5-1.5 MORE THAN TWO RANDOM VARIABLES - Applied Statistics and Probability for Engineers, 6th E

More than two random variables can be defi ned in a random experiment. Results for multiple random variables are straightforward extensions of those for two random variables. A summary is provided here.

The joint probability distribution of random variables X , X , X ,

1 2 3

… , X

can be specifi ed with a method to calculate the probability that X , X , X ,

1 2 3

… , X

assume a value in any region Rof p-dimensional space. For continuous random variables, a joint probability density function f

X X1 2…Xp

( x , x ,

1 2

… , x

) is used to determine the probability that ( X , X , X ,

1 2 3

… , X

)

^∈

^{R by the}

multiple integral of f

X X1 2…Xp

( x , x ,

1 2

… , x

) over the region R.

A joint probability density function for the continuous random variables X , X , X ,

1 2 3

… , X ,

denoted as f

X X1 2…Xp

( x , x ,

1 2

… , x ,

) satisfi es the following properties:

(1) f

X X1 2_…Xp

( x , x ,

1 2

… , x

) ^≥ 0

(2)

???

f

X X1 2_…Xp

x , x ,

1 2

, x

dx dx

1 2

dx

1

∞

∫ ∫ ∫ ( … ) … ⁼

2 2 2

(3) For any region B of p-dimensional space,

P _⎡⎣ ( X , X ,

1 2

… , X

) ^∈ B _⎤⎦

⁵

∫ ∫

f

X X1 2^???X^p

( x , x ,

1 2

… , x

) dx dx

1 2

… dx

(5-8) Typically, f

X X1 2???Xp

( x , x ,

1 2

^… , x

) is defi ned over all of p-dimensional space by assuming that

f

X X1 2_???Xp

( x , x ,

1 2

^… , x

) ⁼ 0 for all points for which f

X X1 2???Xp

( x , x ,

1 2

^… , x

) is not specifi ed.

Machined Dimensions Many dimensions of a machined part are routinely measured during production. Let the random variables, X , X , X

1 2 3

, and X

denote the lengths of four dimensions of a part. Then at least four random variables are of interest in this study.

Example 5-13

Component Lifetimes In an electronic assembly, let the random variables X , X , X , X

2 3 4

denote the lifetime of four components, respectively, in hours. Suppose that the joint probability density function of these variables is

f

X X X X

x , x , x , x e

x x x

1 2 3 4

1 2

2 3 4

9 10

12 0 001 0 002 0 0015

( )

^{5 3} ² ^{2 ?} ^{2 ?} ^{2 ?} ³³ ^{0 003}⁴

0

2 ? x

x , x , x , x for ≥ ≥ ≥ ≥

What is the probability that the device operates for more than 1000 hours without any failures? The requested probability is P X (

> 1000 , X

> 1000 ) , which equals the multiple integral of f

X X X X_{1 2 3 4}

( x , x , x , x

2 3 4

)

over the region x

> 1000 , x

> 1000. The joint probability density function can be written as a product of exponential functions, and each integral is the simple integral of an exponential function. Therefore,

P X > , X > , X > , X >

e

1 2 3 4

1 2 1 5 3

1000 1000 1000 1000

0 00055

( )

=

^{− − − . −}

= .

Example 5-14

Suppose that the joint probability density function of several continuous random variables is a constant c over a region R (and zero elsewhere). In this special case,

… … ∫ ^f

^{X X}^{1 2}^…^X^p

( ^{x , x ,}

¹ ²

^{, x}

) ^{dx dx}

¹ ²

^… ^dx

^{= ×} ^c ( volume of region

) ⁼ 1 1

∫

by property (2) of Equation 5-8. Therefore, c = 1/(volume of region R). Furthermore, by property (3) of Equation 5-8, P _⎡⎣ ( X , X ,

1 2

… , X

) ^∈ B _⎤⎦

= ∫ ∫ ^… ∫ ^f

^{X X}^…^X

( ^{x , x ,} … ^{, x}

) ^{dx dx} ^… ^dx

^{= ×} ^c ( ^B

) ⁼

1 2 1 2 1 2

volume volu

m me

volume

( )

B ( )

R R

When the joint probability density function is constant, the probability that the random variables assume a value in the region B is just the ratio of the volume of the region B

to the volume of the region R for which the probability is positive.

