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ProbabilityProbability
©
Sample SpaceSample Space
The possible outcomes of a random experiment are called the basic outcomes, and the set of all basic outcomes is called the sample space.sample space. The
symbol SS will be used to denote the sample space.
Sample SpaceSample Space- An Example -- An Example -
What is the sample space for a roll of a single six-sided die?
S = [1, 2, 3, 4, 5, 6]
Mutually ExclusiveMutually Exclusive
If the events A and B have no common basic outcomes, they are mutually exclusive mutually exclusive and their intersection A B is said to be the empty set indicating that A B cannot
occur.
More generally, the K events E1, E2, . . . , EK are said to be mutually exclusive if every pair of them is a
pair of mutually exclusive events.
Venn DiagramsVenn Diagrams
Venn DiagramsVenn Diagrams are drawings, usually using geometric shapes, used to depict basic
concepts in set theory and the outcomes of random experiments.
Intersection of Events A and BIntersection of Events A and B
A B A BAB
(a) AB is the striped area
S S
(b) A and B are Mutually Exclusive
Collectively ExhaustiveCollectively Exhaustive
Given the K events E1, E2, . . ., EK in the sample space S. If E1 E2 . . . EK = S,
these events are said to be collectively collectively exhaustiveexhaustive.
ComplementComplement
Let A be an event in the sample space S. The set of basic outcomes of a random experiment
belonging to S but not to A is called the complement complement of A and is denoted by A.
Venn Diagram for the Venn Diagram for the Complement of Event AComplement of Event A
AA
S
Unions, Intersections, and Unions, Intersections, and ComplementsComplements
A die is rolled. Let A be the event “Number rolled is even” and B be the event “Number rolled is at least 4.” Then
A = [2, 4, 6] and B = [4, 5, 6]
3] 2, [1, B and 5] 3, [1, A 6] [4, BA
6] 5, 4, [2, BA
S 6] 5, 4, 3, 2, [1, AA
Classical ProbabilityClassical Probability
The classical definition of probabilityclassical definition of probability is the proportion of times that an event will occur,
assuming that all outcomes in a sample space are equally likely to occur. The probability of an
event is determined by counting the number of outcomes in the sample space that satisfy the
event and dividing by the number of outcomes in the sample space.
Classical ProbabilityClassical ProbabilityThe probability of an event A is
where NA is the number of outcomes that satisfy the condition of event A and N is the total number of outcomes in the sample space. The important idea here is that one can develop a probability from fundamental reasoning
about the process.
N
N P(A) A
CombinationsCombinations
The counting process can be generalized by using the following equation to compare the number of combinations of n things
taken k at a time.
1!0)!(!
!
knk
n C n
k
Relative FrequencyRelative Frequency
The relative frequency definition of probabilityrelative frequency definition of probability is the limit of the proportion of times that an event A occurs in a large number of trials, n,
where nA is the number of A outcomes and n is the total number of trials or outcomes in the population. The probability is the limit as n becomes large.
n
n P(A) A
Subjective ProbabilitySubjective Probability
The subjective definition of probabilitysubjective definition of probability expresses an individual’s degree of belief about
the chance that an event will occur. These subjective probabilities are used in certain
management decision procedures.
Probability PostulatesProbability Postulates
Let S denote the sample space of a random experiment, O i, the basic outcomes, and A, an event. For each event A of the sample space S, we assume that a number P(A) is defined and we have the postulates
1. If A is any event in the sample space S
2. Let A be an event in S, and let Oi denote the basic outcomes. Then
where the notation implies that the summation extends over all the basic outcomes in A.
