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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 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.