Marginal Probability Density

Function

If the joint probability density function of continuous random variables X , X ,

1 2

… , X

is f

X X1 2…Xp

( x , x ,

1 2

… , x ,

) the marginal probability density function of X

is

f

( ) x

^{= …} ∫ ∫ ∫ f

X X^{1 2}^…Xp

( x , x ,

^…

, x

) dx dx

¹ ²

^… dx

i⁻¹

^… dx

i⁺¹

dx

^(5-9)

where the integral is over all points in the range of X , X ,

1 2

… , X

for which X

= . x

E X ( )

⁼ ^… x f

i X X^…Xp

( x , x , ^… , x

) dx dx ^… dx

⁼ x f

i Xi

x

−∞

∞

∫

^{1 2} ¹ ² ¹ ²

^{( )} ^d ^dx

ⁱ

−∞

∞

−∞

∞

−∞

∞

∫ ∫ ∫

and (5-10) V X ( )

⁼ ^… ( x

^{− μ}

) f

X X^…Xp

( x , x , ^… , x

) dx dx ^… dx

−∞

∞

−∞

∞

−∞^∞^∞

∫ ∫

² ^{1 2} ¹ ² ¹ ²

−∞

∫ ⁼

∞

∫ ⁽ ^x

ⁱ

⁻ ^μ

^Xⁱ

⁾

^f

^Xⁱ

^{( )} ^{x dx}

ⁱ ⁱ

As for two random variables, a probability involving only one random variable, for example, P a < X < b , (

) can be determined from the marginal probability distribution of X

or from the joint probability distribution of X , X ,

1 2

… , X

. That is,

P a < X < b P < X < , , < X < , a < X < b ,

< X < , ,

i i i

( ) ⁼ ( ^−∞ ^{∞ … − ∞} ^∞

−∞ ∞ … − ∞

− +

1 1

< < X <

∞ )

Furthermore, E X ( )

^and ^{V X} ( )

ⁱ

^for ⁱ ^{= …} ^{1 2} ^{, ,} ^{, p} can be determined from the marginal probability distribution of X

or from the joint probability distribution of X , X ,

…

, X

as follows.

Mean and Variance from Joint Distribution

Probability as a Ratio of Volumes Suppose that the joint probability density function of the continuous random variables X and Y is constant over the region x

+ y

≤ . 4 Determine the probability that X

+ Y

≤ . 1

The region that receives positive probability is a circle of radius 2. Therefore, the area of this region is 4 π . The area of the region x

+ y

≤ 1 is π . Consequently, the requested probability is π ^{/ 4} ( ) π ⁼ ^{1 4} ^/ ^.

Example 5-15

Points that have positive probability in the joint probability distribution of three random variables X , X , X

1 2 3

are shown in Fig. 5-11. Suppose the 10 points are equally likely with probability 0.1 each. The range is the non-negative integers with x x x

+

+ =

3 . The marginal probability distribution of X

is found as follows.

P X (

= 0 ) ⁼ f

X X X_{1 2 3}

( 3 0 0 , , ) ⁺ f

X X X_{1 2 3}

( 0 0 3 , , ) ⁺ f

X X X_{1 2 3}

( 1 0 2 , , ) ⁺⁺ ( ) ⁼

( = ) ⁼ ( ) ⁺

f , ,

P X f , , f ,

X X X

X X X X X X

1 2 3

1 2 3 1 2 3

2 0 1 0 4

1 2 1 0 0 1

.

,, f , ,

P X f , , f

X X X

X X X X X X

( ) ⁺ ( ) ⁼

( = ) ⁼ ( ) ⁺

2 1 1 1 0 3

2 1 2 0

1 2 3

1 2 3 1 2

.

3 3

1 2 3

0 2 1 0 2

3 0 3 0

, ,

P X f

X X X

, ,

( ) ⁼

( = ) ⁼ ( )

.

= 0.1 Also, ( E X

₂

) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1

Example 5-16

Section 5-1/Two or More Random Variables 169

With several random variables, we might be interested in the probability distribution of some subset of the collection of variables. The probability distribution of X , X ,

…

, X ,

k < p can be obtained from the joint probability distribution of X , X ,

…

, X

as follows.

If the joint probability density function of continuous random variables X , X ,

1 2

… , X

is f

X X1 2…Xp

( x , x ,

1 2

… , x ,

) the probability density function of X , X ,

1 2

… , X ,

k < p , is f

X X1 2…Xk

( x , x ,

…

, x

) ^{= …} ∫ ∫ ∫ f

X X1 2…Xp

( x , x ,

1 2

^… , x

) dx

k+1

dx

k+2

^… dx

^p^p

^(5-11) where the integral is over all points R in the range of X , X ,

1 2

… , X

for which X

= = x , X

1 2

x ,

… = . , X

x

Distribution of a Subset of Random Variables

FIGURE 5-11 Joint probability distribution of X X1, 2,andX3. Points

are equally likely. ⁰⁰ ¹ ² ³ ^x¹

1 2 3 x₃

x₂ 2

3 1

Conditional Probability Distribution

Conditional probability distributions can be developed for multiple random variables by an extension of the ideas used for two random variables. For example, the joint conditional probability distribution of X

₁

, X

₂

, and X

₃

given (X

₄

= x

₄

, X

₅

= x

₅

) is

f x , x , x f x , x , x , x , x

X X X x x

f

X X X X X X 1 2 3 4 5

1 2 3 4 5

1 2 3

1 2 3 4 5

( ) ⁼ ( )

4 4 5

4 5

0 x , x f

X X

x , x >

( ) ^for ( ) ^.

