (drawing a statistical job) on a
single trial equal to 0.80.
• Example 5.9 Test for impurities commonly found in drinking
water from private wells showed that 30% of all wells in a
particular country have impurity A. If 20 wells are selected at
random then it would be analogous to tossing an unbalanced
coin 20 times, with the probability of observing a head
(selecting a well with impurity A) on a single trial equal to
0.30.
Model (or characteristics) of a binomial random
variable
•
•
•
•
•
The experiment consists of n identical trials
There are only 2 possible outcomes on each trial.
We will denote one outcome by S (for Success) and
the other by F (for Failure).
The probability of S remains the same from trial to
trial. This probability will be denoted by p, and the
probability of F will be denoted by q ( q = 1-p).
The trials are independent.
The binomial random variable x is the number of S’
in n trials.
The probability distribution, mean and variance
for a binomial random variable:
The probability distribution:
p(x) = Cnx p x q n  x
(x = 0, 1, 2, ..., n),
where
p = probability of a success on a single trial, q=1-p
n = number of trials, x= number of successes in n trials
Cnx =
n!
x!(n - x)!
=combination of x from n.
The mean:
μ = np
The variance:
σ 2 = npq
5.5 The Poisson distribution
Characteristics defining a Poisson random variable
• The experiment consists of counting the number x of
times a particular event occurs during a given unit of
time
• The probability that an event occurs in a given unit of
time is the same for all units.
• The number of events that occur in one unit of time
is independent of the number that occur in other
units.
• The mean number of events in each unit will be
denoted by the Greek letter 
The probability distribution, mean and variance for a
Poisson random variable x:
•
•
The probability distribution:
λ xe λ
p(x)=
x!
Where :
 = mean of events during the given time periode
e = 2.71828...(the base of natural logarithm).
•
•
( x = 0, 1, 2,...),
The mean:
The variance:
μ= λ
σ =λ
2
5.6 Continuous random variables: distribution
function and density function
• The distinction between discrete random
variables and continuous random variables is
usually based on the difference in their
cumulative distribution functions.
Definition 5.7
• Let ξ be a continuous random variable assuming any
value in the interval (-  ,+ ). Then the cumulative
distribution function F(x) of the variable ξ is defined
as follows:
F(x)= P(ξ  x)
• i.e., F(x) is equal to the probability that the variable ξ
assumes values , which are less than or equal to x.
Properties for continous random
variable ξ
1. 0  F ( x)  1
2. F ( x) is a monotonica lly non - decreasing function,
that is a  b then F(a)  F(b)
3. P( a    b)  F(b) - F(a)
4. F(a)  0 as x  - and F(x)  1 as x  
x
F ( x) 

f (t )dt

The cumulative area under the curve between -∞ and a point x0 is equal to F(x0).
5.7 Numerical characteristics of a continuous
random variable
• Definition 5.8
Let  be a continuous random variable
with density function f ( x), the mean:
E ( ) 

 xf ( x)dx

• Defintion 5.9
Let  be a continuous random variable
with densty function f ( x)
g ( x) is a function of x, then the mean :
E  g ( x)  

 g ( x) f ( x)dx

• Definition 5.10
Let  be a continuous random variable
with the expected value E ( )  
The variance of  is :
2

  E     


The standard deviation of  is the possitive
2
square root of the variance:
 
2
Normal Distribution
• The most well known distribution for Discrete
Random Variable is Binomial Distribution
• The most well known distribution for Continous
Random Variable is
Normal Distribution
• We say x that has a normal distribution if its
values fall into a smooth (continuous) curve with
a bell-shaped, symmetric pattern, meaning it
looks the same on each side when cut down the
middle. The total area under the curve is 1.
Normal Distributions
5.8 Standard Normal Distribution
• General density function:
• For standardized normal
distribution:
1
( x   ) 2 / 2 2
f ( x) 
e
 2
f ( x) 
1  ( x )2 /2
e
2
Standard Normal Distribution
• It’s also called z-distribution
• It has a mean of 0 and standard deviation of 1
• Transforming normal random variable x to standard normal
random variable z
z
x

Find x from z-distribution
Example: (P < -2.13) = 0.0166
“Finding x from z”
Homework 1:
• Chapter 5, Exercise 5.10, number 1
(page 16)
Homework 2:
How to do Homework 2:
1. Get the copy of z-table and use it to
compute the result.
2. Draw the distribution of fish-length
and shade representative areas as stated
in problems 1, 2, and 3.
3. Use “Finding x from z” to solve
problem 1, 2, and 3.
Download All You Need Here:
http://te.ugm.ac.id/~wibirama/notes/
Due date: Friday, 14/10/2010
@class
Thank You