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p187_exercices_5-4

p187_exercices_5-4 - 54 BIVARIATE NORMAL DISTRIBUTION 187...

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Unformatted text preview: 54 BIVARIATE NORMAL DISTRIBUTION 187 In general, zero correlation does not imply independence. But in the special case that X and Yhaveabivariatenormal distribution, ifp = 0,Xand Yare independent. The detailsare left as an exercise. For Bivariate Normal Random Variables Zero - elation Implies Independence An important use of the bivariate normal distribution is to calculate probabilities involving two correlated normal random variables. ‘I'LE 5-31 Suppose that the X and Y dimensions of an injection-molded part have a bivariate normal distribution _--'n— with ox = 0.04, Cy = 0.08, "ax = 3.00, a, = 7.70, and p = 0.8. Then, the probability that apart satis- '._ . Part fies both specifications is H235 < X < 3.05, 7.60 < Y < 7.80) This probability canbe obtainedby integrating fn(x, y; 0'1, 0’17, 14.1 m, p) over the region 2.95 < x < 3.05 and 7.60 < y < 7.80, as shown in Fig. 5-7. Unfortunately, there is often no closed-form solution to prob- abilities involving bivariate normal distributions. In this case, the integration must be done numerically. ‘ SES FOR SECTION 5-4 product. SupposeXand Yhavcabivatiate normal _thithD'x= 4,0,1“: 1, "IX: 2,3.nd "DY: LDI'aW . contour plot of the joint probability density function of the following values for p: ' (b) P = 0-3 ‘ .. - ..~ ' Xand Yhaveabivsriatenormal distribution 0.04, or = 0.08, 1.1.; = 3.00, M = 7.70, and p = 0. the following: 5 < X < 3.05) fit: < 1’ < 7.80) . an acid-base titration, a base or acid is gradually ' the other until they have completely neutralized . LetX and Y denote the milliliters of acid andbase Lu- equivalence, respectively. Assume X and Yhave a normal distribution with ox = 5 mL, (1-,. = 2 mL, the following: -'. cc betweenXsnd Y _- ' ; probability distribution of X 1- 116) ‘- 11'6|Y= 102) 5-49. In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifi- cations regarding the final color and intensity of light are to be met. Let X and 1’ denote the thickness of two difi'erent layers of ink. It is known thatX is normally distribumd with a mean of 0.1 millimeter and a standard deviation of 0.0003] millimeter, and Y is normally distfibmed with a mean of 0.23 millimeter and a standard deviation 010000.17 millimeter. The value of p for these variables is equal to zero. Specifications call for a lamp to have a thickness of the ink corresponding ml? in the range of 0.099535 to 0.100465 millimeters and Y in the range of 0.22966 to 0.23034 mil- limeters. What is the probability that a randomly selected lamp will conform to specifications? 5-50. Suppose that Xand Yhave a bivariate normal dishi- bution with joint probability density ftmctionfxfix, y; ex, 03,, an My, 9)- (a) Show that the conditional distribution of Y, given that X = x is normal. 0:) Determine E(YIX = x). (c) Determine VU’IX = 1:). 5-51. IfX and Yhave a bivariate normal distribution with p = 0, show thatde Yare independent. 5—52. Show that the probability density function fxy(x, y; ox, 0y, n.1, try, p) of a bivariate normal distribution integrates to one. [Hint Complete the square in the exponent and use the ...
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