Engineering Probability and Statistics - SE-205 -Chap 4

By

S. O. Duffuaa

Introduction to Probability Density Function

x

y

)()1( xF

Density function of loading on a long, thin beam

x

Loa

ding

Introduction to Probability Density Function

x

y

)()1( xF

Density function of loading on a long, thin beam

x

f(x)

a b

P(a < X < b)

Probability Density Function

For a continuous random variable X, a probability density function is a function such that

b

a

bandaanyforbtoafromxfunderareadxxfbXaP

dxxf

xf

)()()()3(

1)()2(

0)()1(

Probability for Continuous Random Variable

If X is a continuous variable, then for any x1 and x2,

)()()()21( 212121 xXxPxXxPxXxPxXxP

ExampleLet the continuous random variable X denote the diameter of a hole drilled in a sheet metal component. The target diameter is 12.5 millimeters. Most random disturbances to the process result in larger diameters. Historical data show that the distribution of X can be modified by a probability density function f(x) = 20e-20(x-12.5) , x 12.5.If a part with a diameter larger than 12.60 millimeters is scrapped, what proportion of parts is scrapped ? A part is scrapped if X 12.60. Now,

6.12 6.12

6.12)5.12(20)5.12(20 135.0|20)()60.12( xx edxedxxfXP

What proportion of parts is between 12.5 and 12.6 millimeters ? Now,

6.12

5.12

6.125.12

)5.12(20 865.0|)()6.125.12( xedxxfXP

Because the total area under f(x) equals one, we can also calculate P(12.5<X<12.6) = 1 – P(X>12.6) = 1 – 0.135 = 0.865

Cumulative Distribution Function

The cumulative distribution function of a continuous random variable X is

x

xforduufxXPxF )()()(

Example for Cumulative Distribution Function

For the copper current measurement in Example 5-1, the cumulative distribution function of the random variable X consists of three expressions. If x < 0, then f(x) = 0. Therefore,

F(x) = 0, for x < 0

x

xforxduufxF0

200,05.0)()(

Finally,

x

xforduufxF0

20,1)()(

Therefore,

x

xx

x

xF

201

20005.0

00

)(

The plot of F(x) is shown in Fig. 5-6

Mean and Variance for Continuous Random Variable

Suppose X is a continuous random variable with probability density function f(x). The mean or expected value of X, denoted as or E(X), is

dxxxfXE )()(

The variance of X, denoted as V(X) or 2, is

2222 )()()()( dxxfxdxxfxXV

The standard deviation of X is = [V(X)]1/2

Uniform DistributionA continuous random variable X with probability density function

bxaabxf ,)/(1)(

has a continuous uniform distribution

Uniform DistributionThe mean and variance of a continuous uniform random variable X over a x b are

12/)()(2/)()( 22 abXVandbaXE

Applications:

• Generating random sample

• Generating random variable

Normal DistributionA random variable X with probability density function

xforexfx

2

2

2

)(

2

1)(

has a normal distribution with parameters , where - < < , and > 0. Also,

2)()( XVandXE

Normal Distribution

9973.0)33(

9545.0)22(

6827.0)(

XP

XP

XP

68%

- 3 - 2 - - - 2 - 3 x

95%

99.7%

f(x)

Probabilities associated with normal distribution

Standard NormalA normal random variable with = 0 and 2 = 1 is called a standard normal random variable. A standard normal random variable is denoted as Z.

The cumulative distribution function of a standard normal random variable is denoted as

)()( zZPz

StandardizationIf X is a normal random variable with E(X) = and V(X) = 2, then the random variable

X

Z

is a normal random variable with E(Z) = 0 and V(Z) = 1. That is , Z is a standard normal random variable.

StandardizationSuppose X is a normal random variable with mean and variance 2 . Then,

)()( zZPxX

PxXP

where, Z is a standard normal random variable, andz = (x - )/ is the z-value obtained by standardizing X.

The probability is obtained by entering Appendix Table II with z = (x - )/.

Applications:

• Modeling errors

• Modeling grades

• Modeling averages

Binomial ApproximationIf X is a binomial random variable, then

)1( pnp

npXZ

is approximately a standard normal random variable. The approximation is good for

np > 5 and n(1-p) > 5

Poisson ApproximationIf X is a Poisson random variable with E(X) = and V(X) = , then

X

Z

is approximately a standard normal random variable. The approximation is good for

> 5

Do not forget correction for continuity

Exponential DistributionThe random variable X that equals the distance between successive counts of a Poisson process with mean > 0 has an exponential distribution with parameter . The probability density function of X is

xforexf x 0,)(

If the random variable X has an exponential distribution with parameter , then

E(X) = 1/ and V(X) = 1/ 2

Lack of Memory Property

For an exponential random variable X,

)()|( 2121 tXPtXttXP

Applications:

• Models random time between failures

• Models inter-arrival times between customers

Erlang DistributionThe random variable X that equals the interval length until r failures occur in a Poisson process with mean > 0 has an Erlang distribution with parameters and r. The probability density function of X is

...,2,10,)!1(

)(1

randxforr

exxf

xrr

Erlang DistributionIf X is an Erlang random variable with parameters and r, then the mean and variance of X are

= E(X) = r/ and 2 = V(X) = r/ 2

Applications:

• Models natural phenomena such as rainfall.

• Time to complete a task

Gamma FunctionThe gamma function is

0

1 0,)( rfordxexr xr

Gamma DistributionThe random variable X with probability density function

0,)(

)(1

xforr

exxf

xrr

has a gamma distribution with parameters > 0 and r > 0. If r is an integer, then X has an Erlang distribution.

Gamma DistributionIf X is a gamma random variable with parameters and r, then the mean and variance of X are

= E(X) = r/ and 2 = V(X) = r/ 2

Applications:• Models natural phenomena such as rainfall.• Time to complete a task

Weibull DistributionThe random variable X with probability density function

0,)( )/(1

xforex

xf x

has a Weibull distribution with scale parameters > 0 and shape parameter > 0

Applications:

• Time to failure for mechanical systems

• Time to complete a task.

Weibull DistributionIf X has a Weibull distribution with parameters and , then the cumulative distribution function of X is

x

exF 1)(

If X has a Weibull distribution with parameters and , then the mean and variance of x are

1

1

and2

222 11

21