Solution

Introduction to Probability and StatisticsProbability & Statistics for Engineers & Scientists, 9th Ed.2009Handout #5Lingzhou XueThe pdf file for this class is available on the class web page.1

Chapter 6Some Continuous Probability Distributions2

Name Notation P.D.F. f(x)Uniform U[a, b]1b−a , a ≤ x ≤ b.Normal N (µ, σ 2 )ExponentialExp(β)√ 1exp{− (x−µ)22πσ 2σ2 }, −∞ < x < ∞.1β e−x/β , x > 0.Gamma Γ(α, β)1β α Γ(α) xα−1 e −x/β , x > 0.Chi-Squared χ 2 υ = Γ ( υ2, 2 ) 12 υ/2 Γ(υ/2) xυ/2−1 e −x/2 , x > 0.Lognormal LogN(µ, σ 2 )1xσ √ 12π e− 2σ 2[ln(x)−µ]2 , x > 03

Continuous Uniform Distribution (Rectangular Distribution)The density function of the continuous uniform random variableX on the interval [a, b] isf(x; a, b) ={ 1b−a, a ≤ x ≤ b;0, elsewhere.E(X) = a + b2(b − a)2, and Var(X) = .124

Probability density function for a continuous uniform randomvariable.5

Example 1Suppose that a large conference room for a certain company canbe reserved for no more than 4 hours. However, the use of theconference room is such that both long and short conferencesoccur quite often. In fact, it can assumed that length X of aconference has a uniform distribution on the interval [0, 4].1. What is the probability density?2. What is the probability that any given conference lasts atleast 3 hours?6

Solution:• The appropriate density function for the uniformly distributedrandom variable X in this situation isf(x) ={ 14, 0 ≤ x ≤ 4;0, elsewhere.•P (X ≥ 3) =∫ 4314 dx = 1 4 . 7

Normal Distribution8

The most important continuous probability distribution in theentire field of statistics is the normal distribution. Its graph,called the normal curve, is the bell-shaped curve, which describesapproximately many phenomena that occur in nature,industry, and research.• Physical measurements in areas such as meteorological experiments,rainfall studies, and measurements of manufacturedparts are often more than adequately explained with anormal distribution.• Errors in scientific measurements are extremely well approximatedby a normal distribution.9

Probability density function for a normal random variable.10

Normal DistributionThe density function of the normal random variable X, withmean µ and variance σ 2 , isf(x; µ, σ 2 ) = √ 1 (x − µ)2exp{− 2πσ 2σ 2 }, −∞ < x < ∞.where π = 3.12159 . . ., and e = 2.71828 . . . .E(X) = µ, and Var(X) = σ 2 .Standard Normal DistributionThe distribution of a normal random variable with mean 0 andvariance 1 is called a standard normal distribution.11

Example 2Let the probability density function of a random variable X bef(x) = √ 1 exp{− x2}, if −∞ < x < ∞.10π 101. Find the mean and variance of the random variable X.2. Use Chebyshev’s inequality to show thatP (|X| ≥ √ 10) ≤ 1 2 . 12

Solution:1.E(X) = 0, Var(X) = 5.2.P (|X| ≥ √ 10) ≤ 1 2 .P (|X| ≥ √ 10) = 1 − P (|X| ≤ √ 10)= 1 − P (|X| ≤ √ 2 √ 5)≤ 1 − 1 √2 2 = 1 2 . 13

Normal curves with µ 1 < µ 2 and σ 1 = σ 2 .14

Normal curves with µ 1 = µ 2 and σ 1 < σ 2 .

Normal curves with µ 1 < µ 2 and σ 1 < σ 2 .

