Probabilistic models for dynamical systems, second edition (1)

Give three examples of certain events from fluids engineering.Solution: Examples include a faster incompressible flow in a smaller diameter pipe, denser fluid sinking,flow exerting a for

Probabilistic Models for Dynamical Systems

S E C O N D E D I T I O N

Haym Benaroya Seon Mi Han Mark Nagurka

Benaroya Han Nagurka

CIVIL & MECHANICAL ENGINEERING

“The book presents a nice balance of theory and pragmatic application It begins by

setting the foundations of probability theory as it is applied to engineering systems

and then develops fundamental engineering concepts that integrate these probabilistic

concepts … The book also presents a wonderful historic perspective of some of

the greatest scientific and engineering minds who have made the most significant

contributions to the fields of both probability and dynamics … This is an excellent first

text for integrating mechanical system design and probability/reliability concepts.”

—DR THOMAS R KU R FE SS, Georgia Institute of Technology, Atlanta, USA

“Probabilistic Models for Dynamical Systems, Second Edition is an excellent

resource for faculty to engage students in pursuit of knowledge about physical

systems which are naturally complex, nonlinear, and non-deterministic The book is

well written and complete with a plethora of worked out examples in each chapter.”

—DR ROB E RT H B I S HOP, Marquette University, Milwaukee, USA

Now in its second edition, Probabilistic Models for Dynamical Systems expands

its coverage of probability theory for engineering This new revised version introduces

students to the randomness in variables and time-dependent functions and allows

them to solve governing equations

This book provides a suitable resource for self-study and can be used as an all-inclusive

introduction to probability for engineering It introduces basic concepts, and highlights

applied probability in a practical manner With updated information and over 300

illustrations, this edition includes new sections, problems, applications, and examples

Biographical summaries spotlight relevant historical figures, providing life sketches,

major contributions, and relevant quotes from their works A new chapter on control

and mechatronics rounds out the coverage

solutions MAnuAl FoR

Probabilistic Models for

dynaMical

systeMs

s e c o n d e d i t i o n

Haym Benaroya seon Mi Han Mark nagurka

by

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solutions MAnuAl FoR

Probabilistic Models for

dynaMical

systeMs

s e c o n d e d i t i o n

Haym Benaroya seon Mi Han Mark nagurka

by

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Solutions Manual - Probabilistic Models for Dynamical Systems,

2 Events and Probability 3

3 Random Variable Models 16

4 Functions of Random Variables 76

1 Introduction

1.1 Identify engineering and scientific applications where uncertainties can be ignored Explain

Solution: Generally, uncertainties can be ignored if their magnitudes are significantly smaller that the inal (average) magnitudes For example, for a circular cylindrical column of diameter D, with uncertainties

nom-in the diameter of 0.001D, then it is likely that we can ignore the uncertanom-inty nom-in the nomnom-inal diameter whencalculating the quantities based on diameter

1.2 Identify engineering and scientific applications where uncertainties cannot be ignored Explain

Solution: If the magnitude of the uncertainty is significant compared to the nominal (average) value then

it cannot be ignored For example, for a circular cylindrical column of diameter D, with uncertainties inthe diameter of 0.1D, then we cannot ignore the uncertainty in the nominal diameter when calculating thequantities based on diameter It may also be that the application/the design requires a very tight toleranceand therefore uncertainties must always be incorporated into the design

1.3 Discuss how an engineer may ascertain whether uncertainties are important or can be ignored in

an analysis and design Use examples in your discussion

Solution: A design engineer will perform the design using alternate values of parameters that have tainties and then ascertain whether the final design is significantly different In the simple example of thedesign of a column, various diameters will be used in order to determine the necessary buckling strength.Similarly, if there are uncertainties in the forces on a structure, then high and low values might be used to becertain that the design will resist those forces The use of probability makes this process more streamlined

uncer-1.4 How can Miner’s rule for fatigue damage be extended to cases where the order of loading cycles isimportant? Explain with or without using equations

Solution: If the order of the loading cycles is important then higher weighting can be given to a particularordering of stress cycles The summation has to be identical to the ordering For example,

1.6 Which variables or parameters in Equations 1.3 and 1.4 are better assumed to be random variables?Explain your choices.

