STAT 509: Statistics for Engineers Dr. Dewei Wang. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

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1 STAT 509: Statistics for Engineers Dr. Dewei Wang

2 Applied Statistics and Probability for Engineers Sixth Edition Douglas C. Montgomery George C. Runger

3 4 Continuous CHAPTER OUTLINE 4-1 Continuous Random Variables 4-2 Probability Distributions and Probability Density Functions 4-3 Cumulative Distribution Functions 4-4 Mean and Variance of a Continuous Random Variable 4-5 Continuous Uniform Distribution 4-6 Normal Distribution Random Variables and Probability Distributions 4-8 Exponential Distribution 4-9 Gamma Distributions 4-10 Weibull Distribution 4-11 Lognormal Distribution 4-12 Beta Distribution Chapter 4 Title and Outline 3

4 Learning Objectives for Chapter 4 After careful study of this chapter, you should be able to do the following: 1. Determine probabilities from probability density functions. 2. Determine probabilities from cumulative distribution functions, and cumulative distribution functions from probability density functions, and the reverse. 3. Calculate means and variances for continuous random variables. 4. Understand the assumptions for continuous probability distributions. 5. Select an appropriate continuous probability distribution to calculate probabilities for specific applications. 6. Calculate probabilities, means and variances for continuous probability distributions. 7. Standardize normal random variables. 8. Use the table for the cumulative distribution function of a standard normal distribution to calculate probabilities. Chapter 4 Learning Objectives 4

5 Continuous Random Variables A continuous random variable is one which takes values in an uncountable set. They are used to measure physical characteristics such as height, weight, time, volume, position, etc... Examples 1. Let Y be the height of a person (a real number). 2. Let X be the volume of juice in a can. 3. Let Y be the waiting time until the next person arrives at the server. Sec 4-1 Continuos Radom Variables 5

6 Probability Density Function PlacXebl 11 PY cbl a fad FEE pl Xex Sec 4-2 Probability Distributions & Probability Density Functions 6

7 St Example 4-1: Electric Current Let the continuous random variable X denote the current measured in a thin copper wire in milliamperes(ma). Assume that the range of X is 4.9 x 5.1 and f(x) = 5. What is the probability that a current is less than 5mA? 1st Answer: A 5 5 P X < 5 = f ( x) dx = 5dx = 0.5 ( ) ò P 4.9 Exes 5 1 g ò another eagle ( ) P 4.95 < X < 5.1 = f ( x) dx = ò 4.95 Figure 4-4 P(X < 5) illustrated. Sec 4-2 Probability Distributions & Probability Density Functions 7

8 Cumulative Distribution Functions The cumulative distribution function is defined for all real numbers. Sec 4-3 Cumulative Distribution Functions 8

9 Example 4-3: Electric Current For the copper wire current measurement in Exercise 4-1, the cumulative distribution function consists of three expressions. 49 EX E5 I f 17 5 for 49exesI PIXex 11 0 x < 4.9 F (x ) = 5x x x The plot of F(x) is shown in Figure 4-6. Figure 4-6 Cumulative distribution function For 4.9 EXESL Fix PIX Ex P 4.9 EX ex Sec 4-3 Cumulative Distribution Functions 9 ItPIX x p Xx I ff.ge du fgg5du 5u I.g 5x 5x49 5X 24.5

10 Probability Density Function from the Cumulative Distribution Function The probability density function (PDF) is the derivative of the cumulative distribution function (CDF). The cumulative distribution function (CDF) is the integral of the probability density function (PDF). ( ) df x Given F( x), f ( x) = as long as the derivative exists. dx Sec 4-3 Cumulative Distribution Functions 10

11 Exercise 4-5: Reaction Time The time until a chemical reaction is complete (in milliseconds, ms) is approximated by this cumulative distribution function: plxexj 0 for x < 0 = x 1 - e - for 0 x ( ) { 0.01 F x What is the Probability density function? df ( x) d ì ( ) { x < f x = = -0.01x = -0.01x dx 0 0 for 0 í dx î1 - e 0.01 e for 0 x What proportion of reactions is complete within 200 ms? r.ph E2o9 ( ) ( ) 2 P X < 200 = F 200 = 1- e - = Sec 4-3 Cumulative Distribution Functions 11

12 Mean & Variance If X is discrete f Exfix or Sec 4-4 Mean & Variance of a Continuous Random Variable 12

13 Example 4-6: Electric Current For the copper wire current measurement, the PDF is f(x) = 0.05 for 0 x 20. Find the mean and variance x ò o ( ) ( ) E X = x f x dx = = 10 0 text 1 5 Ext II xfa.mx ( x - ) V ( X ) = ò ( x - 10) f ( x) dx = = VIX J ex pig Ix fo 0 05CxIoi'd Sec 4-4 Mean & Variance of a Continuous Random Variable f xfixidx

