chapter four continuous random variables & probability distributions

76
Chapter Four Continuous Random Variables & Probability Distributions

Upload: galvin

Post on 25-Feb-2016

76 views

Category:

Documents


0 download

DESCRIPTION

Chapter Four Continuous Random Variables & Probability Distributions. Continuous Probability Distribution (pdf) Definition: b P(a  X  b) =  f(x) dx a For continuous RV X & a  b. Two Conditions for a pdf: 1) f(x)  0 for all x + 2)  f(x) dx = 1 - . - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter Four Continuous Random Variables & Probability Distributions

Chapter FourContinuous Random

Variables & Probability

Distributions

Page 2: Chapter Four Continuous Random Variables & Probability Distributions

Continuous Probability Distribution (pdf) Definition: b

P(a X b) = f(x)dx a

For continuous RV X & a b.Two Conditions for a pdf:

1) f(x) 0 for all x +

2) f(x)dx = 1 -

Page 3: Chapter Four Continuous Random Variables & Probability Distributions
Page 4: Chapter Four Continuous Random Variables & Probability Distributions

Cumulative Distribution FunctionDefinition: a

F(a) = P(X a) = f(x)dx -

For continuous RV X.

Page 5: Chapter Four Continuous Random Variables & Probability Distributions

PDF ExampleGiven pdf: f(x) = cx2, for 0 x 2 f(x) = 0, elsewhere.

Find the value of c for which f(x) will be a valid pdf.

Find the probability that 1 x 2.

Page 6: Chapter Four Continuous Random Variables & Probability Distributions

Cumulative Distribution Function

Plot the CDF for RV X:

F(x) = 0 for x < 0F(x) = x for 0 x 1F(x) = 1 for x > 1.

Page 7: Chapter Four Continuous Random Variables & Probability Distributions

Expected Values for Continuous RV + E(X) = x = x f(x)dx - Variance

+ V(X) = 2

x = (x-)2 f(x)dx=E[(X-)2]-

Shortcut: V(X) = E(X2) – [E(X)]2

Page 8: Chapter Four Continuous Random Variables & Probability Distributions

Expected Value of RV X ExampleFor a RV X, the pdf is given by:

f(x) = 0 for x < 50f(x) = 1/20 for 50 x 70f(x) = 0 for x > 70.

What is the expected value & variance of RV X?

Page 9: Chapter Four Continuous Random Variables & Probability Distributions

Expected Value & Variance ExampleAn electronic component is tested to failure. Let continuous RV X measure the time to failure for this component. The pdf for this electronic component is shown to be in hours by the following:

f(t) = 0 for t < 0f(t) = e-t/1,000 for t 0.

1,000What is the expected value & variance of RV X?

Page 10: Chapter Four Continuous Random Variables & Probability Distributions

Continuous Probability DistributionsUniformNormalGammaExponentialChi-SquaredWeibullLognormalBeta

Page 11: Chapter Four Continuous Random Variables & Probability Distributions

Uniform Distribution

f(x; a, b) = 1 a x b b-a

Page 12: Chapter Four Continuous Random Variables & Probability Distributions

Uniform pdf ExampleA telephone call is equally likely to occur any time between Noon &

1 PM. Let X be a RV that is 0 at Noon & is the time of the call in fractions of a hour past Noon for any other outcome.Find P(0 X 1.0).Find P(X = 0.2).Find P(X 0.5).Find P(0.2 < X < 0.8).

Page 13: Chapter Four Continuous Random Variables & Probability Distributions

Uniform Probability DistributionResistors are produced that have a nominal value of 10 ohms & are +/- 10% resistors where any possible value of resistance is equally likely.Find the pdf & cdf of the RV X, which represents resistance.Find the probability that a resistor, selected at random, is between 9.5 & 10.5 ohms.