SymmetryA function f(x) is sometimes said to be symmetric about theorigin at x-axis iff(−x) = f(x).A function f(x) is sometimes said to be symmetric about the µat x-axis iff(µ − x) = f(µ + x).15

Properties of Normal Distribution1. The mode (the mode is the value that occurs the most frequentlyin a data set or a probability distribution.) occurs atx = µ.2. The curve is symmetric about a vertical axis through themean µ.3. The curve has its points of inflection at x = µ ± σ, is concavedownward if µ − σ < X < µ + σ, and is concave upwardotherwise.4. The normal curve approaches the horizontal axis asymptoticallyas we proceed in either direction away from the mean.16

5. The total area under the curve and above the horizontal axisis equal to 1.

Mean of a Normal Random Variablef(x; µ, σ 2 ) = 1 √2πσe −(x−µ)2 2σ 2 , −∞ < x < ∞.E(X) =∫ ∞−∞ xf(x; µ, σ2 )dx= 1 √2πσ∫ ∞−∞ e−(x−µ)2 2σ 2dx(setting z = x − µ and dx = σdz)σ= √ 1 ∫ ∞(µ + 2π −∞ σz)e−z2 2 dz= µ √2π∫ ∞= µ.−∞ e−z2 2 dz +σ √2π∫ ∞−∞ ze−z2 2 dz17

Variance of a Normal Random Variablef(x; µ, σ 2 ) = 1 √2πσe −(x−µ)2 2σ 2 , −∞ < x < ∞.Var(X) = E[(X − µ) 2 ]∫ ∞= (x − −∞ µ)2 f(x; µ, σ 2 )dx(setting z = x − µ and dx = σdz)σ= √ σ2 ∫ ∞2π −∞ z2 e −z2 2 dz= σ2 √2π(−ze −z2 2 | ∞ −∞ += σ 2 .∫ ∞−∞ ze−z2 2 dz)18

Areas Under Normal Curve19

Areas Under the Normal CurveThe curve of any continuous probability distribution or densityfunction is constructed so that the area under the curve boundedby the two ordinates x = x 1 and x = x 2 equals the probabilitythat the random variable X assumes a value between x 1 and x 2 .Thus, for the normal curveP (x 1 < X < x 2 ) =∫ x2x 11√ e −(x−µ)2 2σ 2 dx2πσis represented by the area of the shaded region.20

P (x1 < X < x2)= area of the shaded region.21

The difficulty encountered in solving integrals of normal densityfunctions necessitates the tabulation of normal curve areas forquick reference. However, it would be a hopeless task to attemptto set up separate tables for every conceivable values of µ andσ. Fortunately, we are able to transform all the observations ofany normal random variable X to a new set of observation of anormal random variable Z with mean 0 and variance 1. This canbe done by means of the transformationZ = X − µ .σWhenever X assumes a values x, the corresponding value of Zis given by z = x−µσ. Therefore, if X falls between the valuesx = x 1 and x = x 2 , the random variable Z will fall between the22

corresponding values z 1 = x 1−µσand z 2 = x 2−µσ. Consequently,P (x1 < X < x2) =∫ x2x 1(n(x; µ, σ)dx= P z 1 = x 1 − µ( σ= P z 1 = x 1 − µσ=∫ z2z 1n(x; 0, 1)dz< X − µ < z 2 = x 2 − µσσ< Z < z 2 = x )2 − µσwhere Z is seen to be a normal random variable with mean 0and variance 1.)

Example 3Given a standard normal distribution, find the area under thecurve that lies1. to the right of z = 1.84, and2. between z = −1.97 and z = 0.86.23

Solution:1. The area in Figure (a) to the right of z = 1.84 is equal to1 minus the area in Table to the left of z = 1.84, namely,1-0.9671=0.0329.2. The area in Figure (b) to between z = −1.97 and z = 0.86is equal to the area to the left of z = 0.86 minus the area tothe left of z = -1.97. From Table we find the desired areato be 0.8051-0.0244=0.7807.24

Area of Example 3.25

Example 4Given a standard normal distribution, find the value of k suchthat1. P (Z > k) = 0.3015 and2. P (k < Z < −0.18) = 0.4917.26

Solution:1. In Figure (a) we see that the k value leaving an area of 0.3015to the right must then leave an area of 0.6985 to the left.From the Table it follows that k = 0.52.2. From Table we note that the total area to the left of -0.18 isequal to 0.4286. In Figure (b) we see that the area betweenk and -0.18 is 0.4197 so that the area to the left of k mustbe 0.4286-0.4197=0.0089. Hence, from Table, we have k =-2.37.27

Area of Example 4.28

Example 5Given a random variable X having a normal distribution withµ = 50 and σ = 10, find the probability that X assumes a valuebetween 45 and 62.Solution:( 45 − 50P (45 < X < 62) = P< X − 50)62 − 50