3. P(S) = 1
1)(0 AP
)()( A
iOPAP
Probability RulesProbability Rules
Let A be an event and A its complement. The the complement rule iscomplement rule is:
)(1)( APAP
Probability RulesProbability Rules
The Addition Rule of ProbabilitiesThe Addition Rule of Probabilities:
Let A and B be two events. The probability of their union is
)()()()( BAPBPAPBAP
Probability RulesProbability RulesVenn Diagram for Addition RuleVenn Diagram for Addition Rule
)()()()( BAPBPAPBAP P(AB)
A B
P(A)
A B
P(B)
A B
P(AB)
A B+ -
=
Probability RulesProbability Rules
Conditional ProbabilityConditional Probability:
Let A and B be two events. The conditional probabilityconditional probability of event A, given that event B has occurred, is denoted by the
symbol P(A|B) and is found to be:
provided that P(B > 0).
)(
)()|(
BP
BAPBAP
Probability RulesProbability Rules
Conditional ProbabilityConditional Probability:
Let A and B be two events. The conditional probabilityconditional probability of event B, given that event A has occurred, is denoted by the
symbol P(B|A) and is found to be:
provided that P(A > 0).
)(
)()|(
AP
BAPABP
Probability RulesProbability Rules
The Multiplication Rule of ProbabilitiesThe Multiplication Rule of Probabilities:
Let A and B be two events. The probability of their intersection can be derived from the
conditional probability as
Also,
)()|()( BPBAPBAP
)()|()( APABPBAP
Statistical IndependenceStatistical Independence
Let A and B be two events. These events are said to be statistically independent if and only if
From the multiplication rule it also follows that
More generally, the events E1, E2, . . ., Ek are mutually statistically independent if and only if
)()()( BPAPBAP
0)P(B) if(P(A)B)|P(A 0)P(A) if(P(B)A)|P(B
)P(E)P(E )P(E)EEP(E K21K21
Bivariate ProbabilitiesBivariate Probabilities
B1 B2 . . . Bk
A1P(A1B1) P(A1B2) . . . P(A1Bk)
A2P(A2B1) P(A2B2) . . . P(A2Bk)
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
AhP(AhB1) P(AhB2) . . . P(AhBk)
Outcomes for Bivariate Events
Joint and Marginal ProbabilitiesJoint and Marginal Probabilities
In the context of bivariate probabilities, the intersection probabilities P(Ai Bj) are called joint joint
probabilities.probabilities. The probabilities for individual events P(Ai) and P(Bj) are called marginal probabilitiesmarginal probabilities.
Marginal probabilities are at the margin of a bivariate table and can be computed by summing the
corresponding row or column.
Probabilities for the Television Probabilities for the Television Viewing and Income ExampleViewing and Income Example
Viewing Frequency
High Income
Middle
Income
Low Income
Totals
Regular 0.04 0.13 0.04 0.21
Occasional 0.10 0.11 0.06 0.27
Never 0.13 0.17 0.22 0.52
Totals 0.27 0.41 0.32 1.00
Tree DiagramsTree Diagrams
P(A3 )
= .52
P(A1 B1) = .04
P(A2) = .27
P(A 1
) = .2
1 P(A1 B2) = .13
P(A1 B3) = .04
P(A2 B1) = .10
P(A2 B2) = .11
P(A2 B3) = .06
P(A3 B1) = .13
P(A3 B2) = .17
P(A3 B3) = .22
P(S) = 1
Probability RulesProbability Rules
Rule for Determining the Independence of AttributesRule for Determining the Independence of Attributes
Let A and B be a pair of attributes, each broken into mutually exclusive and collectively exhaustive event
categories denoted by labels A1, A2, . . ., Ah and
B1, B2, . . ., Bk. If every Ai is statistically independentstatistically independent of every event Bj, then the attributes A and B are
independent.