Solution: CD, CI, u, ˙u are all likely candidates to be random variables The coefficients CD, CI are imentally determined and have significant scatter associated with their values, which depend on the fluidspeed Similarly, fluid speed and acceleration u, ˙u are also random

exper-2 Events and Probability

Section 2.1: Sets

2.1 Give three examples of impossible events from everyday life

Solution: Examples may include the sun rising in the West and setting in the East, water flowing upstream,and time going backwards

2.2 Give three examples of impossible events from fluids engineering

Solution: Examples may include fluid with no viscosity, higher density fluid rising, and pressure headdecreasing with increasing column height

2.3 Give three examples of impossible events from materials engineering

Solution: Examples may include temperature dipping below 0 degrees Kelvin, massless material, sible metal

inexten-2.4 Give three examples of impossible events from strength of materials

Solution: Examples may include the proportional limit of a material exceeding the yield stress, fracturebeing a reversible event, plastic deformation that is recoverable and not permanent, the sum of the forces in

a static structure being not zero, and a linear isotropic material with Poisson’s ratio greater than 0.5.2.5 Give three examples of impossible events from mechanical vibration

Solution: Examples may include a structure that oscillating without any forces or initial conditions on it,for the same structure, a stiffer structure oscillating slower, a structure oscillating with unlimited speeds, adamped oscillator having a non-decaying amplitude in free vibration, a spring with an infinite stiffness, afree pendulum swinging indefinitely without the addition of energy, eliminating low frequency vibration in

a baseball bat when hitting a ball outside the sweet zone

2.6 Give three examples of impossible events from thermal engineering

Solution: Examples may include decreasing entropy without energy, increasing energy in a closed system,and radiation without temperature

2.7 Give three examples of certain events from everyday life

Solution: Examples may include the sun rising tomorrow, having to fill out tax forms every year, the suncausing the evaporation of 1 trillion tons of water daily, being summer time in the northern hemisphere when

it is winter time in the southern hemisphere, applying a force on an object results in an equal and oppositeforce, and all creatures needing energy to survive

2.8 Give three examples of certain events from fluids engineering.

Solution: Examples include a faster incompressible flow in a smaller diameter pipe, denser fluid sinking,flow exerting a force on impeller

2.9 Give three examples of certain events from materials engineering

Solution: All materials have imperfections, a higher pressure results in a higher boiling point, a constrainedbeam has a higher buckling load

2.10 Give three examples of certain events from strength of materials

Solution: An axial force on a column leads to compression of that column; an elastic body will alwaysdeform if forces act on it; any beam subjected to an axial force will eventually fail; in polymer composites,nanoparticles do not create large stress concentrations and thus do not compromise the ductility of thepolymer

2.11 Give three examples of certain events from mechanical vibration

Solution: If a body is oscillating, then there must be inertia; if a body is oscillating, then there is a restoringforce; there is an energy exchange mechanism between the kinetic and potential energies; when hit at thesweet zone, the baseball bat will not be excited at its natural frequencies and therefore there will not beenergy loss due to these frequencies vibrating, resulting in more energy being transmitted to the ball.2.12 Give three examples of certain events from thermal engineering

Solution: Heating an object results in a higher temperature Heating a metal object results in expansion.Objects expand and contract at different rates

2.13 Give an example of complementary events from everyday life

Solution: winning/losing a game; catching/missing a train; night/day;

2.14 Give an example of complementary events from fluids engineering

Solution: The flow may be considered laminar or turbulent

2.15 Give an example of complementary events from materials engineering

Solution: The material extends linearly or nonlinearly In other words, the material may or may not obeyHooke’s law