Using the Normal Curve in ReverseOccasionally, we are required to find the value of z correspondingto a specified probability that falls between values. For convenience,we shall always choose the z value corresponding to thetabular probability that comes closest to the specified probability.We reverse the process and begin with a known area or probability,find the z value, and then determine x by rearranging theformulaz = x − µ to give x = σz + µ.σ30

Example 6Given a normal distribution with µ = 40 and σ = 6, find thevalue of x that has1. 45% of the area to the left, and2. 14% of area to the right.31

Solution:1.P (X < x) = 0.45( X − 40⇒P < x − 40 )= 0.45(6 6⇒P Z < x − 40 )= 0.456⇒P (Z < −0.13) = 0.45⇒x = 6 − (−0.13) + 40 = 39.2232

2.P (X > x) = 0.14 ⇒P (X < x) = 0.86( X − 40⇒P < x − 40 )= 0.86(6 6⇒P Z < x − 40 )= 0.866⇒P (Z < 1.08) = 0.86⇒x = 6 · (1.08) + 40 = 46.48

Area of Example 6.33

Applications of the Normal Distribution34

Example 7A certain type of storage battery lasts, on average, 3.0 years witha standard deviation of 0.5 year. Assuming that the battery livesare normally distributed, find the probability that a given batterywill last less than 2.3 years.Solution:Let X represent the life of the storage battery. 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 the areato the left of the corresponding z value. Hence we find that( X − 2.3P (X < 2.3) = P< 2.3 − 3 )= P (Z < −1.4) = 0.0808.0.5 0.535

Example 8An electrical firm manufactures light bulbs that have a life, beforeburn-out, that is normally distributed with mean equal to 800hours and a standard deviation of 40 hours. Find the probabilitythat a bulb burns between 778 and 834 hours.Solution:Let X be the life of the light bulb. X ∼ N(µ = 800, σ 2 = 40 2 ).( 778 − 800P (778 < X < 834) = P< X − 800)834 − 800

Example 9In an industrial process the diameter of a ball bearing is an importantcomponent part. The buyer sets specifications on the diameterto be 3.0 ± 0.01 cm. The implication is that no part fallingoutside these specifications will be accepted. It is known that inthe process the diameter of a ball bearing has a normal distributionwith mean µ = 3.0 and standard deviation σ = 0.005. Onaverage, how many manufactured ball bearings will be scrapped?37

Solution:The values corresponding to the specification are x 1 = 2.99 andx 2 = 3.01. Hence( 2.99 − 3P (2.99 < X < 3.01) = P0.005 < X − 30.005 < 3.1 − 3 )0.005= P (−2.0 < Z < 2.0)From Table, P (Z < −2.0) = 0.0228.normal distribution, we find thatDue to symmetry of the1 − P (−2.0 < Z < 2.0) = 1 − P (Z < 2.0) + P (Z < −2.0)= P (Z > 2.0) + P (Z < −2.0)= 2(0.0228) = 0.0456.As a result, it is anticipated that on the average, 4.56% of manufacturedball bearings will be scrapped.38

Example 10The average grade for an exam is 74, and the standard deviationis 7. If 25% of the class is given A’s, and the grades are curvedto follow a normal distribution what is the lowest possible A andthe height possible B?Solution:In this example we begin with a known area of probability, findthe z value, and the determine x from the formula x = zσ + µ.We require a z value that leaves 0.25 of the area to the right andhence, an area of 0.75 to the left. P (Z < 0.67) has the closet to0.75. Hence x = 7 × 0.67 + 74 = 78.69. Therefore, the lowestA is 79. and the highest B is 78.39

Normal Approximation to the Binomial40

TheoremIf X is a binomial random variable with mean µ = np and varianceσ = npq = np(1 − p), then the limiting form of the distribution ofT = X − np√ npqas n → ∞, is the standard normal distribution.Normal approximation of b(x; 15, 0.4).41

Normal Approximation to the Binomial-ILet X be a binomial random variable with parameters n and p.Then X has approximately a normal distribution with µ = np andσ 2 = npq = np(1 − p) andP (X ≤ x) =x∑k=0b(k; n, p) (1)≈ area under normal curve to the left of x + 0.5 (2)()x + 0.5 − np≈ P Z ≤ √ (3)npqwhere Z ∼ N(0, 1), and the approximation will be good if np andn(1 − p) are greater than or equal to 5.42

