Applied Statistics and Probability for Engineers

5-3 TWO CONTINUOUS RANDOM VARIABLES 165

5-34. Determine the value of csuch that the function

f(x,y)cxyfor 0x3 and 0y3 satisfies the

properties of a joint probability density function.

5-35. Continuation of Exercise 5-34. Determine the

following:

(a) (b)

(c) (d)

(e)E(X) (f)P 1 X0, Y 42

P 11 Y2.5 2 P 1 X1.8, 1Y2.5 2

P 1 X2, Y 32 P 1 X2.5 2

fact that these variables are not independent can be determined quickly by noticing that the

range of (X, Y), shown in Fig. 5-8, is not rectangular. Consequently, knowledge of Xchanges

the interval of values for Ythat receives nonzero probability.

EXAMPLE 5-20 Suppose that Example 5-15 is modified so that the joint probability density function of Xand Y

is Show that Xand Yare independ-

ent and determine

The marginal probability density function of Xis

The marginal probability density function of yis

Therefore, fXY(x,y)fX(x) fY(y) for all xand yand Xand Yare independent.

To determine the probability requested, property (4) of Equation 5-21 and the fact that

each random variable has an exponential distribution can be applied.

Often, based on knowledge of the system under study, random variables are assumed to be in-

dependent. Then, probabilities involving both variables can be determined from the marginal

probability distributions.

EXAMPLE 5-21 Let the random variables Xand Ydenote the lengths of two dimensions of a machined part, re-

spectively. Assume that Xand Yare independent random variables, and further assume that the

distribution of Xis normal with mean 10.5 millimeters and variance 0.0025 (millimeter)^2 and

that the distribution of Yis normal with mean 3.2 millimeters and variance 0.0036 (millime-

ter)^2. Determine the probability that 10.4X10.6 and 3.15Y3.25.

Because Xand Yare independent,

where Zdenotes a standard normal random variable.

EXERCISES FOR SECTION 5-3

P 1  2 Z 22 P 1 0.833Z0.833 2 0.566

P a

10.410.5

0.05

Z

10.610.5

0.05

b P a

3.153.2

0.06

Z

3.253.2

0.06

b

P 1 10.4X10.6, 3.15Y3.25 2 P 1 10.4X10.6 2 P 1 3.15Y3.25 2

P 1 X1000, Y 10002 P 1 X 10002 P 1 Y 10002 e^1 11 e^22 0.318

fY 1 y 2 



0

2  10 ^6 e0.001x0.002y dx0.002 e0.002y for y 0

fX 1 x 2 



0

2  10 ^6 e0.001x0.002y dy0.001 e0.001x for x 0

P 1 X1000, Y 10002.

fXY 1 x, y 2  2  10 ^6 e0.001x0.002y for x0 and y0.

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