Page 14: Chapter Four Continuous Random Variables & Probability Distributions
Page 15: Chapter Four Continuous Random Variables & Probability Distributions

Uniform pdfThe time X in hours for an engineering student to complete a homework assignment in ISE 261 that receives a grade above 85% is uniformly distributed with a minimum effort of 3 hours & a maximum effort of 10 hours. What is the probability that preparation time exceeds 5 hours?What is the probability that the preparation time is within 2 hours of the mean time?

Page 16: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution 2 2

f(x;,) = e-(x-) / 2

(2) -< x <+

Abbreviated: X N(,2)

Page 17: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution Interval b 2 2

P(a<x<b) = e-(x-) / 2 / (2) dx a

Page 18: Chapter Four Continuous Random Variables & Probability Distributions

Properties of Normal pdf’sBell ShapedSymmetrical about the MeanUnimodalMean = Median = ModeOne Standard Deviation is at the inflection points on the curveThe Standard Deviation defines the area under the pdf

Page 19: Chapter Four Continuous Random Variables & Probability Distributions
Page 20: Chapter Four Continuous Random Variables & Probability Distributions

Standard Normal Distribution 2 f(z;0,1) = e-z / 2

(2) -< z <+

Denote: N(0,1) by Z CDF(Z) by (z)

Page 21: Chapter Four Continuous Random Variables & Probability Distributions

Z TransformationFor RV X with N(,2):

Z = X -

has a standard Normal pdfThusP(aXb)=P[ a- Z b- ]

Page 22: Chapter Four Continuous Random Variables & Probability Distributions
Page 23: Chapter Four Continuous Random Variables & Probability Distributions
Page 24: Chapter Four Continuous Random Variables & Probability Distributions

Transformation Example

Find the probability that RV X is greater than 5 if X is N(3,4).

Page 25: Chapter Four Continuous Random Variables & Probability Distributions
Page 26: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution ExampleThe time to wear out of a cutting tool edge is distributed normally with = 2.8 hours & = .60 hour. What is the probability that the tool will wear out in less than 1.5 hour?How often should the cutting edges be replaced to keep the failure rate less than 10% of the tool’s?

Page 27: Chapter Four Continuous Random Variables & Probability Distributions

Normal DistributionYou are an engineering manager of a PCB operation where the mean thickness of a board should be 65mm with a SD of 2mm. Previous studies indicate that this operation can be assumed to be normally distributed.One of the boards inspected has just produced a thickness of 68mm. What is the probability that a board greater than this thickness can be produced from this process?

Page 28: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution ExampleThe achievement scores for a college entrance exam are normally distributed with a mean equal to 75 & standard deviation equal to 10. What fraction of the scores are expected to lie between 70 & 90?

Page 29: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution ExampleThe width of a slot on a duralumin forging is normally distributed with mean equal to 0.90 inch & SD equal to 0.002 inch. The specification limits are given as 0.90 +/- 0.005 inch. What percentage of forgings will be defective?

Page 30: Chapter Four Continuous Random Variables & Probability Distributions

Normal Distribution ExampleA process manufactures ball bearings whose diameters (in cm) are N(2.505, 0.0082). Specifications call for the diameter to be in the interval 2.5 +/- 0.01 cm. What proportion of the ball bearings will meet the specification?

Page 31: Chapter Four Continuous Random Variables & Probability Distributions

Normal DistributionA remote computer sends a 1 or 0 by sending either p(t) or –p(t), respectively… where the pulse p(t) is given by:

p(t) = 1, 0 t T 0, otherwise

When the pulse is received it is corrupted by Gaussian noise which is added to the signal. Assume that this noise has zero mean and variance of 1. Suppose that you wish to measure a single value within a received pulse and use that to determine if a 1 or 0 was sent. Then the value that you measure would be a Gaussian RV Z modeled as follows:

Z = ± 1 + VWhere V is a Gaussian RV with zero mean and variance of 1. You decide that a 1 was sent if the measured Z is such that Z > 0 and otherwise will decide that a 0 was sent. Consider the case where a 1 was sent (thus Z = +1 +V). What is the probability of making an error?