Normal Approximation to the Binomial-IILet X be a binomial random variable with parameters n and p.Then X has approximately a normal distribution with µ = np andσ 2 = npq = np(1 − p) andP (x 1 ≤ X ≤ x 2 ) =x 2 ∑k=x 1b(k; n, p)≈ area under normal curve tothe right of x 1 − 0.5 and left of x 2 + 0.5.(x1 − 0.5 − np≈ P √ ≤ Z ≤ x )2 + 0.5 − np√ npqnpqwhere Z ∼ N(0, 1), and the approximation will be good if np andn(1 − p) are greater than or equal to 5.43

Example 11The probability that a patient recovers from a rare blood diseaseis 0.4. If 100 people are known to have contracted this disease,what is the probability that less than 30 survive?Solution:Let the binomial variable X represent the number of patientsthat survive. Since n = 100, we should obtain fairly accurateresults using the normal-curve approximation with µ = np = 40and σ = √ npq = 4.8994.P (X < 30) ≈ P (Z < −2.14) = 0.0162.44

Example 12A multiple-choice quiz has 200 questions each with 4 possibleanswers of which only 1 is the correct answer. What is theprobability that sheer guesswork yields from 25 to 30 correctanswers for 80 of the 200 problems about which the student hasno knowledge?Solution:The probability of a correct answer for each of the 80 questionsis p = 1 4. If X represents the number of correct answers due toguesswork, thenP (25 ≤ X ≤ 30) =30∑x=25b(x; 80, 1/4)Using the normal-curve approximation with µ = np = 20 andσ = √ npq = 3.873.P (25 ≤ X ≤ 30)) ≈ P (1.16 < Z < 2.71) = 0.1196.45

Gamma and Exponential Distributions46

Exponential and Gamma DistributionsExponential Distribution Describing the time until the occurrenceof a Poisson event (or the time between Poisson events).Gamma Distribution Describing the time (or space) occurringuntil a specified number of Poisson events occur.47

Gamma and Exponential DistributionsAlthough the normal distribution can be used to solve manyproblems in engineering and science, there are still numeroussituations that require different types of density functions. Twosuch density functions, the gamma and exponential distributions.The exponential and gamma distributions play an important rolein both queuing theory and reliability problems. Time betweenarrivals at service facilities, and time to failure of componentparts and electrical systems.48

Gamma functionThe gamma function is defined byΓ(α) =∫ ∞0 xα−1 e −x dx, for α > 0.Prove: Γ(α) = (α − 1)Γ(α − 1) for α > 1.Integrating by parts with u = x α−1 and dv = e −x dx, we obtain∫ ∞∫ ∞Γ(α) = x α−1 e −x | ∞ 0 + 0 (α−1)xα−2 e −x dx = (α−1)0 xα−2 e −x dxfor α > 1 which yields the recursion formulaΓ(α) = (α − 1)Γ(α − 1).PropertyΓ( 12)= √ π.49

Gamma DistributionThe continuous random variable X has a gamma distribution,with parameters α and β, if the density function given byf(x; α, β) =where α > 0 and β > 0.{1β α Γ(α) xα−1 e −x/β , x > 0;0, elsewhere.E(X) = αβ and Var(X) = αβ 2 .50

Gamma DistributionGamma distributions.51

Exponential DistributionThe continuous random variable X has an exponential distribution,with parameter β, if the density function given bywhere β > 0.f(x; β) ={ 1βe −x/β , x > 0;0, elsewhere.E(X) = β and Var(X) = β 2 .52

Example 13What is the probability of an item surviving until t = 100 units ifthe item is exponentially distributed with a mean time betweenfailure of 80 units? Given that the item survived to 200 units,what is the probability of survival until t = 300units?Solution:Let X denote the amount of time units that the item survives.X ∼ Exp(β = 80)1.P (X > 100) = e −100/80 .2.P (X > 300|X > 200) = P (X > 100) = e −100/80 .53

Relationship to the Poisson ProcessLet X be the amount of time until the first arrival in a Poissonprocess with rate λ. Then X ∼ Exp(λ).Proof:Note that the number of arrivals in [0, t] is Poi(λt).F (x) = P (X ≤ t) (4)= 1 − P (no arrivals in [0, t]) (5)= 1 − eλt (λt) 00!(6)= 1 − e λt . (7)54