2.16 Give an example of complementary events from strength of materials.

Solution: Examples: A material deforming elastically/plastically; a tensile test yielding expected/unexpectedresults; a part failing/passing a drop test

2.17 Give an example of complementary events from mechanical vibration

Solution: An increase in stiffness and lower frequency

2.18 Give an example of complementary events from thermal engineering

Solution: The entropy of the system is increasing/decreasing

2.19 Given the three events:

X = {odd numbers}

Y = {even numbers}

Z = {negative numbers} ,obtain the following:

(a) X ∪ Y (b) X ∩ Y(c) X (d) Y(e) Z (f) Y ∩ Z

Solution: We are given the following:

X = {odd numbers}

Y = {even numbers}

Z = {negative numbers} Then,

X ∪ Y = {All numbers (odd numbers + even numbers) }

2.20 Extend Example 2.5 to the case where there are three shafts connecting two clutches (instead oftwo shafts connected to one clutch) The shafts are numbered from left to right as 1, 2, 3 The failure ofthe drive train is defined as the failure of either of the three shafts, with events defined by E1, E2, and

E3, respectively Assuming the clutches will not fail, find the following events:

(a) failure of the drive train,

(b) no failure of the drive trains, and

(c) show an illustration of de Morgan’s rule

Solution: Define the following events:

Section 2.2: Probability

2.21 Consider Figure 2.8 where dav= 50 mm and 50 shafts are manufactured From the measurements

we observe that

25 have the diameter dav

10 have the diameter 1.01 dav

6 have the diameter 1.02 dav

5 have the diameter 0.99 dav

4 have the diameter 0.98 dav.Sketch the frequency diagram showing appropriate numbers along the axes Using the frequency inter-pretation for probability, calculate the probability of occurrence for each shaft size and verify that thesum of these probabilities equals 1

Solution: We are given that dav= 50 mm and the number of shafts manufactured is 50 Also, the followingmeasurements are observed:

25 have the diameter dav

10 have the diameter 1.01 · dav

6 have the diameter 1.02 · dav

5 have the diameter 0.99 · dav

4 have the diameter 0.98 · dav.From this, we can draw a frequency diagram, as shown in Figure 1 Then,

Figure 1: Frequency Diagram of the Number of Shafts

2.23 Suppose Pr(E1) = 0.20, and Pr(E2) = 0.30.

(a) If E1and E2 relate to a particular process, are any events not accounted for here? Why?

(b) If Pr(E1∪ E2) = 0.90 are these processes mutually exclusive? Why?

(c) If Pr(E1∪ E2) = 0.50, then what is the value of Pr(E1E2)?

Solution: The following data is given

Pr{E1} = 0.20Pr{E2} = 0.30

(a) We know that the probability of any certain event is 1 If E1and E2 are the only two events occurring,then the probability that any event will definitely occur is given by

Pr{E1∪ E2} = Pr{E1} + Pr{E2} − Pr{E1∩ E2}

Even if E1and E2are assumed to be mutually exclusive, that is E1∩ E2= 0, the probability that E1or E2

or both occur is given by

Pr{E1∪ E2} = Pr{E1} + Pr{E2}

= 0.20 + 0.30

= 0.50

Here even if both events occur, the maximum probability is 0.50, which indicates that some event has beenleft out

(b) If Pr{E1∪ E2} = 0.90, then there is an inconsistency, as follows:

Pr{E1∪ E2} = Pr{E1} + Pr{E2} − Pr{E1∩ E2}

So, Pr{E1∩ E2} = Pr{E1} + Pr{E2} − Pr{E1∪ E2}

= 0.20 + 0.30 − 0.90

= −0.40, which is impossible

The maximum probability of the union is Pr(E1∪E2) = 0.50 Then the two events are mutually exclusivebecause

