3 3 Stu

 

Topic 3-3

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Solved problems Problem 3-3-1

The profit (in thousand dollars) of a construction job is described by the following probability density function: f X(x) 0.02

0.01

Profit, x ($1000) -10

(a)

0

10

20

30

40

50

60

70

What is is the probabil probability ity that that the contrac contractor tor will will lose money money on this this job? job? (ans. 0.2) 0.2)

(b) Suppose Suppose the contract contractor or declares declares that he has has made money money on this job; job; what is the the probability probability that that his  profit was more than 40 thousand thousand dollars? (ans. 0.1875) Solution: Let X be the profit (in $1000) from the construction job. (a) P(lose money) = P( X < 0) = Area under the PDF where x is negative = 0.02× 10 = 0.2 (b) Given event is  X > 0 (i.e. money was made), hence the conditional probability, probability, X > 40 | X > 0) P(  X  = P( X > 40 ∩   X  X > 0) ÷ P(  X  X > 0) = P( X > 40) ÷ P(  X  X > 0) Let's first calculate P( X > 40): comparing similar triangles formed by the PDF and the  x-axis (with vertical edges at x = 10 and at x = 40, respectively), we see that P( X > 40) = Area of smaller triangle 2 70 − 40     = × Area of larger triangle    70 − 10  

= 0.52 [(70 - 10)× (0.02) ÷ 2] = 0.015 Hence the required probability P( X > 40 | X > 0) is 0.015 ÷ [10× 0.02 + (70 - 10)× (0.02) ÷ 2] = 0.015 ÷ 0.08 = 0.1875

 

Problem 3-3-2 The storm run-off X (in cubic meters per second, cms) from a subdivision can be modeled by a random variable with the following probability density function: f X(x) =

c( x  x -

0

 x 2

for 0 ≤  x  x ≤ 6

) ;

6

;

otherwise

Determine the constant c and sketch the density function. (ans. 1/6)

(a)

(b) The run-off run-off is carried carried by a pipe pipe with a capacity capacity of 4 cms. cms. Overflow Overflow occurs occurs when the the run-off exceeds exceeds the pipe capacity. If overflow occurs after a storm, what is the probability that the run-off in this storm is less than 5 cms? (ans. 0.714) (c) An engineer engineer conside considers rs repla replacing cing the the current current pipe by a larger larger pipe having having a capacity capacity of of 5 cms. Suppose Suppose there is a probability of 60% that the replacement would be completed by the next storm. What is the  probability of overflow in in the next storm? (ans. 0.148) Solution: ∞

Applying the normalization condition

∫  f  ( x)dx = 1  X 

−∞

6

 x 2

∫ c( x − 6

⇒

0

⇒  c =

6

 x 2  x 3  )dx = 1 ⇒  c  −  =1  2 18  0

18 9 × 36 − 6 3

= 1/6

(a) To avoid repeating integration, integration, let’s work with the CDF of X, which is FX( x  x) =

1  x 2



6 2 2

−

 x 3 



18 

3

 x )/108 = (9 x –  x

(for x  between 0 and 6 only)

Since overflow already occurred, the given event is X > 4 (cms), hence the conditional probability P(X < 5 | X > 4) = P(X < 5 and X > 4) / P(X > 4) = P(4 < X < 5) / [1 – P(X ≤ 4)] = [FX(5) - FX(4)] / [1 - F X(4)] = [(9× 52 – 53) – (9× 42 – 43)]/ [108 – (9 × 42 – 43)] = (100 – 80) / (108 – 80) = 20/28 = 5/7 ≅  0.714 (b) Let C denote “completio “completion n of pipe replaceme replacement nt by the next storm”, where where P(C) = 0.6. If C indeed indeed occurs, overflow means X > 5, whereas if C did not occur then overflow would correspond to X > 4. Hence the total  probability of overflow is is (with ’ denoting com compliment) pliment) P(overflow) = P(overflow | C)P(C) + P(overflow | C’)P(C’) = P(X > 5)× 0.6 + P(X > 4) × (1 – 0.6)

 

= [1 - FX(5)]× 0.6 + [1 - FX(4)]× 0.4 = (1 – 100/108)× 0.6 + (1 – 80/108)× 0.4 ≅  0.148 Problem 3-3-3 Severe snow storm is defined as a storm whose snowfall exceeds 10 inches. Let X be the amount of snowfall in a severe snow storm. The cumulative distribution function (CDF) of X for a given town is:  F X( x  x) = 1 - (

10  x

)4

for  x  x ≥ 10

and the CDF is zero for   xx < 10. (a)

Determ Determine ine the the medi median an of of X. (ans. (ans. 11.9 11.9))

(b)

What is the expected amount of snowfall in a severe snow storm? ((ans. ans. 13.3)

Suppose a disastrous snow storm is defined as a storm with over 15 inches of snowfall. What  percentage of the severe snow storms storms are disastrous? (ans. 0.2) (c)

