Lc05 SL Some Continuous Probability Distributions.pdf

8/10/2019 Lc05 SL Some Continuous Probability Distributions.pdf

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Some Continuous Probability Distributions

I Wayan Mustika, Ph.D.Hanung Adi Nugroho, Ph.D.


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Outline

! Continuous Uniform Distribution! Normal Distribution! Areas Under the Normal Curve! Applications of the Normal Distributions! Gamma and Exponential Distributions! Chi-Squared Distribution! Lognormal Distribution! Weibull Distribution


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Continuous Uniform Distribution

" The density function of the continuous uniform randomvariable X on the interval [A, B] is

f (x ; A, B ) = 1B A A x B0, elsewhere


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" The mean and variance of the uniform distribution are

=A + B

2

2

= (B A )2

12


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Normal Distribution " The most important continuous probability distribution inthe entire field of statistics

" Its graph, called the normal curve, is the bell- shapedcurve

" The normal distribution is often referred to as theGaussian distribution , in honor of Karl Friedrich Gauss(1777-1855), who also derived its equation from a study

of errors in repeated measurements of the same quantity


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" The density of the normal random variable X,with mean and variance ! 2, is

where ! = 3.14159 and e = 2.71828

n (x ; , ) = 1 2 e [( x ) / ]

22 < x <


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Various Normal Curves


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Areas Under the Normal Curve

" Fortunately, we are able to transform all the observationsof any normal random variable X to a new set ofobservations of a normal random variable Z with mean 0and variance 1. This can be done by means of thetransformation

P (x 1 < X < x 2 ) = ?

Z =X


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where Z is seen to be a normal random variable withmean 0 and variance 1

P (x 1 < X < x 2 ) = 1 2 Z

x2

x 1e

[( x ) / ]22

= 1 2 Z z2

z1e z 2 / 2 dz

= Z z2

z1n (z ; 0, 1)dz

= P (z 1 < Z < z 2 )


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Example 1 Given a standard normal distribution, find the area

under the curve that lies(a) to the right of z = 1.84, and

(b) between z = " 1.97 and z = 0.86.Solution

(a) The area to the right of z = 1.84 is equal to 1 minusthe area in Table Z to the left of Z = 1.84, namely,1" 0.9671 = 0.0329.

(b) The area between z = " 1.97 and z = 0.86 is equal tothe area to the left of z = 0.86 minus the area to theleft of z = " 1.97. From Table Z we find the desiredarea to be 0.8051 " 0.0244 = 0.7807.


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Z Table


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Example 2

" Given a random variable X having a normaldistribution with = 50 and ! = 10, find theprobability that X assumes a value between 45and 62.


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Solution

The z values corresponding to x 1 = 45 and x 2 = 62 are

Therefore, P(45 < X < 62) = P( " 0.5 < Z < 1.2). This area may be found by subtracting the area to the left ofthe ordinate z = " 0.5 from the entire area to the left of z = 1.2.Using Z Table, we have

z 1 = 45 50

10 = 0. 5 and z 2 =

62 5010 = 1

. 2

P (45 < X < 62) = P ( 0.5 < Z < 1.2)= P (Z < 1.2) P (Z < 0.5)= 0 .8849 0.3085 = 0 .5764


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Applications of the NormalDistribution

" A certain type of storage battery lasts, on

average, 3.0 years with a standard deviation of

0.5 year. Assuming that the battery lives arenormally distributed, find the probability that agiven battery will last less than 2.3 years.


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" Solution

" First construct a diagram such as above, showing the givendistribution of battery lives and the desired area. To find theP(X < 2.3), we need to evaluate the area under the normalcurve to the left of 2.3. This is accomplished by finding thearea to the left of the corresponding z value. Hence we findthat

Z = (2.3 " 3)/0.5 = " 1.4Using Z table, we have

P(X < 2.3) = P(Z < " 1.4) = 0.0808


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Gamma and Exponential

Distribution " The continuous random variable X has a gammadistribution, with parameters # and $, if its densityfunction is given by

f (x ) = 1

( ) x 1

e x/

x > 00, elsewherewhere # > 0 and $ > 0


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" The continuous random variable X has an exponential

distribution, with parameter $, if its density function isgiven by

f (x ) = 1

e x/ x > 0

0, elsewhere

where $ > 0


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Chi-Squared Distribution

" A special case of the gamma distribution which isobtained by letting # = v /2 and $ = 2, where v is apositive integer.

" The distribution has a single parameter, v, called thedegrees of freedom .

" The continuous random variable X has a chi-squared

distribution, with v degrees of freedom, if its densityfunction is given by

f (x ) = 1

2v/ 2 (v/ 2) xv/ 2 1e x/ 2 x > 0

0, elsewhere

where v is a positive integer


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Lognormal Distribution

" The lognormal distribution applies in cases where anatural log transformation results in a normaldistribution.

" The continuous random variable X has a lognormaldistribution if the random variable Y = ln(X) has anormal distribution with mean and standard deviation! . The resulting density function of X is

f (x ) = ( 1 2 x e [ln( x ) ]2 / (2 2 ) x 0

0, x < 0


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Example

" Concentration of pollutants produced by chemical plantshistorically are known to exhibit behavior that resemblesa lognormal distribution. This is important when oneconsiders issues regarding compliance to governmentregulations. Suppose it is assumed that theconcentration of a certain pollutant, in parts per million,has a lognormal distribution with parameters = 3.2and ! = 1. What is the probability that the concentrationexceeds 8 parts per million?


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Example

SolutionLet the random variable X be pollutant concentration

P(X > 8) = 1 " P(X % 8).Since ln( X ) has a normal distribution with mean = 3.2and standard deviation ! = 1

P (X 8) = ln(8) 3.2

1 = ( 1.12) = 0 .1314


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Weibull Distribution

" The continuous random variable X has a Weibulldistribution, with parameters # and $ if its densityfunction is given by

f (x ) = x 1 e x

x > 0

0, elsewherewhere # > 0 and $ > 0

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Lc05 SL Some Continuous Probability Distributions.pdf