Pr{E1∪ E2} = Pr{E1} + Pr{E2} − Pr{E1∩ E2}

So, Pr{E1∩ E2} = Pr{E1} + Pr{E2} − Pr{E1∪ E2}

= 0.20 + 0.30 − 0.50

= 0, which indicates that E1and E2 are mutually exclusive

If the union is less than Pr = 0.5, the intersection will have a positive probability, indicating that thetwo events are not mutually exclusive

(c) To find Pr(E1E2) given Pr(E1∪ E2) = 0.50,

Pr{E1∪ E2} = Pr{E1} + Pr{E2} − Pr{E1∩ E2}

So, Pr{E1∩ E2} = Pr{E1} + Pr{E2} − Pr{E1∪ E2}

= 0.20 + 0.30 − 0.50

= 0.00

2.24 Suppose Pr(A) = 0.5, Pr(A|B) = 0.3, and Pr(B|A) = 0.1, calculate Pr(B).Solution: We are given the following:

2.25 Suppose Pr(A) = 0.5, Pr(A|B) = 0.5, and Pr(B|A) = 0.1, calculate Pr(B) What can be concludedabout the statistical relationship, if any, between A and B.

Solution: We are given the following:

Since Pr {A ∩ B} = Pr {A} Pr {B} , the two events are statistically independent

2.26 Continuing Example 2.17, let G be the event of rolling a 3 and A1,A2, A3be the events of selectingdie 1, 2, or 3, respectively.

(a) Find the probability of rolling a 3 if one die is selected at random Use the theorem of total probability.(b) Determine the probability that die 2 was chosen if a 3 was rolled with the randomly selected die

Solution: It is given that G is an event of rolling a 3 and A1, A2, A3 equal the event of selecting die 1,2,

or 3, respectively

(a) Since the die is selected at random,

Pr(A1) = Pr(A2) = Pr(A3) = 1/3

From the table,

Pr(G|A1) = 1/12, Pr(G|A2) = 1/6, Pr(G|A3) = 1/6

Substituting these into the total probability theorem, we obtain the probability of rolling a 3,

Pr(G) = Pr(G|A1) Pr(A1) + Pr(G|A2) Pr(A2) + Pr(G|A3) Pr(A3)

=

1

12·13

+

1

6·13

+

1

6· 13



= 5

36 = 0.139(b) We need to evaluate Pr(A2|G) Using Bayes’ rule,

Pr(A2|G) = Pr(G|A2) Pr(A2)

= 2

5= 0.4.

2.27 Two cables are used to lift load W Normally, only cable A will be carrying the load; cable B isslightly longer than A, so it does not participate in carrying the load But if cable A breaks, then Bwill have to carry the full load until A is replaced The probability that cable A will break is 0.02 Theprobability that B will fail if it has to carry the load by itself is 0.30.

A = cable A breaks

B = cable B breaksPr(A) = 0.02

Pr(B|A) = 0.30 = probability B breaks if A already broke

(a) What is the probability that both cables will fail?

(b) If the load remains lifted, what is the probability that none of the cables have failed?

(a) The probability that both cables will fail is

Pr(AB) = Pr(BA) = Pr(B|A) Pr(A) = 0.3 · 0.02 = 0.006

(b) The condition that the load remains lifted is expressed as A ∪ B, but if A happened then B woulddefinitely occur, that is, Pr(AB) = Pr(A),

Pr(AB|A ∪ B) = Pr(A|A ∪ B) = Pr[A ∩ (A ∪ B)]

2.28 Consider the drive train Example 2.5 from a different perspective The drive train consists of a rotor

R and turbine blades B How well the system operates depends on the precision of the manufacturedcomponents Testing of the individual components by the manufacturer yields the following information:

0.1% of R have imperfections0.01% of B fail

Also, it is determined that if R has imperfections, then the blades B are 50% more likely to faildue to the additional vibration forces that result Determine the probability that the system will passinspection

Solution: Let ER = failure event (imperfection) for R, EB = failure event for blade We are given thefollowing:

Pr(ER) = 0.001Pr(EB) = 0.0001Pr(EB|ER) = 1.5 Pr(EB|ER)

Then,

Pr(failure) = Pr(ER∪ EB)

= Pr(ER) + Pr(EB) − Pr(ER∩ EB)

= Pr(ER) + Pr(EB) − Pr(EB|ER) Pr(ER)

Use Theorem of Total Probability:

Pr(EB) = Pr(EB|ER) Pr(ER) + Pr(EB|ER) Pr(ER)0.0001 = Pr(EB|ER) × 0.001 + 1

1.5Pr(EB|ER)(1 − 0.001)Pr(EB|ER) = 0.00015

3 Random Variable Models

Section 3.1: Probability Distribution Function

3.1 Can the following functions in Figures 3.41 (a), (b), (c) represent possible probability distributionfunctions? Explain

Figure 3.41: Possible probability distribution functions

Solution: Figures (b) and (c) can be possible distributions, but Figure (a) cannot be because F (x) cannotdecrease in value as x increases

3.2 Show that the expected value of a constant equals that constant.

Solution: Let the random variable, X, take a constant value, c Then, the probability density function isthe Dirac delta function,

fX(x) = δ (x − c) Then, the expected value is given by

3.3 Random variable X is mixed, containing continuous ranges as well as discontinuities Its cumulativeprobability distribution function is

(d) Pr(2 < X ≤ 4) = Pr (X ≤ 4) − Pr (X ≤ 2) = 1 − 11/12 = 1/12

Figure 2: The cumulative density function for Problem 3.3

3.4 Suppose X is distributed as indicated in Figure 3.42 All lines are straight except for the exponentialcurve in the range 1 < x < 3.

Figure 3.42: Generic probability distribution function FX(x)

Express FX(x) algebraically and then find the following probabilities:

(a) Pr(X = 1/3) (b) Pr(X = 1) (c) Pr(X < 1/3)

(e) Pr(X < 1) (f) Pr(X ≤ 1)(g) Pr(1 < X ≤ 2) (h) Pr(1 ≤ X ≤ 2) (i) Pr(X = 1 or 1/4 < X < 1/2)

Solution: FX(x) is a straight line except between 1 < x < 3, where it may be assumed to be parabolic.Then,

8x +3 8

1

for 0 ≤ x < 1for 1 ≤ x < 3for 3 ≤ x < 5for x ≥ 5From the figure we find:

Section 3.2: Probability Density Function

3.5 (a) Verify and sketch the following probability density function,

(b) Find and sketch F (x)

Solution: The probability density function is shown below It is a valid probability density function because

f (x) is non-negative for all valid x and the area under f (x) is 1 This can be seen in Figure 3(a) To findthe functional form of the cumulative density function, we can express the probability density as

1

4x −ˆ 14 dˆx = 18x2−14x +18 1 ≤ x < 3x

Figure 3: Problem 3.5

3.6 Random variable X is continuous and its values are governed by fX(x) or FX(x) If random variable

Y = 2X, derive an expression for fY(y) and FY(y)

Solution: Consider a simple case where fX(x) = 1 for 0 ≤ x < 1 The corresponding cumulative distribution

is FX(x) = x for 0 ≤ x < 1 and FX(x) = 1 for x ≥ 1 We wish to use another random variable, Y, that isrelated to our original random variable, X, by Y = 2X

The probability does not change even when the random variables change That is,

Fy(y) = FX(2x) For example, Pr (x < 1/2) = 1/2 = FX(x = 1/2) = FY (y = 2x = 1) We apply the same principle for theincremental probability,

3.7 Continuous random variable X is governed by the probability density function

For Y = X2, derive fY(y) and FY(y)

Solution: The cumulative distribution is

is dropped because F (x) is defined for positive x only

3.8 For each of the following functions, state why they are valid or invalid probability density functions.

If arbitrary constants are present, evaluate them Sketch the probability density functions Then, findthe cumulative distribution functions and sketch

1

10dx = 1.