Bayes’ TheoremBayes’ Theorem
Let A and B be two events. Then Bayes’ TheoremBayes’ Theorem states that:
and
P(A)
B)P(B)|P(A)|( BAP
P(B)
A)P(A)|P(B)|( BAP
Bayes’ TheoremBayes’ Theorem(Alternative Statement)(Alternative Statement)
Let E1, E2, . . . , Ek be mutually exclusive and collectively exhaustive events and let A be some other event. The
conditional probability of Ei given A can be expressed as
Bayes’ TheoremBayes’ Theorem:
))P(EE|P(A))P(EE|P(A))P(EE|P(A
))P(EE|P(AA)|P(E
KK2211
iii
Bayes’ TheoremBayes’ Theorem- Solution Steps -- Solution Steps -
1. Define the subset events from the problem.
2. Define the probabilities for the events defined in step 1.
3. Compute the complements of the probabilities.
4. Apply Bayes’ theorem to compute the probability for the problem solution.
Discrete Random Variables and Discrete Random Variables and Probability DistributionsProbability Distributions
©
Random VariablesRandom Variables
A random variablerandom variable is a variable that takes on numerical values determined by the outcome
of a random experiment.
Discrete Random VariablesDiscrete Random Variables
A random variable is discrete discrete if it can take on no more than a countable
number of values.
Discrete Random VariablesDiscrete Random Variables(Examples)(Examples)
1. The number of defective items in a sample of twenty items taken from a large shipment.
2. The number of customers arriving at a check-out counter in an hour.
3. The number of errors detected in a corporation’s accounts.
4. The number of claims on a medical insurance policy in a particular year.
Continuous Random VariablesContinuous Random Variables
A random variable is continuous continuous if it can take any value in an interval.
Continuous Random VariablesContinuous Random Variables(Examples)(Examples)
1. The income in a year for a family.
2. The amount of oil imported into the U.S. in a particular month.
3. The change in the price of a share of IBM common stock in a month.
4. The time that elapses between the installation of a new computer and its failure.
5. The percentage of impurity in a batch of chemicals.
Discrete Probability DistributionsDiscrete Probability Distributions
The probability distribution function (DPF),probability distribution function (DPF), P(x), of a discrete random variable expresses the
probability that X takes the value x, as a function of x. That is
. of valuesallfor ),()( xxXPxP
Discrete Probability DistributionsDiscrete Probability Distributions
Graph the probability distribution function for the roll of a single six-sided die.
1 2 3 4 5 6
1/6
P(x)
x
Required Properties of Probability Required Properties of Probability Distribution Functions of Discrete Distribution Functions of Discrete
Random VariablesRandom VariablesLet X be a discrete random variable with probability distribution function, P(x). Then
i. P(x) 0 for any value of xii. The individual probabilities sum to 1; that is
Where the notation indicates summation over all possible values x.
x
xP 1)(
Cumulative Probability FunctionCumulative Probability Function
The cumulative probability function,cumulative probability function, F(x0), of a random variable X expresses the probability
that X does not exceed the value x0, as a function of x0. That is
Where the function is evaluated at all values x0
)()( 00 xXPxF
Derived Relationship Between Probability Derived Relationship Between Probability Function and Cumulative Probability Function and Cumulative Probability
FunctionFunction
Let X be a random variable with probability function P(x) and cumulative probability function F(x0). Then it
can be shown that
Where the notation implies that summation is over all possible values x that are less than or equal to x0.
0
)()( 0xx
XPxF
Derived Properties of Cumulative Derived Properties of Cumulative Probability Functions for Discrete Probability Functions for Discrete
Random VariablesRandom Variables
Let X be a discrete random variable with a cumulative probability function, F(x0). Then we can show that
i. 0 F(x0) 1 for every number x0
ii. If x0 and x1 are two numbers with x0 < x1, then F(x0) F(x1)
Expected ValueExpected Value
The expected value, E(X),expected value, E(X), of a discrete random variable X is defined
Where the notation indicates that summation extends over all possible values x.