(b) Suppose the probabi probability lity that the town will experience 0, 1 and 2 severe snow storm in in a year is 0.5, 0.4 and 0.1 respectively. Determine the probability that the town will not experience a disastrous snow storm in a given year. Assume the amounts of snowfall between storms are statistically independent. (ans. 0.885) Solution: X , xm, is obtained by solving the equation (a) The median of   X  equation that defines xm, P(  X  X  ≤   xxm) = 0.5 F X( x  xm) = 0.5 ⇒  F  ⇒ 1- (10 ÷  x  xm)4 = 0.5 ⇒  xxm  ≈ 11.9 (inches)

(b) First, obtain the PDF of   X  X :  f X( x  x) = d F X/d x  = 4× 104 x-5 for  x  x ≥ 10, and zero elsewhere.

Hence the expected amount of snowfall in a severe snow storm is ∞

E(  X  X ) =

∫  xf

 X 

( x )dx  

−∞ ∞

=

∫ 4 × 10  x 4

−4

dx

10

=

−

4 3

∞

10 4 [ x − 3 ] 10  

= 4÷ 3× 10 ≈  13.3 (inches) (c) P(disastrous severe snowstorm) = P( X > 15)

 

= 1 - P(  X  X  ≤ 15) = 1 - [1 - (10/15) 4] = (2/3)4 ≈  0.2 ∴About 20% of snow storms are disastrous. (d) Let N denote "No disastrous snow storm in a given year". Also, let  E 0, E 1, E 2 denote the respective events of 

experiencing 0,1,2 severe snow storms in a year. In each event, the (conditional) probability of   N  N can be computed: P( N | E 0) = 1 (if there's no severe snow storm to begin with, definitely there won't be any disastrous one); P( N | E 1) = P(  X  X  ≤ 15) = 1 - (10/15)4 = 1 - (2/3)4; P( N | E2) = [P(  X  X  ≤ 15)]2 = [1 - (2/3)4 ]2 (statistical independence between storms);  Noting that E 0, E 1 and E 2 are m.e. and c.e., the total probability of   N  N can be computed: P( N ) = P( N   N | E 0)P(  E  E 0) + P( N | E 1)P( E 1) + P( N | E 2)P(  E  E 2) 4 4 2 = 1× 0.5 + [1 - (2/3) ]× 0.4 + [1 - (2/3) ] × 0.1 ≈  0.885

Problem 3-3-4 Suppose X is a random variable defined as X=

final cost of  project estimated cost of  project

Which has a probability density function (PDF) as follows: f ollows:    x < 1  0  2  f   X  ( x) = 3 /  x   1 ≤  x ≤ a  0    x > 1.5 

(a) Determ Determine ine the the value value of a (ans. (ans. 1.5) (b) What is the probability that the final cost of a project will exceed its estimated estimated cost by 25%? (ans. 0.4) 0.143) 143) (c) Determine the mean value and standard deviation of X. (ans. 1.216, 0. Solution: Let F be the final cost (a random variable), and C be the estimated cost (a constant), hence  X = F / C 

is a random variable. (a) To satis satisfy fy the normalizat normalization ion conditi condition, on, a

hence

a

 − 3 = 3 − 3 = 1 dx = , 2   1  x a   x 1

∫ 

3

 

a = 3/2 =1.5 (b) The given event is “ F exceeds C  by more than 25%”, i.e. “ F > 1.25C”, i.e. “ F / C > 1.25”, whose

 probability is ∞

∫  f  ( x)dx  

P(  X  X > 1.25) =

 X 

1.25 1.5

1.5

∫   x3 dx =  − x3 

=

 

2

1.25

1.25

= -2 – (- 2.4) = 0.4 (c (c)) The The mea mean, n, 1.5

E(  X  X ) =

3

∫   x  x

2

dx = [ 3 ln x] 11.5  ≅  1.216,

1

while 1.5 2

E(  X  X  ) =

∫ 

 x 2

1

3  x 2

dx = 3(1.5 – 1) = 1.5,

with these, we can determine the variance Var(  X  X ) = E( X 2) – [E( X ))]]2 = 1.5  – 1.2163953242 = 0.020382415

∴σ X  =

≅  0.143

0.020382415

Problem 3-3-5 The duration of a rainstorm at a given location is described by the ffollowing ollowing probability densi density ty function

 x / 8   f  X  ( x) = 2 / x 2 0 

0 2)  P ( X  > 2)

=

<  X  < 3)  P ( X  > 2)

 P (2

≅  3.106

=

2

∫  x

2

2

2

1−

 x

∫ 8 0

−

dx

= dx

1−

2  x

3

2

 x 2 

2

 16  0

=

1− 2/3 1 − 1/ 4

= 4/9

 

Exercises Exercise 3-3-1 The annual maximum snow load X (in lb/ft 2) on buildings with a flat roof in a northern location can be modeled  by a random variable with the the following cumulative cumulative probability dist distribution ribution function

0 4   F  X  ( x) =   10   1 −   x    

for  x ≤ 10 for  x > 10

(a) (a)

De Dete term rmin inee tthe he me mean an valu valuee an and d med media ian n of of X. X.