Relationship to the Poisson ProcessThe Poisson distribution is used to compute the probability ofspecific numbers of ”events” during a particular period of time orspace. In many applications, the time period or span of space isthe random variable. For example, an industrial engineer may beinterested in modeling the time T between arrivals at a congestedintersection during rush hour in a large city. An arrival representsthe Poisson event.The poisson distribution was developed as a single-parameterdistribution with parameter λ, where λ may be interpreted as themean number of events per unit time. Consider now the randomvariable described by the time required for the first event tooccur. Using the Poisson distribution, we find that the probabilityof no events occurring in the span up to time t is given byp(0; λt) = e−λt (λt) 00!= e −λt .55

We can now make use of the above and let X be the time tothe first Poisson event. The probability that the length of timeuntil the first event will exceed x is the same as the probabilitythat no Poisson events will occur in x. The latter, of course, isgiven by e −λt . As a result,P (X > x) = e −λx .Thus the cumulative distribution function for X is given byF (x) = P (0 ≤ X ≤ x) = 1 − e −λx .Now, in order that we recognize the presence of the exponentialdistribution, we may differentiate the cumulative distributionfunction above to obtain the density functionf(x) = ddx F (x) = λe−λx ,which is the density function of the exponential distribution withλ = 1/β.

Example 14Let X equal the number of bad records in each 50 feet of a usedcomputer tape. Assume that X has a Poisson distribution withmean 1. Let W equal the number of feet before the first badrecord is found.1. Give the mean and variance of the number of flaws per foot.2. How is W distributed?3. Give the mean and variance of W .4. Find P (W > 1001|W > 951).56

Solution:1. Let Y be the number of bad records per foot.E(Y ) = Var(Y ) = 150 .2.W ∼ Exp(β = 50).That is, W has a probability density functionf(w) = 150 e−w/50 , ifw > 0.3.E(Y ) = 50, Var(Y ) = 2500.

4. Using memoryless property of an exponential distribution, wehaveP (W > 1001|W > 951) = P (W > 50) = e −1 .

Application of Gamma and Exponential Distributions57

Application of Gamma and Exponential DistributionsThe application of the exponential distribution are in ”time toarrival” or time to Poisson event problems. Notice that the meanof the exponential distribution is the parameter β, the reciprocalof the parameter in the Poisson distribution. Recall: the Poissondistribution has no memory, implying the occurrences in successivetime periods are independent. The important parameter βis the mean time between events. In reliability theory, whereequipment failure often conforms to this Poisson process, β iscalled mean time between failures. Many equipment breakdownsdo follow the Poisson process, and thus the exponential distributiondoes apply. Other applications include survival times inbiomedical experiments and computer response time.58

Example 15Suppose that a system contains a certain type of componentwhose time, in years, to failure is given by T . The randomvariable T is modeled nicely by the exponential distribution withmean time to failure β = 5. If 5 of these components are installedin different systems, what is the probability that at least 2 arestill functioning at the end of 8 years?Solution:The probability that a given component is still functioning after8 years is given by∫ ∞P (T > 8) =8 e−t/5 dt = e −8/5 ≈ 0.2.If X represent the number of components functioning after 8years. Then using the binomial distributionP (X ≥ 2) =5∑x=2b(x; 5, 0.2) = 0.2627.59

Memoryless Property for Exponential DistributionA nonnegative random variable X is called memoryless if for alls, t ≥ 0,P (X ≥ t) = P (X ≥ s + t|X ≥ s)The probability of lasting an additional t time units is the sameas the probability of lasting t time units. The fact that it hasn’thappened yet, tells us nothing about how much longer it willtake before it does happen.60

To show that an exponential distribution is memoryless.Solution:P (X ≥ s + t|X ≥ s) ==P (X ≥ s + t, X ≥ s)P (X ≥ s)P (X ≥ s + t)P (X ≥ s)= e−(s+t)λe −sλ= e −tλ= P (X ≥ t)61