The expected value of a random variable is called its meanmean and is denoted xx.
x
xxPXE )()(
Expected Value: Functions of Expected Value: Functions of Random VariablesRandom Variables
Let X be a discrete random variable with probability function P(x) and let g(X) be some
function of X. Then the expected value, E[g(X)], of that function is defined as
x
xPxgXgE )()()]([
Variance and Standard DeviationVariance and Standard Deviation
Let X be a discrete random variable. The expectation of the squared discrepancies about the mean, (X - )2,
is called the variancevariance, denoted 2x and is given by
The standard deviationstandard deviation, x , is the positive square root of the variance.
x
xxx xPxXE )()()( 222
VarianceVariance(Alternative Formula)(Alternative Formula)
22
222
)(
)(
xx
xx
xPx
XE
The variance of a discrete random variable X can be Expressed as
Expected Value and Variance for Expected Value and Variance for Discrete Random Variable Using Discrete Random Variable Using
Microsoft ExcelMicrosoft Excel
Sales P(x) Mean Variance0 0.15 0 0.5703751 0.3 0.3 0.270752 0.2 0.4 0.00053 0.2 0.6 0.22054 0.1 0.4 0.420255 0.05 0.25 0.465125
1.95 1.9475
Expected Value = 1.95 Variance = 1.9475
Summary of Properties for Linear Summary of Properties for Linear Function of a Random VariableFunction of a Random Variable
Let X be a random variable with mean x , and variance 2x
; and let a and b be any constant fixed numbers. Define the random variable Y = a + bX. Then, the meanmean and variancevariance
of Y are
and
so that the standard deviation of Ystandard deviation of Y is
XY babXaE )(
XY bbXaVar 222 )(
XY b
Summary Results for the Mean and Summary Results for the Mean and Variance of Special Linear FunctionsVariance of Special Linear Functions
a) Let b = 0 in the linear function, W = a + bX. Then W = a (for any constant a).
If a random variable always takes the value a, it will have a mean a and a variance 0.
b) Let a = 0 in the linear function, W = a + bX. Then W = bX.
0)()( aVarandaaE
22)()( XX baVarandbbXE
Mean and Variance of ZMean and Variance of Z
Let a = -X/X and b = 1/ X in the linear function Z = a + bX. Then,
so that
and
X
XXbXaZ
01
X
XX
X
X
XXE
11 2
2
X
XX
XXVar
Bernoulli DistributionBernoulli Distribution
A Bernoulli distributionBernoulli distribution arises from a random experiment which can give rise to just two possible outcomes. These
outcomes are usually labeled as either “success” or “failure.” If denotes the probability of a success and the
probability of a failure is (1 - ), the the Bernoulli probability function is
)1()1()0( PandP
Mean and Variance of a Bernoulli Mean and Variance of a Bernoulli Random VariableRandom Variable
The meanmean is:
And the variancevariance is:
)1()1)(0()()( xPxXEX
X
)1()1()1()0(
)()(])[(
22
222
X
XXX xPxXE
Sequences of Sequences of xx Successes in Successes in nn TrialsTrials
The number of sequences with x successes in n independent number of sequences with x successes in n independent trialstrials is:
Where n! = n x (x – 1) x (n – 2) x . . . x 1 and 0! = 1.
)!(!
!
xnx
nC n
x
time. samethe at occur can them of two no since
exclusive,mutually are sequencesC These nx
Binomial DistributionBinomial DistributionSuppose that a random experiment can result in two possible mutually exclusive and collectively exhaustive outcomes, “success” and “failure,” and that is the probability of a success resulting in a single trial. If n
independent trials are carried out, the distribution of the resulting number of successes “x” is called the binomial distributionbinomial distribution. Its
probability distribution function for the binomial random variable X = x is:
P(x successes in n independent trials)=
for x = 0, 1, 2 . . . , n
)()1()!(!
!)( xnx
xnx
nxP
Mean and Variance of a Binomial Mean and Variance of a Binomial Probability DistributionProbability Distribution
Let X be the number of successes in n independent trials, each with probability of success . The x follows a
binomial distribution with meanmean,
and variancevariance,
nXEX )(
)1(])[( 22 nXEX
Binomial ProbabilitiesBinomial Probabilities- An Example –- An Example –
An insurance broker, has five contracts, and he believes that for each contract, the probability of making a sale is 0.40.