Page 32: Chapter Four Continuous Random Variables & Probability Distributions

Gamma Probability Distribution

For RV X 0:

f(x; , ) = x -1 e-x/ ()

Where > 0, > 0.

Page 33: Chapter Four Continuous Random Variables & Probability Distributions

Gamma Function

Definition: () = x-1 e-x dx

0

For > 0.

Page 34: Chapter Four Continuous Random Variables & Probability Distributions

Properties of the Gamma Function

(1) = 1(1/2) =

For any positive integer n:(n) = (n-1)!

For >1: () = (-1) (-1)

Page 35: Chapter Four Continuous Random Variables & Probability Distributions

Gamma pdf Properties Standard Gamma pdf when =1 (Incomplete Gamma function) E(X) = V(X) = 2

Exponential pdf when = 1 Chi-Square pdf when = v/2 &

Page 36: Chapter Four Continuous Random Variables & Probability Distributions
Page 37: Chapter Four Continuous Random Variables & Probability Distributions
Page 38: Chapter Four Continuous Random Variables & Probability Distributions
Page 39: Chapter Four Continuous Random Variables & Probability Distributions

Gamma Transformation

P(X x) = F(x , , ) = F(x/, )

x F(x/, ) = y -1 e-y / () dy

0 x > 0

Page 40: Chapter Four Continuous Random Variables & Probability Distributions
Page 41: Chapter Four Continuous Random Variables & Probability Distributions

Gamma pdf ExampleThe chemical reaction time (in seconds) of a commercial grade of epoxy can be represented with a standard Gamma pdf with

parameter =5.0 seconds. What is the probability that the reaction time will be more than 4 seconds?

Page 42: Chapter Four Continuous Random Variables & Probability Distributions

Gamma DistributionIn simple materials, all molecules are the same and have the same molecular weight. Because polymers contain molecules of different sizes, engineers must use probability distributions to express molecular weight.A commercial grade of polymer has a molecular weight with a standard Gamma distribution with = 2 molar mass. What is the probability that your randomly selected sample has a molar mass less than 3?

Page 43: Chapter Four Continuous Random Variables & Probability Distributions

Gamma pdf ExampleThe arrival time of large comets into our solar system has followed a Gamma distribution with parameters = 4 years & = 15.What is the expected arrival time of large comets?What is the variance in arrival time?What is the probability of a visit by a large comet between the years 2024 & 2039?

Page 44: Chapter Four Continuous Random Variables & Probability Distributions

Gamma DistributionA computer firm introduces a new home computer. Past experience indicates that the time of peak demand (in months) after its introduction, follows a Gamma pdf with Variance = 36.0.If the expected value is 18 months, find and .What is the probability that the peak demand is less than 7 months?

Page 45: Chapter Four Continuous Random Variables & Probability Distributions

Exponential Probability Distribution

f(x; ) = e-x x 0

Where > 0.

Page 46: Chapter Four Continuous Random Variables & Probability Distributions

Exponential pdf Properties

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

= 1 / pdf easily integrated: x

CDF = e-t dt = 1 – e-x

0

Page 47: Chapter Four Continuous Random Variables & Probability Distributions
Page 48: Chapter Four Continuous Random Variables & Probability Distributions

Example Exponential pdfA radioactive mass emits particles according to a Poisson process at a mean rate of 15 per minute. At some point, a clock is started. What is the probability that more than 5 seconds will elapse before the next emission?What is the mean waiting time until the next particle is emitted?

Page 49: Chapter Four Continuous Random Variables & Probability Distributions

Exponential pdf Lack of Memory PropertyThe lifetime of an electrical component has an Exponential pdf with mean = 2 years. What is the probability that the component lasts longer than 3 years?

Assume the component is now 4 years old, and is still functioning. What is the probability that it functions for more than 3 additional years?

