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Chapter 5
Basic Probability Distributions
:: Sunu Wibirama ::
http://te.ugm.ac.id/~wibirama/notes/
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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
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Ch. 5Ch. 2 & 3
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
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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.
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Classification of random variables:
discrete continuous
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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.
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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).
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Table of Probability Distributionfor Random Variable x
x px1 p1
x2 p2
... ...xn pn
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Example
A balanced coin is tossed twice and the number x of heads is observed. Find the probability distribution for x.
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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 H1H2 0.25 2
E2 H1T2 0.25 1
E3 T1H2 0.25 1
E4 T1T2 0.25 0
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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
x P(x)
0 0.25
1 0.50
2 0.25
Probability distribution for x, the number of heads in two tosses of a coin
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Properties of the probability distribution for a discrete random variable x
0 ( ) 1p x
( ) 1all x
p x x P(x)
0 0.25
1 0.50
2 0.25
Table 5.2.
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5.3 Numerical characteristics of a discrete random variable
• Mean or expected value
• Variance and standard deviation
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Definition 5.4• Let x be a discrete random variable with probability distribution p(x).
Then the mean or expected value of x is :
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 :
5.3.1 Mean or expected value
x all
xp(x)=E(x)=μ
xall
g(x)p(x)=E[g(x)]
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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
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Example cont’dCalculating the mean of these 400,000 values of x, we obtain
Thus, the mean of x is 1
100 000 0 200 000 1 100 000 2
400 000
1 1 1 0 1 2
4 2 4 all x
x , ( )+ , ( )+ , ( )μ =
n ,
= ( )+ ( )+ ( )= p(x)x
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Definition 5.6• Let x be a discrete random variable with probability distribution p(x). Then
the variance of x is
• The standard deviation of x is the positive square root of the variance of x:
5.3.2 Variance and standard deviation
]μ)E[(=σ 22 -x
2σ=σ
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Example 5.7
• Refer to the two-coin tossing experiment and the probability distribution for x. Find the variance and standard deviation of x.
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• In Example 5.6 we found the mean of x is 1. Then
and
Solution
22 2 2
0
2 2 2
x- x-
1 1 1 1 0 1 1 1 2 1
4 2 4 2
x=
σ = E[( μ) ] = ( μ) p(x)
( ) +( ) +( ) =
0.7072
12 =σ=σ
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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.
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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.
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The probability distribution:
The probability distribution, mean and variance for a binomial random variable:
xnxxn qpC=p(x)
x)!-(nx!
n!=C x
n
(x = 0, 1, 2, ..., n),
wherep = probability of a success on a single trial, q=1-pn = number of trials, x= number of successes in n trials
=combination of x from n.
The variance:
The mean: np=μ
npq=σ 2
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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
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• The probability distribution:
• Where := mean of events during the given time periode
• The mean:
• The variance:
The probability distribution, mean and variance for a Poisson random variable x:
λ=μ
x!
eλ=p(x)
λx
( x = 0, 1, 2,...),
e = 2.71828...(the base of natural logarithm).
λ=σ 2
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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.
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Definition 5.7• Let be a continuous random variable assuming any
value in the interval (- ,+ ). Then the cumulative distribution function F(x) of the variable is defined as follows:
• i.e., F(x) is equal to the probability that the variable assumes values , which are less than or equal to x.
ξ
x)P(ξ=F(x)
ξ
ξ
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Properties for continous random variable ξ
1)(0 .1 xF
F(a)-F(b) b)a P( 3.
xas 1F(x) and - xas 0F(a) 4.
F(b)F(a) then ba isthat
function, decreasing-nonlly monotonica a is )( .2
xF
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( ) ( )x
F x f t dt
The cumulative area under the curve between -∞ and a point x0 is equal to F(x0).
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5.7 Numerical characteristics of a continuous random variable
• Definition 5.8Let be a continuous random variable
with density function ( ), the mean:
( ) ( )
f x
E xf x dx
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• Defintion 5.9
Let be a continuous random variable
with densty function ( )
( ) is a function of x, then the mean :
( ) ( ) ( )
f x
g x
E g x g x f x dx
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• Definition 5.10
22
2
Let be a continuous random variable
with the expected value ( )
The variance of is :
The standard deviation of is the possitive
square root of the variance:
E
E
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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.
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Normal Distributions
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5.8 Standard Normal Distribution
• General density function:
• For standardized normal distribution:
22 2/)(
2
1)(
xexf
2( ) /21( )
2xf x e
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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 zx
z
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Find x from z-distribution Example: (P < -2.13) = 0.0166
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“Finding x from z”
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Homework 1:
• Chapter 5, Exercise 5.10, number 1(page 16)
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Homework 2:
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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.
How to do Homework 2:
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Download All You Need Here:
http://te.ugm.ac.id/~wibirama/notes/
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Due date: Friday, 14/10/2010@class
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Thank You