What is the probability that he makes at most one sale?
P(at most one sale) = P(X 1) = P(X = 0) + P(X = 1)
= 0.078 + 0.259 = 0.337
259.0)6.0()4.0(1!4!
5! P(1) sale) P(1
0.078 )6.0()4.0(0!5!
5! P(0) sales) P(no
41
50
Binomial Probabilities, n = 100, Binomial Probabilities, n = 100, =0.40 =0.40
Sample size 100Probability of success 0.4Mean 40Variance 24Standard deviation 4.898979
Binomial Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)
36 0.059141 0.238611 0.179469 0.761389 0.82053137 0.068199 0.30681 0.238611 0.69319 0.76138938 0.075378 0.382188 0.30681 0.617812 0.6931939 0.079888 0.462075 0.382188 0.537925 0.61781240 0.081219 0.543294 0.462075 0.456706 0.53792541 0.079238 0.622533 0.543294 0.377467 0.45670642 0.074207 0.69674 0.622533 0.30326 0.37746743 0.066729 0.763469 0.69674 0.236531 0.30326
Poisson Probability DistributionPoisson Probability Distribution
Assume that an interval is divided into a very large number of subintervals so that the probability of the occurrence of an event in any subinterval is very small. The assumptions of a Poisson probability distributionPoisson probability distribution are:
1) The probability of an occurrence of an event is constant for all subintervals.
2) There can be no more than one occurrence in each subinterval.
3) Occurrences are independent; that is, the number of occurrences in any non-overlapping intervals in independent of one another.
Poisson Probability DistributionPoisson Probability Distribution
The random variable X is said to follow the Poisson probability distribution if it has the probability function:
where
P(x) = the probability of x successes over a given period of time or space, given
= the expected number of successes per time or space unit; > 0
e = 2.71828 (the base for natural logarithms)
The mean and variance of the Poisson probability distribution aremean and variance of the Poisson probability distribution are:
1,2,... 0,xfor,!
)(
x
exP
x
])[()( 22 XEandXE xx
Partial Poisson Probabilities for Partial Poisson Probabilities for = 0.03 = 0.03 Obtained Using Microsoft Excel Obtained Using Microsoft Excel
Poisson Probabilities TableX P(X) P(<=X) P(<X) P(>X) P(>=X)0 0.970446 0.970446 0.000000 0.029554 1.0000001 0.029113 0.999559 0.970446 0.000441 0.0295542 0.000437 0.999996 0.999559 0.000004 0.0004413 0.000004 1.000000 0.999996 0.000000 0.0000044 0.000000 1.000000 1.000000 0.000000 0.000000
Poisson Approximation to the Poisson Approximation to the Binomial DistributionBinomial Distribution
Let X be the number of successes resulting from n independent trials, each with a probability of success, . The distribution of the number of successes X is binomial, with mean n. If the number of trials n is large and n is of only moderate size (preferably n 7), this distribution can be approximated by the Poisson distribution
with = n. The probability function of the approximating distribution is then:
1,2,... 0,xfor,!
)()(
x
nexP
xn
CovarianceCovariance
Let X be a random variable with mean X , and let Y be a random variable with mean, Y . The expected value of (X -
X )(Y - Y ) is called the covariance covariance between X and Y, denoted Cov(X, Y).
For discrete random variables
An equivalent expression is
x y
yxYX yxPyxYXEYXCov ),())(()])([(),(
x y
yxyx yxxyPXYEYXCov ),()(),(
CorrelationCorrelation
Let X and Y be jointly distributed random variables. The correlation between X and Y is:
YX
YXCovYXCorr
),(
),(
Covariance and Statistical Covariance and Statistical IndependenceIndependence
If two random variables are statistically statistically independentindependent, the covariance between them is 0. However, the converse is not necessarily true.