Page 50: Chapter Four Continuous Random Variables & Probability Distributions

Exponential ExampleSome types of equipment have been observed to fail according to an Exponential pdf. The time to failure T for these types has a mean time to failure (MTTF) equal to 1/.The mean time to failure of a light bulb is 100 hours, what is the probability that a light bulb will last for more than 150 hours?

Page 51: Chapter Four Continuous Random Variables & Probability Distributions

Exponential DistributionA computer component has a constant failure rate of = .02/hour. The time to failure follows an Exponential probability distribution. What is the probability that the component will fail during the first 10 hours of operation? Suppose that the device has been successfully operated for 100 hours. What is the probability that it will fail during the next 10 hours of operation?

Page 52: Chapter Four Continuous Random Variables & Probability Distributions

Exponential pdf ExampleA National Guard unit is supplied with 20,000 rounds of ammunition for a new model rifle. After 5 years, 18,200 rounds remain unused. From these, 200 rounds are chosen randomly and test-fired. Twelve of them misfire. Assuming that the misfires are random failures of the ammunition caused by storage condition, estimate the MTTF.

Page 53: Chapter Four Continuous Random Variables & Probability Distributions

Chi-Squared Distribution

f(x; v) = 2 = x (v/2)-1 e-x/2 x 0 2v/2 (v/2)

v = number degrees of freedomProperties: Gamma pdf with = v/2 & = 2 E(2) = v V(2) = 2v

Page 54: Chapter Four Continuous Random Variables & Probability Distributions
Page 55: Chapter Four Continuous Random Variables & Probability Distributions
Page 56: Chapter Four Continuous Random Variables & Probability Distributions
Page 57: Chapter Four Continuous Random Variables & Probability Distributions

Chi-Square pdf ExampleFor a Chi-Square probability distribution with 10 degrees of freedom, what is the probability that the RV X is greater than 18.307?

Page 58: Chapter Four Continuous Random Variables & Probability Distributions

Chi-Square pdf ExampleAn unmanned space probe is directed to land on Mars. The lateral and forward distances of the point of impact from the target location are independent normally distributed RVs with zero mean and variance 5. The squared distance between the landing point and the target value, divided by the variance, has a Chi-Square distribution with v = 2. What is the probability that the distance between the measured point and the desired location is less than 4.80 units?

Page 59: Chapter Four Continuous Random Variables & Probability Distributions

Chi-Square pdf ExampleMost galaxies take the form of a flattened disc with the major part of the light coming from this thin fundamental plane. The degree of flattening differs from galaxy to galaxy. In the Milky Way Galaxy most gases are concentrated near the center of the fundamental plane. Let X denote the perpendicular distance from this center to a random gaseous mass. RV X is normally distributed with zero mean & = 100 parsecs. ( A parsec is equal to approximately 19.2 trillion miles.) The squared distance between a scientifically measured location of a gaseous mass and its actual location, divided by the variance, has a Chi-Square distribution with v = 3. What is the probability that the distance between the measured point and the actual location is greater than 144.25 parsecs?

Page 60: Chapter Four Continuous Random Variables & Probability Distributions

Weibull Distribution

f(x; , ) = x -1 e-(x/) x 0

Parameters: > 0 & >0

Page 61: Chapter Four Continuous Random Variables & Probability Distributions

Weibull pdf Properties

Exponential when = 1 (constant failure rate) E(X)= (1+1/) V(X)= 2{(1+2/)–[(1+1/)]2}

CDF= 1 – e-(x/) For x 0

Page 62: Chapter Four Continuous Random Variables & Probability Distributions
Page 63: Chapter Four Continuous Random Variables & Probability Distributions

Weibull pdf ExampleThe lifetime measured in years of a brand of TV picture tubes follows a Weibull probability distribution with parameters = 2 years & = 13. What is the probability that a TV tube fails before the expiration of the two-year warranty?