The concept of independence can be extended to multiple random variables.

Independence

Random variables X , X ,

1 2

… , X

are independent if and only if

f

X X1 2…Xp

( x , x

1 2

… , x

) ⁼ f

( ) ( ) x

f

x

^… f

( ) x

^for all x , x ,

1 2

^… , ^xx

^(5-12) Similar to the result for only two random variables, independence implies that Equation 5-12 holds for all x , x ,

1 2

… , x

. If we fi nd one point for which the equality fails, X , X ,

1 2

… , X

are not independent. It is left as an exercise to show that if X , X ,

1 2

… , X

are independent,

P X (

∈ ∈ A , X

1 2

A ,

… ∈ , X

A

) ⁼ P X (

^∈ A P X

) (

^∈ A

) ^… P X (

^∈ A

)

for any regions A , A ,

…

, A

in the range of X , X ,

…

, X ,

respectively.

In Chapter 3, we showed that a negative binomial random variable with parameters p and r can be represented as a sum of r geometric random variables X , X ,

1 2

… , X

. Each geometric random variable represents the additional trials required to obtain the next success. Because the trials in a binomial experiment are independent, X , X ,

1 2

… , X

are independent random variables.

Example 5-17

5-1. Show that the following function satisfi es the properties of a joint probability mass function.

x y fXY(x, y)

1.0 1 1 4

1.5 2 1 8

1.5 3 1 4

2.5 4 1 4

3.0 5 1 8

Determine the following:

(a) P X <

(

2 5. ,Y <3

)

(b) P X <

(

2 5.

)

(

)

(d) P X >

(

1 8. .,Y >4 7

)

(e) E X

( )

, E Y ,V X ,

( ) ( )

^andV Y

( )

(f) Marginal probability distribution of X

(g) Conditional probability distribution of Y given that X=1.5 (h) Conditional probability distribution of X given that Y=2 (i) E Y X

(

| = .^{1 5}

)

^(j) Are^X^{and Y} independent?

5-2. Determine the value of c that makes the function f x, y

( )

⁼c x

(

⁺y

)

a joint probability mass function over the nine points with x=1 2 3, , and y=1 2 3, , .

Determine the following:

(a) P X

(

= ¹,Y <⁴

)

^{(b) P X}

(

⁼¹

)

(

=²

)

(d) P X < ,Y <

(

2 2

)

(e) E X ,

( )

^{E Y ,}

( )

^{V X ,}

( )

^{and V Y}

( )

(f) Marginal probability distribution of X

(g) Conditional probability distribution of Y given that X=1 (h) Conditional probability distribution of X given that Y=2 (i) E Y X

(

| =¹

)

^(j) Are^X^{and Y} independent?

5-3. Show that the following function satisfi es the properties of a joint probability mass function.

x y fXY(x, y)

–1.0 –2 1 8

–0.5 –1 1 4

0.5 1 1 2

1.0 2 1 8

Determine the following:

(a) P X <

(

0 5. .,Y <1 5

)

(b) P X <

(

0 5.

)

(

)

(d) P X >

(

0 25. .,Y <4 5

)

(e) E X ,

( )

^{E Y ,}

( )

^{V X ,}

( )

^{and V Y}

( )

(f) Marginal probability distribution of X

(g) Conditional probability distribution of Y given that X=1 (h) Conditional probability distribution of X given that Y=1 (i) E X y

(

| =¹

)

^(j) Are^X^{and Y} independent?

5-4. Four electronic printers are selected from a large lot of damaged printers. Each printer is inspected and classifi ed as containing either a major or a minor defect. Let the random variables X and Y denote the number of printers with major FOR SECTION 5-1

EXERCISES

Problem available in WileyPLUS at instructor’s discretion.

Tutoring problem available in WileyPLUS at instructor’s discretion.

Layer Thickness Suppose that X , X ,

1 2

and X

represent the thickness in micrometers of a substrate, an active layer, and a coating layer of a chemical product, respectively. Assume that X , X ,

1 2

and X

are independent and normally distributed with μ =

10000 , μ =

1000 , μ =

80, σ =

250 , σ =

20 , and σ =

4, respectively. The specifi cations for the thickness of the substrate, active layer, and coating layer are 9200 < x <

10 800 950 , , < x <

1050 , and 75 < x <

85 , respectively. What proportion of chemical products meets all thickness specifi cations? Which one of the three thicknesses has the least probability of meeting specifi cations?