14 Mean of a Function of a Continuous Random Variable If X is a continuous random variable with a probability density function f x, E[ h( x) ] = ò h( x) f ( x) dx - O (4-5) Example 4-7: Let X be the current measured in ma. The PDF is f(x) = 0.05 for 0 x 20. What is the expected value of power when the resistance is 100 ohms? Use the result that power in watts P = 10 6 RI 2, where I is the current in milliamperes and R is the resistance in ohms. Now, h(x) = X 2. If 10 x'f dig x E éëh ( x) ùû = 10 ò x dx = = watts Fix ( ) Sec 4-4 Mean & Variance of a Continuous Random Variable 14

15 Summery continuous pdf random variable X fan cdf Feel How to get Fixi from fix How to get fan from Fla Probability Calculation Mean Variance Expectation of HIX

16 Continuous Uniform Distribution This is the simplest continuous distribution and analogous to its discrete counterpart. A continuous random variable X with probability density function f(x) = 1 / (b-a) for a x b Figure 4-8 Continuous uniform Probability Density Function Sec 4-5 Continuous Uniform Distribution 15

17 Mean & Variance Mean & variance are: ( ) µ = E X = 2 s and ( ) = V X = ( a+ b) 2 ( b- a) 2 12 Sec 4-5 Continuous Uniform Distribution 16

18 Example 4-9: Uniform Current The random variable X has a continuous uniform distribution on [4.9, 5.1]. The probability density function of X is f(x) = 5, 4.9 x 5.1. What is the probability that a measurement of current is between 4.95 & 5.0 ma? Sais 54 The mean and variance formulas can be applied with a = 4.9 and b = 5.1. Therefore, ( ) V ( X) ( 0.2) 2 2 µ = E X = 5 ma and = = ma 12 Figure 4-9 Sec 4-5 Continuous Uniform Distribution 17

19 Cumulative distribution function of Uniform distribution ( ) F x x 1 x- a = ò du = b-a b-a a ( ) The Cumulative distribution function is ì ï 0 x< a F( x) = í( x-a) ( b- a) a x< b ïî 1 b x PHET Ff xt fs.fi Figure 4-6 Cumulative distribution function Sec 4-5 Continuous Uniform Distribution 18 0 Xc4.9 g4.9excbplxex IT p I X25 fix pixel PlXE495 Fts a cab

20 Let X be a continuous pdf of X is FKF random variable 3 2 Os Xc 1 0 o chemise cal Find Fix b Calculate Pl Xc 0.2 Cc E Ix VG and Pl xco.si X 0 3 Cal Fixt PHExk 3 1 For p x PC o.cl ex fluid J 3h due b p Xue 2 Flo or p Oc Xco 2 3x2dx X plalbi pl xco.se Xso 3 P X o SAX 03 PlX70.3 MAMI PC o 3cXco 5 Plotaxeas PCB PIX o s PG 344

21 S 3x'dy x'l as 0.53 o 323 cnn.elxt f.fm d I as or J X 3X'DX Kk f ex Mifixidx f f x'fixidx pi I O O 27 ex 5 3x2dxe fo x 3x'dx fu f 3x4dx 5 Exit i Is 5

22 Normal Distribution Sec 4-6 Normal Distribution 19

23 Empirical Rule For any normal random variable, P(μ σ < X < μ + σ) = P(μ 2σ < X < μ + 2σ) = P(μ 3σ < X < μ + 3σ) = Figure 4-12 Probabilities associated with a normal distribution Sec 4-6 Normal Distribution 20

24 Standard Normal Random Variable A normal random variable with μ = 0 and σ 2 = 1 me 2 n NCO l is called a standard normal random variable and is denoted as Z. The cumulative distribution function of a standard normal random variable is denoted as: Φ(z) = P(Z z) Values are found in Appendix Table III and by using Excel and Minitab. Sec 4-6 Normal Distribution 21

25 Example 4-11: Standard Normal Distribution Assume Z is a standard normal random variable. Find P(Z 1.50). Answer: Figure 4-13 Standard normal Probability density function Z Nco l normalcdf LB UB 14,6 Find P(Z 1.53). Answer: Find P(Z 0.02). Answer: normal calf 1099,0 02,011 µ o normulcdff 1099,1 53,0 1 NOTE : The column headings refer to the hundredths digit of the value of z in P(Z z). For example, P(Z 1.53) is found by reading down the z column to the row 1.5 and then selecting the probability from the column labeled 0.03 to be Sec 4-6 Normal Distribution 22