The sketch of probability density function is shown in Figure 4

Figure 4: Probability Density Function f (x) = 1

10, 0 ≤ x ≤ 10The cumulative distribution function is found by integrating the density functions as follows:

1

10dξ =

x

10.The sketch of the cumulative distribution function is shown in Figure 5

(b) The probability density function

Figure 5: Cumulative Distribution Function F (x) = x

10, 0 ≤ x ≤ 10

 ∞

−∞

f (x)dx = 1c

The sketch of probability density function for λ = 1 is shown in Figure 6

Figure 6: Probability Density Function f(x) = ce−λx, x ≥ 0, λ = 1

The cumulative distribution function is found by integrating the density functions as follows

λe−λξdx

= 1 − e−xλ, x > 0

The sketch of cumulative distribution function for λ = 1 is shown in Figure 7

Figure 7: Cumulative Distribution Function F (x) = 1 − e−λx, x > 0, λ = 1

(c) The probability density function

 ∞

−∞

f(x)dx = 1

 10 0



c + 120



dx = 1

cx + x20

 10

0 = 1

c = 1

20.The shape of the sketch of the probability density function and distribution function is same as for (a)

3.9 For each of the following applications, sketch an appropriate probability density function for therespective random variable Explain your reasoning.

(a) A box is at rest on a horizontal table A force is applied on the end of the box and the boxeventually moves as the force is increased, overcoming the static friction force Sketch a reasonableprobability density function for the friction force

(b) A ball is dropped from rest very high in the atmosphere As it falls it is subjected to wind gustsfrom all directions The gust magnitudes comprise a random variable Based on the acceleration of theball, sketch a reasonable probability density function of the speed

Solution: (a) We expect a range of values of the friction force of uneven probabilities

Figure 8: Problem 3.9(a)(b) We know that the ball will accelerate from zero to a peak value Without the fluid drag force, thetheoretical maximum value is vmax=√

2gh With fluid drag force, the ball will reach its terminal velocity

Figure 9: Problem 3.9(b)

3.10 For each of the following applications, sketch an appropriate probability density function for therespective random variable Explain your reasoning.

(a) A turbine (rotating machine) is started from rest and increases its speed to its final operatingspeed The rotational speed of the turbine is a random variable due to very small variability in voltageand friction in the shafts Sketch a reasonable probability density function for the final rotational speed.(b) A fair coin is tossed n times Draw a reasonable probability mass function for n = 1, 10, 1000

Solution:

(a) We expect a final rotation speed that is quite precise, perhaps with very small variance to account forsmall variability in friction and voltage

(b) The coin can show either head or tail Let us find and plot the probability mass function for the number

of heads The horizontal axis is the number of heads, and the vertical axis is the probability of obtainingthat many heads When the coin is tossed once, the possible number of heads is 0 (tail showed) or 1 (headshowed) with respective probabilities of 1/2 and 1/2 This is shown in the first figure below

Figure 10: Problem 3.10(b)

If the coin is tossed twice, the possible number of heads is 0 (TT), 1(TH or HT), or 2 (HH) The respectiveprobabilities are 1/4, 1/2, and 1/4 The probabilities are found by

Pr (r heads) = number of ways to get r number of heads

total number of different outcomes for n toss.For example, the number of ways to get 1 head is 2 (HT, TH), and the total number of outcomes is 4 (HH,

HT, TH, TT) Therefore, the probability of obtaining 1 head when the coin is tossed twice is 2/4 or 1/2.Another way of writing the probability is

Pr (r heads) = nCr

2n The number of heads has a binormal probability mass function That is, the probability of turning r headsis

Pr (r heads) =

nr



pro(1 − po)n−r,where pois the probability of obtaining a head at each toss (po= 1/2), and 1 − po is the probability of notobtaining a head at each toss Note that the binomial coefficient (nr) is same as the combination,nCr Then,

we find

Pr (r heads) =

nk

 12

r12

n−r

= nCr

2n

For n = 2 (two tosses) and r = 1 (1 head), the probability is Pr (1 head) =2C1/(22) = 2/4 = 1/2 See thecorresponding probability mass function in Figure 10.