The requested probability is P ( 9200 < X <

₁

10 800 , , ⁹⁵⁰ < X <

^{1050 75} , < X <

⁸⁵ ) Because the random variables . are independent,

P ( 9200 < X <

₁

10 800 950 , , < X <

₂

1050 75 , < X <

₃

85 ) ⁼ P ⁽ 9200 < X <

₁

10 800 , ⁾

×× ^P ( 950 < X <

₂

1050 ) P ( 75 < X <

₃

85 )

After standardizing, the above equals

P ( − . ^{3 2} < Z < ^{3 2} . ) P ( ^{− .} ^{2 5} < Z < ^{2 5} ^. ) P ( ^{− .} ^{1 25} < Z < ^{1 25} ^. )

where Z is a standard normal random variable. From the table of the standard normal distribution, the requested probability equals

0 99862 0 98758 0 78870 . 0 7778

( ) ( ^. ) ( ^. ) ^{= .}

The thickness of the coating layer has the least probability of meeting specifi cations. Consequently, a priority should be to reduce variability in this part of the process.

Example 5-18

Section 5-1/Two or More Random Variables 171

and minor defects, respectively. Determine the range of the joint probability distribution of X and Y.

5-5. In the transmission of digital information, the probability that a bit has high, moderate, and low distortion is 0.01, 0.04, and 0.95, respectively. Suppose that three bits are transmitted and that the amount of distortion of each bit is assumed to be independent. Let X and Y denote the number of bits with high and moderate distortion out of the three, respectively. Determine:

(a) fXY

( )

x, y^(b)^f^X

( )

^(d)^f^Y|¹

( )

(e) E Y X

(

| =¹

)

^(f)^Are^X^{and Y} independent?

5-6. A small-business Web site contains 100 pages and 60%, 30%, and 10% of the pages contain low, moderate, and high graphic content, respectively. A sample of four pages is selected without replacement, and X and Y denote the number of pages with moderate and high graphics output in the sample. Determine:

(a) fXY

( )

x, y ^(b)^f^X

( )

^(d)^f^Y|³

( )

(e) E Y X

(

| =³

)

^(f)^{V Y X}

(

^{| =}³

)

(g) Are X and Y independent?

5-7. A manufacturing company employs two devices to inspect output for quality control purposes. The first device is able to accurately detect 99.3% of the defective items it receives, whereas the second is able to do so in 99.7% of the cases.

Assume that four defective items are produced and sent out for inspection. Let X and Y denote the number of items that will be identified as defective by inspecting devices 1 and 2, respectively. Assume that the devices are independent. Determine:

(a) fXY

( )

x, y^(b)^f^X

( )

^(d)^f^Y|²

( )

(e) E Y X

(

| =²

)

^(f)^{V Y X}

(

^{| =}²

)

(g) Are X and Y independent?

5-8. Suppose that the random variables X, Y, and Z have the following joint probability distribution.

x y z f x y z

(

, ,

)

1 1 1 0.05

1 1 2 0.10

1 2 1 0.15

1 2 2 0.20

2 1 1 0.20

2 1 2 0.15

2 2 1 0.10

2 2 2 0.05

Determine the following:

(a) P X

(

=²

)

^{(b) P X}

(

^{= =}¹^,Y ²

)

(

)

^{(d) P X}

(

⁼¹ ^or ^Z⁼²

)

(e) E X

( )

^{(f) P X}

(

^{= | =}¹ ^Y ¹

)

(g) P X

(

= = | =¹,Y ¹ Z ^{2 (h) P X}

) (

^{= | = =}¹ ^Y ¹^{, Z} ²

)

(i) Conditional probability distribution of X given that Y=1 and Z=2

5-9. An engineering statistics class has 40 students; 60% are electrical engineering majors, 10% are industrial engineering majors, and 30% are mechanical engineering majors. A sample of four students is selected randomly without replacement for a project team. Let X and Y denote the number of industrial engineering and mechanical engineering majors, respectively.

Determine the following:

(a) fXY

( )

x, y^(b)^f^X

( )

^x ^{(c) E X}

( )

(d) f_Y|³

( )

y (e) E Y X

(

| =³

)

(f) V Y X

(

| =³

)

(g) Are X and Y independent?