(b)

An engineer recommends a design snow of 30 lb/ft 2 for a building. What is the probability of roof failure (design load being exceeded) in a given year? For the first time in the 5th year?

(c)

If roof ffailur ailuree occurs occurs in 2 or or more of of the next next 10 years, years, the the design design eng engineer ineer will will face face a penalty penalty.. What is his chance of going through the next 10 years without a penalty?

Solution:

Exercise 3-3-2 The cumulative distribution function (CDF) of the daily progress in a tunnel excavation construction project is given below.

F (x) X

1.0

0.6 0.4

0

2

6

10

x, length in meters

(a)

What What is the proba probabil bility ity th that at the progr progress ess in tunne tunnell excavat excavation ion will will be betwee between n 2 and 8 meter meterss on a given day?

(b)

Determine the median length of tunnel excavated on a given day.

(c)

Plot the probab probability ility density density functio function n (PDF) (PDF) of the the daily daily progr progress ess in tunnel tunnel excavat excavation. ion.

 

(d)

Determine the mean length of tunnel excavated on a given day.

Solution: CDF, FX(x) = P(X ≤ x) and it is a continuous function. Hence (a)  Note that we are given the CDF, P(2 ≤ X ≤ 8) = FX(8) – FX(2) = 0.8 – 0.4 =

0.4

(b) The median median is defined by where where F(x) = 0.5, which occurs occurs somew somewhere here along along x = 2 and x = 6; the constant constant slope there gives (0.5 – 0.4) / (x m – 2) = (0.6 – 0.4) / (6 – 2) ⇒ xm = 2 + 2 = 4 (c) The PDF is the derivative (i.e. slope) of the CDF CDF,, hence it is the following piecewise constant function:

f X(x) 0.2 0.1 0.05

0.05 2

6

10

x

Exercise 3-3-3 The maximum load S in tons on a structure is modeled by a continuous random variable S whose cumulative distribution function CDF is given as follows:

 s 3 0 s 2  F S  ( s ) = +  8641 48 

for  s ≤ 0 for 0 12

(a)

Determine the mode and mean value of S.

(b) (b)

Th Thee stren strengt gth h R of the struc structu ture re can be mode modele led d by a di disc scre rete te random random varia variabl blee with with the follo followi wing ng  probability mass function. function.

 

PR (r) 0.7

0.3 10

r, tons

13

Determine the probability of failure, i.e., the probability that loading S is greater than the strength R.

Exercise 3-3-4 The cumulative distribution function (CDF) of an engineering variable X is: for  x < 2 for  x ≥ 2

0  F  X  ( x) =  1 − e −0.5( x − 2)  (a) (a)

Det eter erm mine ine tth he med media ian n of of X

(b)

Determ Determine ine the the proba probabil bility ity of X being being betwe between en 1 and 4.

(c)

Determ Determine ine the the proba probabil bility ity densit density y ffunc unctio tion n (PDF) (PDF) of X. X.

Exercise 3-3-5 The annual maximum flood level of a river at a given measuring location has the following cumulative distribution function (CDF):

0    F  X  ( x) = − 0.0025 x 2 + 0.1x  1 

 x ≤ 0

 

0 20

(a)

Determine and sketch the probability density function f  X  (x).

(b)

What is the median annual maximum flood level?

Exercise 3-3-6 The size (in mm) of a crack in a structural weld is described by a random variable x with the following  probability density density function:

 

 x / 8   f  X  ( x) =  1 / 4 0 

     

0 <  x ≤ 2 2 <  x ≤ 5 elsewhere

(a) (a)

Sk Sket etch ch the the P PDF DF aand nd CDF CDF on on a p pie iece ce of of grap graph h pa pape per. r.

(b)

Determine the mean crack size.

(c)

What What is the pro probab babili ility ty that that a cra crack ck will will be be small smaller er th than an 4mm? 4mm?

(d)

Determine the median crack size.

(e)

Suppos Supposee there there are four four cracks cracks in the weld. weld. What What is the probab probabili ility ty that that only 1 of these these 4 cracks cracks is larger  larger  than 4mm? (ans. P(N=1|n=4,p=1/4) = 0.422)

Exercise 3-3-7 The total load, X (in tons), on the roof of a building has the following Probability Density Function

24 / x 3  f  X  ( x) =  0

3 ≤  X  ≤ 6   elsewhere

(a)

Determine and plot the cumulative distribution function of X, i.e., FX (x).

(b (b))

Wh What at is is the the ex expe pect cted ed tot total al load load??

(c)

Determ Determine ine the the coef coeffic ficien ientt of v vari ariati ation on of the the tota totall load. load.

(d)

Suppos Supposee the roof roof could could carry only only 5.5 ton befor beforee collaps collapse. e. What What is the probabi probabilit lity y that the the roof will will collapse?