26 No.mn edf 2B VB M 61 X NC M 62 Pt LB UB If K 447 Mia 5 PL XE 2 X tho 5 4 Nomulchff ,54

27 Standardizing a Normal Random Variable 2 Suppose X is a normal random variable with mean µ and variance s, the random variable for XNNCM.ci ( X - µ ) we Z = Nlo l get s is a normal random variable with EZ ( ) = 0 and VZ ( ) = 1. The probability is obtained by using Appendix Table III ( x - µ ) with z =. s Sec 4-6 Normal Distribution 23

28 Example 4-14: Normally Distributed Current-1 Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution with μ = 10 and σ = 2 ma, what is the probability that the current measurement is between 9 and 11 ma? Xn HC µ lo Answer: æ9-10 x ö P( 9 < X < 11) = Pç < < è ø = P - < < Normal cdf ( 0.5 z 0.5) ( 0.5) P( z 0.5) = P z < - < - = = pl ,52044, Sec 4-6 Normal Distribution 24

29 Example 4-14: Normally Distributed Current-2 Determine the value for which the probability that a current measurement is below Answer: Find an X N In to such that to find x suchthat PlXEx a 0.98 p X ex e 0.98 Fk Fix T y T we use inuerseofanoro.mu cdfinvnorm p M o X invnorm O 98 lo 2 14 I Sec 4-6 Normal Distribution 25

30 First step Lee xnnlo.si 5 2 if pl ex o 4 find x ifplkxcxy o.zf.no PlXcx1 PlXEoltP ocxax f 0.4 Second step x in Norm 0.9 O E X HCM.ci E VIX G Normadcdf inv Norm

31 Example l say XnN i 62 where 62 unknown if Pf X Find 0 standardization X gtt Z NCO.JP X4.5I P X Ict5o I Plz c x P 2 ex where x based on Z X invnorm 0.9 o l a s Lef Xu NCH l where His unknow if Pks Find µ

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33 Exponential Distribution Definition The random variable X that equals the distance between successive events of a Poisson process with mean number of events λ > 0 per unit interval is an exponential random variable with parameter λ. The probability density function of X is: afix pdf f(x) = λe -λx for 0 x < cdf PIX I e Fix x Sec 4-8 Exponential Distribution 26 d

34 Exponential distribution - Mean & Variance If the random variable X has an exponential distribution with parameter l, µ = E( X) = and s = V ( X) = (4-15) 2 l l Note: Poisson distribution : Mean and variance are same. Exponential distribution : Mean and standard deviation are same. Sec 4-8 Exponential Distribution 27

35 Example 4-21: Computer Usage-1 In a large corporate computer network, user log-ons to the system can be modeled as a Poisson process with a mean of 25 log-ons per hour. What is the probability that there are no log-ons in the next 6 minutes (0.1 hour)? per hour Let X denote the time I in hours from the start of the interval until the first log-on. Fix _I e -25x -25( 0.1) P X > 0.1 = 25e dx = e = ( ) The cumulative distribution function also can be used to obtain the same result as follows ( ) F( ) ò P X > 0.1 = = I plxeo.tl Figure 4-23 Desired probability Flo 1 I l e Sec 4-8 Exponential Distribution 28

36 Poisson Process I I lo d time Poisson Random Variables of arrivals time Time Interval Exp R V ish j X time till 1st X Exp d

37 Example 4-21: Computer Usage-2 Continuing, what is the probability that the time until the next log-on is between 2 and 3 minutes (0.033 & 0.05 hours)? x P < X < 0.05 = 25e dx ( ) An alternative solution is -25x 0.05 = - e = ( ) ( ) ( ) ò P < X < 0.05 = F F = Sec 4-8 Exponential Distribution 29

38 H Example 4-21: Computer Usage-3 Continuing, what is the interval of time such that the probability that no log-on occurs during the interval is 0.90? I P X Ext l Fix I CI e -25x ( > ) = = 0.90, - 25 = ln ( 0.90) P X x e x x = = hour = 0.25 minute -25 What is the mean and standard deviation of the time until the next log-in? 1 1 µ = = = 0.04 hour = 2.4 minutes l s = = = 0.04 hour = 2.4 minutes l 25 Sec 4-8 Exponential Distribution 30

39 Lack of Memory Property An interesting property of an exponential random variable concerns conditional probabilities. For an exponential random variable X, P(X<t 1 +t 2 X>t 1 )= P(X < t 2 ) Figure 4-24 Lack of memory property of an exponential distribution. Sec 4-8 Exponential Distribution 31