Portfolio AnalysisPortfolio Analysis
The random variable X is the price for stock A and the random variable Y is the price for stock B. The market value, W, for the portfolio is given by the linear function,
Where, a, is the number of shares of stock A and, b, is the number of shares of stock B.
bYaXW
Portfolio AnalysisPortfolio Analysis
The mean value for Wmean value for W is,
The variance for Wvariance for W is,
or using the correlation,
YX
W
ba
bYaXEWE
][][
),(222222 YXabCovba YXW
YXYXW YXabCorrba ),(222222
Continuous Random Variables Continuous Random Variables and Probability Distributionsand Probability Distributions
©
Continuous Random VariablesContinuous Random Variables
A random variable is continuous continuous if it can take any value in an interval.
Cumulative Distribution FunctionCumulative Distribution Function
The cumulative distribution functioncumulative distribution function, F(x), for a continuous random variable X expresses the
probability that X does not exceed the value of x, as a function of x
)()( xXPxF
Cumulative Distribution FunctionCumulative Distribution Function
0 1
1
F(x)
Cumulative Distribution Function for a Random variable Over 0 to 1
Cumulative Distribution FunctionCumulative Distribution Function
Let X be a continuous random variable with a cumulative distribution function F(x), and let a and
b be two possible values of X, with a < b. The probability that X lies between a and bprobability that X lies between a and b is
)()()( aFbFbXaP
Probability Density FunctionProbability Density Function
Let X be a continuous random variable, and let x be any number lying in the range of values this random variable can take. The probability density probability density functionfunction, f(x), of the random variable is a function with the following properties:
1. f(x) > 0 for all values of x2. The area under the probability density function f(x) over all values of the
random variable X is equal to 1.03. Suppose this density function is graphed. Let a and b be two possible
values of the random variable X, with a<b. Then the probability that X lies between a and b is the area under the density function between these points.
4. The cumulative density function F(x0) is the area under the probability density function f(x) up to x0
where xm is the minimum value of the random variable x.
0
)()( 0
x
xm
dxxfxf
Shaded Area is the Probability That Shaded Area is the Probability That X is Between a and bX is Between a and b
x ba
Probability Density Function for a Probability Density Function for a Uniform 0 to 1 Random VariableUniform 0 to 1 Random Variable
0 1
1
x
f(x)
Areas Under Continuous Probability Areas Under Continuous Probability Density FunctionsDensity Functions
Let X be a continuous random variable with the probability density function f(x) and cumulative distribution F(x). Then the following properties hold:
1. The total area under the curve f(x) = 1.
2. The area under the curve f(x) to the left of x0 is F(x0), where x0 is any value that the random variable can take.
Properties of the Probability Density Properties of the Probability Density FunctionFunction
0 1 xx0
f(x)
0
1Comments
Total area under the uniform probability density function is 1.
Properties of the Probability Density Properties of the Probability Density FunctionFunction
0 1 xx0
f(x)
0
1
Comments
Area under the uniform probability density function to the left of x0 is F(x0), which is equal to x0 for this uniform distribution because f(x)=1.
Rationale for Expectations of Rationale for Expectations of Continuous Random VariablesContinuous Random Variables
Suppose that a random experiment leads to an outcome that can be represented by a continuous
random variable. If N independent replications of this experiment are carried out, then the expected expected valuevalue of the random variable is the average of the
values taken, as the number of replications becomes infinitely large. The expected value of a random
variable is denoted by E(X).E(X).