Page 64: Chapter Four Continuous Random Variables & Probability Distributions

Weibull pdf ExampleThe bake step in the manufacture of a semiconductor follows a Weibull pdf with = 2.0 hours and = 10. What is the expected amount of time needed to bake a semiconductor?

What is the probability that a randomly chosen bake step takes longer than 4 hours?

What is the probability that it takes between 2 & 7 hours?

Page 65: Chapter Four Continuous Random Variables & Probability Distributions

Weibull Threshold ExampleAn extremely hard silicon carbide composite has a tensile strength in MegaPascals that under specified conditions can be modeled by a Weibull distribution after a minimum tensile strength of 2 MPa with = 9 and = 180. What strength value separates the weakest 10% of all specimens from the remaining 90%?

Page 66: Chapter Four Continuous Random Variables & Probability Distributions

Weibull DistributionAn electronic device has decreasing failure rate characterized by the two parameter Weibull pdf with = 0.50 & = 180 years. The device is required to exhibit a design life reliability of 0.90. What is the design life if there is no wear-in period?What is the design life if the device is first subject to a wear-in period of one month? (Hint: conditional probability)

Page 67: Chapter Four Continuous Random Variables & Probability Distributions

Lognormal Distribution 2 2 f(x; , ) = e-[ln(x) - ] / (2)

2 x For x

0

Parameters ln & ln of ln(x)

Page 68: Chapter Four Continuous Random Variables & Probability Distributions

Properties of Lognormal pdfln(x) is a Normal pdf 2 E(X) = e + /2

2 2

V(X) = e 2 + (e - 1) CDF = [(ln(x) - ) / ] note: is CDF of ZIf RVs X & Y are LN than XY is LN

Page 69: Chapter Four Continuous Random Variables & Probability Distributions
Page 70: Chapter Four Continuous Random Variables & Probability Distributions

Lognormal pdf ExampleThe lifetime of a certain insect is Lognormally distributed with parameters ln = 1 week & ln = ½ weeks. What is the expected lifetime of a randomly selected insect?What is the standard deviation of the lifetime?What is the probability that a insect lives longer then 4 weeks?

Page 71: Chapter Four Continuous Random Variables & Probability Distributions

Lognormal ExampleFatigue life data for an industrial rocker arm is fit to a Lognormal pdf. The following parameters are obtained: ln = 16.8 cycles & ln = 2.3. To what value should the design life be set if the probability of failure is not to exceed 1.0%?

Page 72: Chapter Four Continuous Random Variables & Probability Distributions

Beta Probability Distributionf(x; , , A, B) -1 -1

= 1 (+) x-A B-x B-A ()() B-A B-A

For x on an interval A x B. Parameters > 0 & >0.

Page 73: Chapter Four Continuous Random Variables & Probability Distributions

Properties of Beta pdf

E(X) = A + (B - A) ( + )

V(X) = (B - A)2 (+)2(++1)

A = 0 & B = 1 yields standard Beta

Page 74: Chapter Four Continuous Random Variables & Probability Distributions
Page 75: Chapter Four Continuous Random Variables & Probability Distributions

Beta pdf ExampleThe proportion X of a surface area in a randomly selected forest is covered by Lilac trees. The RV X has a standard Beta distribution with = 5 and = 2. What is the expected proportion of the surface area to be covered by this tree?The first quadrant is taken for study. What is the probability that the proportion of surface area to be covered by this tree is less than or equal to 20%?

Page 76: Chapter Four Continuous Random Variables & Probability Distributions

Beta pdf ExampleTime analysis in the construction industry assumes that the time necessary to complete any particular activity once it has been initiated has a Beta pdf with A = the optimistic time (no mistakes are made) and B = the pessimistic time (all possible mistakes are made). If the time T (days) for laying the foundation of a single family home has a Beta pdf with A = 2, B = 5, =2, & =3, what is the probability that it takes at most 3 days to lay the foundation?