5-10. An article in the Journal of Database Management [“Experimental Study of a Self-Tuning Algorithm for DBMS Buffer Pools” (2005, Vol. 16, pp. 1–20)] provided the workload used in the TPC-C OLTP (Transaction Processing Performance Council’s Version C On-Line Transaction Processing) benchmark, which simulates a typical order entry application. See the following table. The frequency of each type of transaction (in the second column) can be used as the percentage of each type of transaction. Let X and Y denote the average number of selects and updates operations, respectively, required for each type transaction. Determine the following:

(a) P X <

(

)

^{(b) E X}

( )

(

6| =Y 0

)

^{(e) E X Y}

(

^{| =}⁰

)

5-11. For the Transaction Processing Performance Coun- cil’s benchmark in Exercise 5-10, let X, Y, and Z denote the average number of selects, updates, and inserts operations required for each type of transaction, respectively. Calculate the following:

(a) fXYZ

(

x, y, z

)

(b) Conditional probability mass function for X and Y given Z=0 (c) P X < ,Y <

(

6 | =6 Z 0

)

^{(d) E X Y}

(

^{| = =}⁰^{, Z} ⁰

)

5-12. In the transmission of digital information, the probability that a bit has high, moderate, or low distortion is 0.01, 0.04, and 0.95, respectively. Suppose that three bits are transmitted and that the amount of distortion of each bit is assumed to be independent. Let X and Y denote the number of bits with high and moderate distortion of the three transmitted, respectively. Determine the following:

Average Frequencies and Operations in TPC-C

Transaction Frequency Selects Updates Inserts Deletes Non-unique Selects Joins

New order 43 23.0 11 12 0 0 0

Payment 44 4.2 3 1 0 0.6 0

Order status 4 11.4 0 0 0 0.6 0

Delivery 5 130.0 120 0 10 0 0

Stock level 4 0 0 0 0 0 1

(a) Probability that two bits have high distortion and one has moderate distortion

(b) Probability that all three bits have low distortion (c) Probability distribution, mean, and variance of X

(d) Conditional probability distribution, conditional mean, and conditional variance of X given that Y=2

5-13. Determine the value of c such that the function f x y

( )

, ⁼cxy^{for 0}^{< <}^x ³^{and 0}^{< <}^y ³ satisfies the properties of a joint probability density function.

Determine the following:

(a) P X < ,Y <

(

2 3

)

^(b)^{P X <}

(

^{2 5}^.

)

(

1 2 5.

)

^(d)^{P X >}

(

1 8 1. ., < Y <2 5

)

(e) E X

( )

^(f)P X < ,Y <

(

0 4

)

(g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given that X=1 5. (i) E Y X

( )

| ^{= .}^{1 5}

)

(j) P Y <

(

2| = .X 1 5

)

(k) Conditional probability distribution of X given that Y=2 5-14. Determine the value of c that makes the function f x, y

( )

⁼c x

(

⁺y

)

a joint probability density function over the range 0< <x 3 and x< < +y x 2.

Determine the following:

(a) P X < ,Y <

(

1 2

)

(b) P < X <

(

1 2

)

( )

^(d)P X < ,Y <

(

2 2

)

(e) E X

( )

^(f)^{V X}

( )

(g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given that X=1 (i) E Y X

(

| =¹

)

^(j)^{P Y >}

(

²^{| =}^X ¹

)

(k) Conditional probability distribution of X given that Y=2 5-15. Determine the value of c that makes the function f x, y

( )

⁼c x y

(

⁺

)

a joint probability density function over the range 0< <x 3 and 0< <y x.

Determine the following:

(a) P X < ,Y <

(

1 2

)

(b) P < X <

(

1 2

)

( )

^(d)P X < ,Y <

(

2 2

)

(e) E X( ) (f) E Y( ) (g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given X=1 (i) E Y X

(

| =¹

)

(j) P Y >

(

2| =X 1

)

(k) Conditional probability distribution of X given Y=2 5-16. Determine the value of c that makes the function f x, y

( )

⁼ce⁻²^x⁻³^y a joint probability density function over the range 0<x and 0< <y x.

Determine the following:

(a) P X < ,Y <

(

1 2

)

(b) P < X <

(

1 2

)

(

)

^(d)

^{P X} ⁽ ^< ² ^, ^Y ^< ² ⁾

(e) E X( ) (f) E Y( ) (g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given X=1 (i) E Y X

(

| =¹

)

(j) Conditional probability distribution of X given Y=2 5-17. Determine the value of c that makes the function f x, y

( )

⁼ce^{− −}²^x³^y, a joint probability density function over the range 0<x and x<y.

Determine the following:

(a) P X < ,Y <

(

1 2

)

(b) P < X <

(

1 2

)

(

)

^(d)P X < ,Y <

(

2 2

)

(e) E X

( )

^(f)^{E Y}

( )

(g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given X=1 (i) E Y X

(

| =¹

)

(j) P Y <

(

2| =X 1

)

(k) Conditional probability distribution of X given Y=2 5-18. The conditional probability distribution of Y given X=x is f_{Y x}_|

( )

y ⁼xe⁻^xy^{for y}^>0, and the marginal probability distribution of X is a continuous uniform distribution over 0 to 10.