40 Example 4-22: Lack of Memory Property Let X denote the time between detections of a particle with a Geiger counter. Assume X has an exponential distribution with E(X) = 1.4 minutes. What is the probability that a particle is detected in the next 30 seconds? ( ) ( ) P X < 0.5 = F 0.5 = 1- e - = 0.30 No particle has been detected in the last 3 minutes. Will the probability increase since it is due? T ( ) P( X ) No, the probability that a particle will be detected depends only on the interval of time, not its detection history. ( ) ( ) F( ) Using Excel = EXPONDIST(0.5, 1/1.4, TRUE) P 3 < X < 3.5 F F P( X < 3.5 X > 3) = = = = 0.30 > Sec 4-8 Exponential Distribution 32

41 Exp.RU X Explx d of arrivals in a time unit µ I f Ix de x o Flxt plxex l e Y.no Locket Meaney Papay p Xcx _o X 2

42 Gamma Distributions The exponential distribution models the interval to the 1 st event, while the Gamma distribution models the interval to the r th event, i.e., a sum of exponentials. r is not required to be an integer. The exponential and gamma distributions are based on the Poisson process. Sec 4-9 Erlang & Gamma Distributions 33

43 Gamma Function The gamma function is the generalization of the factorial function for r > 0, not just non-negative integers. r-1 -x r x e dx r ( ) ò G =, for > 0 (4-17) 0 Properties of the gamma function ( r) ( r ) ( r ) ( r) 0( r ) ( 1) 0! 1 12 ( ) p G = -1 G -1 recursive property G = -1! factorial function G = = G 1 2 = = 1.77 useful if manual Sec 4-9 Erlang & Gamma Distributions 34

44 Gamma Distribution The random variable X with probability density function: r r-1 -lx l x e f ( x) =, for x> 0 (4-18) G ( r) Playas f faadx is a gamma random variable with parameters λ > 0 and r > 0. If r is an integer, then X has an Erlang distribution. Sec 4-9 Erlang & Gamma Distributions 35

45 Mean & Variance of the Gamma If X is a gamma random variable with parameters λ and r, μ = E(X) = r / λ and σ 2 = V(X) = r / λ 2 Sec 4-9 Erlang & Gamma Distributions 36

46 Example 4-24: Gamma Application-1 The time to prepare a micro-array slide for high-output genomics is a Poisson process with a mean of 2 hours per slide. What is the probability that 10 slides require more than 25 hours? Poisson process, X has a gamma distribution with λ = ½, r = 10, and the requested probability is P(X > 25). Using the Poisson distribution, let the random variable N denote the number of slides made in 10 hours. The time until 10 slides are made exceeds 25 hours if and only if the number of slides made in 25 hours is 9. ( > 25) = P( N 9) E( N) = ( ) = P X slides in 25 hours k = 0 Fzslide perhouxittotifisinatui7fexngammald Let X denote the time to prepare 10 slides. Because of the assumption of a ( 12.5) e P( N 9) = å = k! k IET Sec 4-9 Erlang & Gamma Distributions 37 I.r 10 Using Excel = POISSON(9, 12.5, TRUE)

47 Example 4-24: Gamma Application-2 Using the gamma distribution, the same result is obtained ( > 25) = 1- P( X 25) P X = x ò xe G = = ( 10) dx I P O C XE doesthis l fine o.to gi.xx9xe o FX 0.25 What is the mean and standard deviation of the time to prepare 10 slides? V ( ) E X ( X ) ( ) SD X r 10 = = = 20 hours l 0.5 r 10 = = = 40 hours 2 l 0.25 = V ( X ) = 40 = 6.32 hours Sec 4-9 Erlang & Gamma Distributions 38

48 Weibull Distribution The random variable X with probability density function b -1 b x b æ ö -( x d ) f ( x) = ç e, for x> 0 (4-20) d èd ø is a Weibull random variable with scale parameter d > 0 and shape parameter b > 0. The cumulative distribution function is: ( ) F x ( x d ) - = 1 - e (4-21) The mean and variance is given by æ 1 ö µ = E( X) = d G ç 1+ and è b ø b P lax 2 2 é æ 2 öù 2 é æ 1 öù s = V ( X) = d êg ç 1+ ú-d êg ç 1+ ú ë è b øû ë è b øû Sec 4-10 Weibull Distribution 39 2 (4-21a)

