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 531 Suppose that the X and Y dimensions of an injectionmolded 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. 57. Unfortunately, there is often no closedform solution to prob abilities involving bivariate normal distributions. In this case, the integration must be done numerically. ‘ SES FOR SECTION 54 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 = 03 ‘ ..  ..~ ' 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 acidbase 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) ‘ 11'6Y= 102) 549. 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? 550. 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:).
551. 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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 Spring '08
 Wherly
 Statistics

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