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p191_exercices_5-5

# p191_exercices_5-5 - 5.5 LINEAR FUNCTIONS OF RANDOM...

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Unformatted text preview: 5.5 LINEAR FUNCTIONS OF RANDOM VARIABLES 191 | Now, an 14.5) = P[(l’— |Ly)/t]'y> (14.5 — non/0’1] = P(Z > 1.12) = 0.13 Soft-drink cansareﬁlledbyanautomatedﬁllingmachine Themeanﬁll volume 1s 12. l ﬂmdounces and the standard deviation' 1s 0.1 ﬂuid ounce. Assume that the ﬁll volumes of the cans are independent, nor- mal random variables What is the probability that the average volume of 10 cans selected from this process is less than 12 ﬂuid ounces? 2X+ 3Y< 30) (d) P(2X+ 3Y< 40) X and Y are independent, normal random variables (x) = 2, V(X) = 5, E(Y) = 6, and KY) = s. .u ”no the following: + 2r) (b) V(3X+ 21*) 3X + 2Y< 18) (d) P(3X + 2Y< 23) Suppose that the random variable X represents the ofa punched part in centimeters. Let Ybe the length partinmillimeters. IfE(X) = 5 and PIX) = 0.25, what '- mean and variance of-Y? A plastic casing for a magnetic disk is composed of -_ ves. The thickness of each halfis normally distributed '. mean of 2 millimeters and a standard deviation of 'ueterandthehalves areindependent. < , tune the mean and standard deviation ofthe total . ' kness of the two halves. . t is the probability that the total thickness exceeds .3 millimeters? ..- Making handcraﬁed pottery generally takes two , steps: Wheel throwing and ﬁring. The time of wheel -' and the time of ﬁring are normally distributed variableswithmeans of40mjnand60minandstan- deviations of 2 min and 3 min, respectively. LetXl,X2,... ,deenotetheﬁllvolumesofthe10cans.’I'heavc1-ageﬁllvohune(dmotedasf)is a normal random variable with — — 0.11 E(X) = 12.1 and var) = W = 0.001 Consequently, I “II“ _ _ 2—1.; 12—121] P(X< 12) — P[ of < —0.001 = P(Z < -3.16) = 0.00079 CISES FOR SECTION 5-5 . Xandeemdwendenormalmndomvariahleswiﬂr (a) Whatistheprobabilitythatapieoeofpottcrywillheﬁn— =0,V(X)=4,E(Y)=10,audI/(Y)=9. ishedwithin95min? A”... thefollowing: (b) Whatistheprobabilitythatitwilltakelongei'thanllomn? ' X+ 3r) (b) V(2X+ 3y) 5-59. Inthemannfactm’e ofelectroluminescentlamps,sev- emldiﬂ‘eraatlaymofinkaredeposimdontoaplasticsub- mm.ThethiekneuofWhymiscﬂticalifspeciﬁmtiaua regarding the ﬁnal colorand intensity ofﬂght ate—to be met. LetXand chnotetheﬂriekneasoftwommofink. It isknown thatth normally distributednithA-mgﬁﬂtl mﬂhmeterandastandarddevmuonofomﬂﬂ Yis also normally distributed with a mﬁmmllm andastandarddevistionofo.00017tmllimeter, Amtbat thesevariablesareindependent. (a) Ifaparticularlsmpismadeupofthesetwoinksonly, whatistheprobabilitythatthetotalinktbicknessisless than0.2337millimeter? (b) Alamp withstotslinkthichessexmding02405 mil- limeters lacksthetmiformity ofoolordemandedbythe customer Find the probability that a randomly selected lampfailstomeetcnstomerspeciﬁcations. 5,60. The width ofa casing for a door' is normally distrib- uted with a mean of 24 inches and a standard deviation of 1/8 inch. The width ofadoor is normally distributed with a meal of23 and 7/8 inches and a standard deviation of 1/16 ineh.Assumeindependence. (:1) Determine the mean and standard deviation ofthediﬁ'er- encebetweenthewidthofdiecasingandthewidthofthe door. ...
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