(a) Graph f_{Y X}_|

( )

y ⁼xe⁻^xy^{for y}^>0 for several values of x. Determine:

(b) P Y <

(

2| =X 2

)

^{(c) E Y X}

(

^{| =}²

)

(d) E Y X

(

| =x

)

^(e)^f^XY

( )

^{x, y}^(f)^f^Y

( )

5-19. Two methods of measuring surface smoothness are used to evaluate a paper product. The measurements are recorded as deviations from the nominal surface smoothness in coded units. The joint probability distribution of the two measurements is a uniform distribution over the region 0< <x 4, 0<y, and x−1< <y x+1. That is, f_XY

( )

x, y ⁼c for x and y in the region. Determine the value for c such that fXY

( )

x, y is a joint probability density function.

Determine the following:

(a) P X <

(

0 5. .,Y <0 5

)

(b) P X <

(

0 5.

)

( )

^{(d) E Y}

( )

(e) Marginal probability distribution of X

(f) Conditional probability distribution of Y given X=1 (g) E Y X

(

| =1 (h) P Y <

) (

0 5. | =X 1

)

5-20. The time between surface finish problems in a galvanizing process is exponentially distributed with a mean of 40 hours. A single plant operates three galvanizing lines that are assumed to operate independently.

(a) What is the probability that none of the lines experiences a surface finish problem in 40 hours of operation?

(b) What is the probability that all three lines experience a surface finish problem between 20 and 40 hours of operation?

5-21. A popular clothing manufacturer receives Internet orders via two different routing systems. The time between orders for each routing system in a typical day is known to be exponentially distributed with a mean of 3.2 minutes. Both systems operate independently.

(a) What is the probability that no orders will be received in a 5-minute period? In a 10-minute period?

(b) What is the probability that both systems receive two orders between 10 and 15 minutes after the site is officially open for business?

5-22. The blade and the bearings are important parts of a lathe.

The lathe can operate only when both of them work properly. The lifetime of the blade is exponentially distributed with the mean three years; the lifetime of the bearings is also exponentially distributed with the mean four years. Assume that each lifetime is independent.

Section 5-1/Two or More Random Variables 173

(a) What is the probability that the lathe will operate for at least five years?

(b) The lifetime of the lathe exceeds what time with 95%

probability?

5-23. Suppose that the random variables X, Y, and Z have the joint probability density function f x, y, z

(

)

⁼⁸xyz^for 0< <x 1, 0< <y 1, and 0< <z 1. Determine the following:

(a) P X <

(

0 5.

)

(b) P X <

(

0 5. .,Y <0 5

)

(

)

(d) P X <

(

0 5. ^orZ <2

)

(e) E X

( )

(f) P X <

(

0 5. | = .Y 0 5

)

(g) P X <

(

0 5. . | = .,Y <0 5 Z 0 8

)

(h) Conditional probability distribution of X given that Y=0 5. and Z=0 8.

(i) P X <

(

0 5. | = . = .Y 0 5, Z 0 8

)

5-24. Suppose that the random variables X, Y, and Z have the joint probability density function fXYZ

(

x, y z ,

)

⁼c^{over the} cylinder x²+y²<4 and 0< <z 4. Determine the constant c so that fXYZ

(

x, y z ,

)

is a probability density function.

Determine the following:

(a) P X

(

²+Y <² 2

)

(b) P Z <

(

)

( )

(d) P X <

(

1| =Y 1

)

(e) P X

(

²+Y <² 1| =Z 1

)

(f) Conditional probability distribution of Z given that X=1 and Y=1.

5-25. Determine the value of c that makes fXYZ

(

x, y z ,

)

⁼c^a joint probability density function over the region x>0, y>0, z>0, and x+ + <y z 1.

Determine the following:

(a) P X <

(

0 5. . .,Y <0 5, Z <0 5

)

(b) P X <

(

0 5. .,Y <0 5

)

(

0 5.

)

^{(d) E X}

( )

(e) Marginal distribution of X (f) Joint distribution of X and Y

(g) Conditional probability distribution of X given that Y=0 5. and Z=0 5.

(h) Conditional probability distribution of X given that Y=0 5. 5-26. The yield in pounds from a day’s production is normally distributed with a mean of 1500 pounds and standard deviation of 100 pounds. Assume that the yields on different days are independent random variables.

(a) What is the probability that the production yield exceeds 1400 pounds on each of five days next week?

(b) What is the probability that the production yield exceeds 1400 pounds on at least four of the five days next week?

5-27. The weights of adobe bricks used for construction are normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 20 bricks is selected.

(a) What is the probability that all the bricks in the sample exceed 2.75 pounds?

(b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

5-28. A manufacturer of electroluminescent lamps knows that the amount of luminescent ink deposited on one of its products is normally distributed with a mean of 1.2 grams and a

standard deviation of 0.03 gram. Any lamp with less than 1.14 grams of luminescent ink fails to meet customers’ specifications. A random sample of 25 lamps is collected and the mass of luminescent ink on each is measured.

(a) What is the probability that at least one lamp fails to meet specifications?