Rationale for Expectations of Rationale for Expectations of Continuous Random VariablesContinuous Random Variables
(continued)(continued)Similarly, if g(x) is any function of the random variable, X, then the expected value of this function is the average value taken by the function over repeated
independent trials, as the number of trials becomes infinitely large. This expectation is denoted E[g(X)]. By using calculus we can define expected values for
continuous random variables similarly to that used for discrete random variables.
x
dxxfxgxgE )()()]([
Mean, Variance, and Standard Mean, Variance, and Standard DeviationDeviation
Let X be a continuous random variable. There are two important expected values that are used routinely to define continuous probability distributions.
i. The mean of Xmean of X, denoted by X, is defined as the expected value of X.
ii. The variance of Xvariance of X, denoted by X2, is defined as the expectation of the
squared deviation, (X - X)2, of a random variable from its mean
Or an alternative expression can be derived
iii. The standard deviation of Xstandard deviation of X, X, is the square root of the variance.
)(XEX
])[( 22XX XE
222 )( XX XE
Linear Functions of VariablesLinear Functions of Variables
Let X be a continuous random variable with mean X and variance X
2, and let a and b any constant fixed numbers. Define the random variable W as
Then the mean and variance of W are
and
and the standard deviation of W is
bXaW
XW babXaE )(
222 )( XW bbXaVar
XW b
Linear Functions of VariableLinear Functions of Variable(continued)(continued)
An important special case of the previous results is the standardized random variable
which has a mean 0 and variance 1.
X
XXZ
Reasons for Using the Normal Reasons for Using the Normal DistributionDistribution
1. The normal distribution closely approximates the probability distributions of a wide range of random variables.
2. Distributions of sample means approach a normal distribution given a “large” sample size.
3. Computations of probabilities are direct and elegant.
4. The normal probability distribution has led to good business decisions for a number of applications.
Probability Density Function for a Probability Density Function for a Normal DistributionNormal Distribution
(Figure 6.8)(Figure 6.8)
x0.0
0.1
0.2
0.3
0.4
Probability Density Function of Probability Density Function of the Normal Distributionthe Normal Distribution
The probability density function for a normally probability density function for a normally distributed random variable Xdistributed random variable X is
Where and 2 are any number such that - < < and - < 2 < and where e and are physical
constants, e = 2.71828. . . and = 3.14159. . .
xexf x -for 2
1)(
22 2/)(
2
Properties of the Normal Properties of the Normal DistributionDistribution
Suppose that the random variable X follows a normal distribution with parameters and 2. Then the following properties hold:
i. The mean of the random variable is ,
ii. The variance of the random variable is 2,
iii. The shape of the probability density function is a symmetric bell-shaped curve centered on the mean as shown in Figure 6.8.
iii. By knowing the mean and variance we can define the normal distribution by using the notation
)(XE
22 ])[( XXE
),(~ 2NX
Effects of Effects of on the Probability Density on the Probability Density Function of a Normal Random VariableFunction of a Normal Random Variable
x
0.0
0.1
0.2
0.3
0.4
1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5
Mean = 5 Mean = 6
Effects of Effects of 22 on the Probability Density on the Probability Density Function of a Normal Random VariableFunction of a Normal Random Variable
x
0.0
0.1
0.2
0.3
0.4
1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5
Variance = 0.0625
Variance = 1
Cumulative Distribution Function Cumulative Distribution Function of the Normal Distributionof the Normal Distribution
Suppose that X is a normal random variable with mean and variance 2 ; that is X~N(, 2). Then the
cumulative distribution functioncumulative distribution function is
This is the area under the normal probability density function to the left of x0, as illustrated in Figure 6.10. As for any proper density function, the total area under the
curve is 1; that is F() = 1.
)()( 00 xXPxF
Shaded Area is the Probability that X Shaded Area is the Probability that X does not Exceed xdoes not Exceed x00 for a Normal for a Normal
Random VariableRandom Variable
xx0
f(x)
Range Probabilities for Normal Range Probabilities for Normal Random VariablesRandom Variables
Let X be a normal random variable with cumulative distribution function F(x), and let a and b be two
possible values of X, with a < b. Then
The probability is the area under the corresponding probability density function between a and b.