If the coin is tossed ten times (n = 10), the possible number of heads is either 0,1, , or 10 The probability

of obtaining no head is10C0/210 = 1/210 The probability of obtaining 1 head is 10C1/210 = 10/210 Theprobability mass functions for n = 10 and 1000 are obtained in Matlab

Figure 11: Problem 3.10(b) The probability Mass density function when n=10 and 1000

In Section 3.6.1, we found that the mean of the binomial distribution is npoand the variance is npo(1 − po)

In our case, the mean number of heads is half of the total number of toss (if you tossed the coin 10 times,you expect to see about 5 heads) The variance is n/4 It is interesting to note that binomial probabilitymass function approaches the normal distribution (with the same mean and variance) The solid lines inFigure 11 is the corresponding normal distribution,

f (r) = 1

σ√2πexp

Section 3.4: Mathematical Expectation

3.11 Find the expectation and variance of the following functions and data sets:

and the variance is defined as

V ar{X} = E{X2} − (E{X})2.(a) The mathematical expectation is

E{X} =

 4 1

4 1

= 2.5,and the variance is

V ar{X} = E{X2} − (E{X})2

=

 4 1

x21

3dx −

 4 1

1

= 7.0,and the variance is

V ar{X2} = E{X4} − (E{X2})2

=

 4 1

x41

3dx −

 4 1

(c) If xiare the test results and pX(xi) are the probability weights, then using the expected value for discretevariables,

2.5 ×102

+



4 ×101

+

4.3 ×101

+

4.9 × 101

+



7 ×101

+



10 ×102

+

= 10.6

3.12 Derive Equation 3.12.

Solution: This is the equation for the variance of X Expanding the square under the integral and consideringeach term individually, the result is obtained

3.13 Continuous random variable X is governed by the probability density function

exdx

= ex|10= e − 1

3.14 For the random variable X we are given the following probability density function,

f (x) = c

x2, x ≥ 10

Derive the cumulative distribution function, F (x), and sketch both functions Find the expected value

of X and the standard deviation of X What is the coefficient of variation? Is it a large number?

Solution: Given the following probability density function,

f (x) = c

x2, x ≥ 10,first find the value of c :

The probability density function is shown in Figure 12

Figure 12: Probability Density Function f (x) = 10x2, x ≥ 10Then cumulative distribution function is the integral,

F (x) =

 x 10

10

ξ2dξ = 1 −10

x.The cumulative distribution function is shown in Figure 13

f (x) = c

x2, 10 ≤ x ≤ 1000

Figure 13: Cumulative Distribution Function F (x) =x

1010ξ 2dξ = 1 −10

x, x ≥ 10

Then,

 1000 10

10.101

ξ2 dξ = 1 −10.101x Now,

E{X} = µ =

 1000 10

x10.101

x2 dx = 46.517,E{X2} =

 1000 10

x210.101

x2 dx = 10000.0,and so the variance

V ar {x} =

 1000 10

x210.101

x2 dx −

 1000 10

x10.101

x2 dx

2

= 7836.2,with the standard deviation

σ =√7836.2 = 88.522and the coefficient of variation

δ = σ

µ =

88.52246.517 = 1.903.

The large coefficient of variation indicates that the variability relative to the mean is very large

(a) A box is at rest on a horizontal table A force is applied on the end of the box and the boxeventually moves as the force... moves as the force is increased, overcoming the static friction force Sketch a reasonableprobability density function for the friction force

(b) A ball is dropped from rest very high in the... number of different outcomes for n toss.For example, the number of ways to get head is (HT, TH), and the total number of outcomes is (HH,

HT, TH, TT) Therefore, the probability of