(b) What is the probability that five or fewer lamps fail to meet specifications?

(d) Why is the joint probability distribution of the 25 lamps not needed to answer the previous questions?

5-29. The lengths of the minor and major axes are used to summarize dust particles that are approximately ellipti- cal in shape. Let X and Y denote the lengths of the minor and major axes (in micrometers), respectively. Suppose that f xX( )=exp(− , <x)0 x and the conditional distribution fY x|( )y =exp[ (− −y x)], <x y. Answer or determine the following:

(a) That fY x|( ) is a probability density function for any value y of x.

(b) P X( <Y) and comment on the magnitudes of X and Y. (c) Joint probability density function fXY(x y, ).

(d) Conditional probability density function of X given Y=y. (e) P Y( < | =2 X 1 ) (f) E Y X( | =1 )

(g) P X( < , <1Y 1 ) (h) P Y( <2 ) (i) c such that P Y( < = .c) 0 9

(j) Are X and Y independent?

5-30. An article in Health Economics [“Estimation of the Transition Matrix of a Discrete-Time Markov Chain” (2002, Vol.11, pp. 33–42)] considered the changes in CD4 white blood cell counts from one month to the next. The CD4 count is an important clinical measure to determine the severity of HIV infections. The CD4 count was grouped into three distinct categories: 0–49, 50–74, and ≥ 75. Let X and Y denote the (category minimum) CD4 count at a month and the following month, respectively. The conditional probabilities for Y given values for X were provided by a transition probability matrix shown in the following table.

X Y

0 50 75

0 0.9819 0.0122 0.0059

50 0.1766 0.7517 0.0717 75 0.0237 0.0933 0.8830 This table is interpreted as follows. For example, P Y( =50| =X 75)

= .0 0717. Suppose also that the probability distribution for X is P X( =75)= . ,0 9P X( =50)= . ,0 08P X( =0)= .0 02. Determine the following:

(a) P Y( ≤50| =X 50 ) (b) P X( = , =0Y 75 )

5-31. An article in Clinical Infectious Diseases [“Strength- ening the Supply of Routinely Administered Vaccines in the United States: Problems and Proposed Solutions” (2006, Vol.42(3), pp. S97–S103)] reported that recommended vaccines for infants and children were periodically unavailable or

That is, E _⎡⎣

( X Y , ) _⎤⎦ can be thought of as the weighted average of h ( ) x y , for each point in the range of ( X Y , ) . The value of E _⎡⎣

( X Y , ) _⎤⎦ represents the average value of h ( X Y , ) ^{that is}

expected in a long sequence of repeated trials of the random experiment.

5-2 Covariance and Correlation

When two or more random variables are defi ned on a probability space, it is useful to describe how they vary together; that is, it is useful to measure the relationship between the variables. A common measure of the relationship between two random variables is the covariance. To defi ne the covariance, we need to describe the expected value of a function of two random variables

( X Y , ) . The defi nition simply extends the one for a function of a single random variable.

in short supply in the United States. Although the number of doses demanded each month is a discrete random variable, the large demands can be approximated with a continuous probability distribution. Suppose that the monthly demands for two of those vaccines, namely measles–mumps–rubella (MMR) and varicella (for chickenpox), are independently, normally distributed with means of 1.1 and 0.55 million doses and standard deviations of 0.3 and 0.1 million doses, respectively. Also suppose that the inventory levels at the beginning of a given month for MMR and varicella vaccines are 1.2 and 0.6 million doses, respectively.

(a) What is the probability that there is no shortage of either vaccine in a month without any vaccine production?

(b) To what should inventory levels be set so that the probability is 90% that there is no shortage of either vaccine in

a month without production? Can there be more than one answer? Explain.

5-32. The systolic and diastolic blood pressure values (mm Hg) are the pressures when the heart muscle contracts and relaxes (denoted as Y and X, respectively). Over a collection of individuals, the distribution of diastolic pressure is normal with mean 73 and standard deviation 8. The systolic pressure is conditionally normally distributed with mean 1 6. x when X=x and standard deviation of 10. Determine the following:

(a) Conditional probability density function f_Y|73( ) of y Y given X=73

(b) P Y( <115| =X 73 ) (c) E Y X( | =73 )

(d) Recognize the distribution fXY(x y, ) and identify the mean and variance of Y and the correlation between X and Y

E

X,Y

x, y f x, y X,Y

x, y f x,

XY XY

⎡⎣ ( ) ⎤⎦ = ( ) ( )

∑ ( )

∑ ^discrete

( )

⎧ ⎨

⎪

⎩⎪ ∫ ∫ ^{y dx dy} ^X,Y ^continuous ^(5-13)

Expected Value of a Function of Two Random Variables

Expected Value of a Function of Two Random Variables For the joint probability distribution of the two random variables in Example 5-1, calculate E _⎡⎣ ( X − μ

) ( Y ^{− μ}

) _⎤⎦.