)()()( aFbFbXaP
Range Probabilities for Normal Range Probabilities for Normal Random VariablesRandom Variables
xb
f(x)
a
The Standard Normal DistributionThe Standard Normal Distribution
Let Z be a normal random variable with mean 0 and variance 1; that is
We say that Z follows the standard normal distribution. Denote the cumulative distribution function as F(z), and a
and b as two numbers with a < b, then
)1,0(~ NZ
)()()( aFbFbZaP
Standard Normal Distribution with Standard Normal Distribution with Probability for z = 1.25Probability for z = 1.25
z 1.25
0.8944
Finding Range Probabilities for Normally Finding Range Probabilities for Normally Distributed Random VariablesDistributed Random Variables
Let X be a normally distributed random variable with mean and variance 2. Then the random variable Z = (X - )/ has a
standard normal distribution: Z ~ N(0, 1)
It follows that if a and b are any numbers with a < b, then
where Z is the standard normal random variable and F(z) denotes its cumulative distribution function.
aF
bF
bZ
aPbXaP )(
Computing Normal ProbabilitiesComputing Normal Probabilities
A very large group of students obtains test scores that are normally distributed with mean 60 and standard deviation 15. What proportion of the students obtained scores between 85
and 95?
0376.09525.09901.0
)67.1()33.2(
)33.267.1(
15
6095
15
6085)9585(
FF
ZP
ZPXP
That is, 3.76% of the students obtained scores in the range 85 to 95.
Approximating Binomial Probabilities Approximating Binomial Probabilities Using the Normal DistributionUsing the Normal Distribution
Let X be the number of successes from n independent Bernoulli trials, each with probability of success . The number of successes,
X, is a Binomial random variable and if n(1 - ) > 9 a good approximation is
Or if 5 < n(1 - ) < 9 we can use the continuity correction factor to obtain
where Z is a standard normal variable.
)1()1()(
n
nbZ
n
naPbXaP
)1(
5.0
)1(
5.0)(
n
nbZ
n
naPbXaP
CovarianceCovariance
Let X and Y be a pair of continuous random variables, with respective means x and y. The expected value of (x - x)(Y - y) is called the covariancecovariance between X and Y. That
is
An alternative but equivalent expression can be derived as
If the random variables X and Y are independent, then the covariance between them is 0. However, the converse is
not true.
)])([(),( yx YXEYXCov
yxXYEYXCov )(),(
CorrelationCorrelation
Let X and Y be jointly distributed random variables. The correlationcorrelation between X and Y is
YX
YXCovYXCorr
),(
),(
Sums of Random VariablesSums of Random Variables
Let X1, X2, . . .Xk be k random variables with means 1, 2,. . . k and variances 1
2, 22,. . ., k
2. The following properties hold:
i. The mean of their sum is the sum of their means; that is
ii. If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances; that is
However, if the covariances between pairs of random variables are not 0, the variance of their sum is
kkXXXE 2121 )(
222
2121 )( kkXXXVar
),(2)(1
1 1
222
2121 j
K
i
K
ijikk XXCovXXXVar
Differences Between a Pair of Differences Between a Pair of Random VariablesRandom Variables
Let X and Y be a pair of random variables with means X and Y and variances X
2 and Y2. The following properties hold:
i. The mean of their difference is the difference of their means; that is
ii. If the covariance between X and Y is 0, then the variance of their difference is
iii. If the covariance between X and Y is not 0, then the variance of their difference is
YXYXE )(
22)( YXYXVar
),(2)( 22 YXCovYXVar YX
Linear Combinations of Random Linear Combinations of Random VariablesVariables
The linear combination of two random variables, X and Y, is
Where a and b are constant numbers.
The mean for W is,
The variance for W is,
Or using the correlation,
If both X and Y are joint normally distributed random variables then the resulting random variable, W, is also normally distributed
with mean and variance derived above.
bYaXW
YXW babYaXEWE ][][
),(222222 YXabCovba YXW
YXYXW YXabCorrba ),(222222