49 Example 4-25: Bearing Wear The time to failure (in hours) of a bearing in a mechanical shaft is modeled as a Weibull random variable with β = ½ and δ = 5,000 hours. What is the mean time until failure? ( ) = 5000 G ( ) = 5000 G( 1.5) E X = p = 4, hours What is the probability that a bearing will last at least 6,000 hours? ( 6, 000) 1 ( 6, 000) P X > = - F = e = e = enough X X X XX X æ6000 ö - ç è5000 ø Only 33.4% of all bearings last at least 6000 hours. 0.5 Backboards plxsxto.cl 1 I e EP 0.9 e Esp _0.9 ftp.dno 9 E lno 1 gxflao 9 Sec 4-10 Weibull Distribution 40

50 Lognormal Distribution Let W denote a normal random variable with mean θ and variance ω 2, then X = exp(w) is a lognormal random variable with probability density function ( ) ( x) 2 2 2q+ w w ( ) ( 1 ) 2 é 2 (ln -q ) ù -ê ú 2 êë 2w úû 1 f ( x) = e 0 < x< xw 2p The mean and variance of X are q+ w 2 E X = e and I V X = e e - Sec 4-11 Lognormal Distribution 41

51 Example 4-26: Semiconductor Laser-1 The lifetime of a semiconductor laser has a lognormal distribution with θ = 10 and ω = 1.5 hours. What is the probability that the lifetime exceeds 10,000 hours? X exp W ( > ) = - éë ( ) = 1- PéëW ln ( 10,000) æ ( ) - P X 10, P exp W 10, 000ùû Wn All lo 1.5 ln 10, ö = 1-Fç è 1.5 ø = 1-F = = ( ) ùû Sec 4-11 Lognormal Distribution 42

52 PIX p explw toooo p W dog 10,000 Wn Mollo 5 I TI 83 normalcalf log lo 1.5 heµo mp lognoramal Pomobabiteies

53 Example 4-26: Semiconductor Laser-2 What lifetime is exceeded by 99% of lasers? ( > ) = exp( ) > = > ln ( ) æ ( x) - P X x Péë W xùû PéëW x ùû ln 10 ö = 1-F ç = 0.99 è 1.5 ø From Appendix Table III, 1-F = 0.99 when z =-2.33 ( x) ( z) ln -10 Hence = and x = exp( 6.505) = hours 1.5 What is the mean and variance of the lifetime? ( ) ( ) q+ w w ( ) = ( - 1) = ( -1) ( ) 2 2 E X = e = e q+ v = exp = 67, V X e e e e SD X ( ) é ( ) P = exp ëexp ùû = 39, 070, 059,886.6 = 197, Sec 4-11 Lognormal Distribution 43 PIX p tog 6gxto.gg WslogxI o 99plWc6gx7 o.o WnN byx inu lo XI

54 Beta Distribution The random variable X with probability density function ( a b) ( a) ( b) G + f ( x) x ( - x) x G G er a -1 b -1 = 1, for in [0, 1] is a beta random variable with parameters α > 0 and β > 0. Sec 4-12 Beta Distribution 44

55 Example 4-27: Beta Computation-1 Consider the completion time of a large commercial real estate development. The proportion of the maximum allowed time to complete a task is a beta random variable with α = 2.5 and β = 1. What is the probability that the proportion of the maximum time exceeds 0.7? Let X denote the proportion. 1 G ( a + b) a -1 b -1 P ( X > 0.7 ) = ò x ( 1- x) dx G a G b 0.7 ( ) ( ) G( 3.5) ( 2.5) ( 1) = G G ( )( ) ( ) ( ) = = x dx p x =! p 2.5 ò TI 83 Sec 4-12 Beta Distribution 45

56 Mean & Variance of the Beta Distribution If X has a beta distribution with parameters α and β, a µ = E( X) = a + b 2 s ( ) = V X = ab 2 ( a + b) ( a + b + 1) Example 4-28: In the above example, α = 2.5 and β = 1. What are the mean and variance of this distribution? µ = = = s 2.5( 1) ( ) ( ) ( 4.5) = = = Sec 4-12 Beta Distribution 46

57 Important Terms & Concepts of Chapter 4 Beta distribution Chi-squared distribution Continuity correction Continuous uniform distribution Cumulative probability distribution for a continuous random variable Erlang distribution Exponential distribution Gamma distribution Lack of memory property of a continuous random variable Lognormal distribution Mean for a continuous random variable Mean of a function of a continuous random variable Normal approximation to binomial & Poisson probabilities Normal distribution Probability density function Probability distribution of a continuous random variable Standard deviation of a continuous random variable Standardizing Standard normal distribution Variance of a continuous random variable Weibull distribution Chapter 4 Summary 47