The result is obtained by multiplying x − μ

times y − μ

, times f

( X Y , ) for each point in the range of ( X Y , ) ^{. First,}

μ

and μ

were determined previously from the marginal distributions for X and Y : μ = .

2 35

and

μ =

2 49 . Therefore,

E X [( − μ

)( Y − μ

)] = − . ( 1 2 35 1 2 49 0 01 )( − . )( . ) ( + − . 2 2 35 1 2 4 0 02 )( − . )( . 2 3 2 35 1 2 49 0 25 1 2 35 2 2 49 0 02 2 2 35

) ( )( )( )

( )( )( ) ( )

+ − . − . .

+ − . − . . + − . (( )( ) ( )( )( )

( )( )(

2 2 4 0 03 3 2 35 2 2 49 0 2 1 2 35 3 2 49 0 02

− . . + − . − . .

+ − . − . . )) ( )( )( ) ( )( )( )

( )(

+ − . − . . + − . − . .

+ − . −

2 2 35 3 2 4 0 1 3 2 35 3 2 49 0 05

1 2 35 4 2 2 49 0 15 . )( . ) ( + − . 2 2 35 4 2 4 0 1 )( − . )( . + − . ) ( 3 2 35 4 2 49 0 05 )( − . )( . ) = − 0 0 5815 .

Example 5-19

The covariance is defi ned for both continuous and discrete random variables by the same

formula.

Dalam dokumen Applied Statistics and Probability for Engineers, 6th Edition ( PDFDrive.com ) compressed (Halaman 189-199)

5-1.5 MORE THAN TWO RANDOM VARIABLES

More than two random variables can be defi ned in a random experiment. Results for multiple random variables are straightforward extensions of those for two random variables. A summary is provided here.

The joint probability distribution of random variables X , X , X ,

… , X

can be specifi ed with a method to calculate the probability that X , X , X ,

… , X

assume a value in any region Rof p-dimensional space. For continuous random variables, a joint probability density function f

( x , x ,

… , x

) is used to determine the probability that ( X , X , X ,

… , X

)

R by the

multiple integral of f

( x , x ,

… , x

) over the region R.

A joint probability density function for the continuous random variables X , X , X ,

… , X ,

denoted as f

( x , x ,

… , x ,

) satisfi es the following properties:

(1) f

( x , x ,

… , x

) ≥ 0

(2)

f

x , x ,

, x

dx dx

dx

1

∫ ∫ ∫ ( … ) … =

(3) For any region B of p-dimensional space,

P ⎡⎣ ( X , X ,

… , X

) ∈ B ⎤⎦

∫ ∫

f

( x , x ,

… , x

) dx dx

… dx

(5-8) Typically, f

( x , x ,

… , x

) is defi ned over all of p-dimensional space by assuming that

f

( x , x ,

… , x

) = 0 for all points for which f

( x , x ,

… , x

) is not specifi ed.

Machined Dimensions Many dimensions of a machined part are routinely measured during production. Let the random variables, X , X , X

, and X

denote the lengths of four dimensions of a part. Then at least four random variables are of interest in this study.

Example 5-13

Component Lifetimes In an electronic assembly, let the random variables X , X , X , X

denote the lifetime of four components, respectively, in hours. Suppose that the joint probability density function of these variables is

f

x , x , x , x e

9 10

( )

0

0

0

0

x , x , x , x for ≥ ≥ ≥ ≥

What is the probability that the device operates for more than 1000 hours without any failures? The requested prob- ability is P X (

> 1000 , X

> 1000 , X

> 1000 , X

> 1000 ) , which equals the multiple integral of f

( x , x , x , x

)

over the region x

> 1000 , x

^{R by the}

) ^≥ 0

∫ ∫ ∫ ( … ) … ⁼

P _⎡⎣ ( X , X ,

) ^∈ B _⎤⎦

^… , x

^… , x

) ⁼ 0 for all points for which f

^… , x

… … ∫ ^f

( ^{x , x ,}

^{, x}

) ^{dx dx}

^… ^dx

^{= ×} ^c ( volume of region

) ⁼ 1 1

by property (2) of Equation 5-8. Therefore, c = 1/(volume of region R). Furthermore, by property (3) of Equation 5-8, P _⎡⎣ ( X , X ,

) ^∈ B _⎤⎦

= ∫ ∫ ^… ∫ ^f

( ^{x , x ,} … ^{, x}

) ^{dx dx} ^… ^dx

^{= ×} ^c ( ^B

) ⁼

^{= …} ∫ ∫ ∫ f

^…

^… dx

^… dx

^(5-9)

⁼ ^… x f

( x , x , ^… , x

) dx dx ^… dx

⁼ x f

^{( )} ^d ^dx