Example 16The lifetime of a TV tube (in years) is an exponential randomvariable with mean 10. If Jim bought his TV set 10 years ago,what is the probability that its tube will last another 10 years?Solution:Let X be the lifetime of a TV tube. X ∼ Exp(β = 10).P (X > 20|X > 10) = P (X > 10) = e −1 .62

Example 17Suppose that telephone calls arriving at a particular switchboardfollow a Poisson process with an average of 5 calls coming perminute. What is the probability that up to a minute will elapseuntil 2 calls have come in to the switchboard?Solution:The poisson process applies with time until Poisson events followinga gamma distribution with β = 1/5 and α = 2. Denoteby X the time in minutes that transpires before 2 calls come. Inother words, X ∼ Γ(2, 1/5). The required probability is given byP (X ≤ 1) =∫ 101β e−x/β dx = 1 − e −5 ≈ 0.96.63

Example 18Based on extensive testing it is determined that the time Yin years before a major repair is required for a certain washingmachine is characterized by the density function{ 14, y ≥ 0;f(x) =0, elsewhere.Note this ia na exponential with µ = 4 years. The machine isconsidered a bargain if it is unlikely to require a major repairbefore the sixth year. Thus, what is the probability P (Y > 6)?Also, what is the probability that a major repair occurs in thefirst year?64

Solution:Consider the cumulative distribution function F (y) for the exponentialdistributionSoF (y) = 1 β∫ y0 e−t/β = 1 − e −y/β .P (Y > 6) = e −3/2 = 0.2331.Thus, the probability that it will require major repair after yearsix is 0.233. Of course, it will require the repair before year sixwith probability 0.777. Thus, one might conclude the machine isnot really a bargain. The probability that a major repair occursin the first year isP (Y < 1) = 1 − e −1/4 = 0.221.65

Chi-Squared Distribution66

Chi-Squared DistributionThe continuous random variable X has a chi-squared distribution,with υ degrees of freedom, if its density function is given byf(x; α, β) =⎧⎨⎩where υ is a positive integer.12 υ/2 Γ(υ/2) xυ/2−1 e −x/2 , x > 0;0, elsewhere.This is very important special case of the gamma distributionis obtained by letting α = υ/2 and β = 2, where υ is a positiveinteger.E(X) = υ and Var(X) = 2υ.67

Lognormal Distribution68

Lognormal DistributionThe continuous random variable X has a lognormal distributionif the random variable Y = ln(X) has a normal distribution withmean µ and standard deviation σ. The resulting density functionof X isf(x; µ, σ) =⎧⎨⎩1σx √ 12π e−2σ 2[ln(x)−µ]2 , x ≥ 0;0, x < 0.E(X) = e µ+σ2 /2and Var(X) = e 2µ+σ2 (e σ2 − 1).69

Example 19Concentration of pollutants produced by chemical plants historicallyare known to exhibit that resembles a lognormal distribution.This is important when one considers issues regardingcompliance to government regulations. Suppose it is assumedthat the concentration of a certain pollutant, in parts per million,has a lognormal distribution with parameters µ = 3.2 and σ = 1.What is the probability that the concentration exceeds 8 partsper million?70

Solution:Let the random variable X be pollutant concentration per million.Then X ∼ LogN(µ = 3.2, σ 2 ), so the probability of interest isP (X > 8) = 1 − P (X ≤ 8)Since ln(X) has a normal distribution with mean µ = 3.2 andstandard deviation σ = 1,(ln(X) − 3.2P (X ≤ 8) = P≤ 8 − 3.2)11= P (Z ≤ −1.12)= Φ(−1.12)= 0.1314.Here, we use the Φ notation to denote the cumulative distributionfunction of the standard normal distribution. As a result,the probability that the pollutant concentration exceeds 8 partsper million is 0.1314.71

Example 20The life, in thousands of miles, of a certain type of electroniccontrol for locomotives has an approximate lognormal distributionwith µ = 5.149 and σ = 0.737. Find the 5th percentile ofthe life of such locomotive.Solution:Since P (Z < −1.645) = 0.05. Denote by X the life of of sucha locomotive. Since ln(X) has a normal distribution with meanµ = 5.149 and σ = 0.737, the 5th percentile of X can be calculatedln(x) = 5.149 + 0.737 · (−1.645) = 3.937.Hence, x = 51.265. This means that only 5% of the locomotiveswill have lifetime less than 51,265 mile.72