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# Intro Methods Stats I MATH 461

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This 16 page Class Notes was uploaded by Henderson Lind II on Tuesday September 8, 2015. The Class Notes belongs to MATH 461 at University of Oregon taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/187190/math-461-university-of-oregon in Mathematics (M) at University of Oregon.

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Date Created: 09/08/15

HW MATH 461561 Lecture Notes 11 1 Example 121 In order to guarantee a lower failure rate of one important su pervisory duty a military control center wants to install sev eral same kind of computers for the supervisory Suppose that each computer will fail independently with the same probability p 001 The supervisory is successful if there is at least one computer is in working order Find a minimum 71 of computers which are required to be installed such that the failure rate is equal or less than 00001 Solution Suppose that there are 71 computers installed We need to nd this number 71 Let X be the number of computers in working order According to the question we need to nd 71 such that the following inequality IPltX 0 3 00001 satis ed Since IPX 0 0611 mop p it just nds 71 such that p 3 00001 Since 001 g 00001ltgt n 2 2 Therefore 71 2 De nition Let X be a rv de ned on a sample space Q For each real number 6 E R we de ne RX 3 3 Thus in this way we have de ned a function FltgtZR gt 01 This function is called the cumulative distribution function or simply distribution function of rv X HW MATH 461561 Lecture Notes 11 2 Example A company has ve applicants for three positions three women and two men Suppose that the ve applicants are equally quali ed and that no preference is given for choosing either gender Let X equal the number of women chosen to ll the three positions Find the probability distribution table and cumulative distribution table of X respectively Solution The Prob Dist table of rv X 6 1 2 3 3 2 3 2 3 2 0102 1 0201 i 030 i 05 10 of 10 of 10 The Cumulative Dist table of rv X 6 lt11 lt22 0 13 0 lt3 3lt S y l Example Let rv X have a density function 6331 6 if 6 E 01 0 if 5 g 0 1 Find the cumulative dist function of X and its graph Solution Since for 0 g y g 1 y Fy 65131 adc 312 2113 0 fltgt we have Fltygt 6140 y gt 0 Theorem If is the cumulative distribution of rv X then a IPX gt1 F for any HSER HW MATH 461561 Lecture Notes 11 3 b Ma ltX g b Fb Fa for any ab ER and a lt b c If X is a continuous rv then IPX 5 O for any 6 E R d If X is a continuous rv then FM N where is the density function of X We have introduced single rv and its density and distribution In many situations one rv can t handle the problems We need to introduce random vectors De nition Suppose that X Q gt R and Y Q gt R are two discrete rv s if there is a two variable nonnegative function fcy Z 0 such that 1 fcy gt O for any 3611 E A C R2 Zzy A 3 m ygt PltX x Y ygt for any ltsc w e R2 Then XY is called a discrete random vector fy is called the probability density function of XY and A is called the range space of X Y Remark Here each component say X is a single rv Example The experiment consists of ipping a fair coin and rolling a fair die The sample space is Q H71 ltHgt2gtgtltHgt3gtgtltHgt4gt7ltH75gt7ltH76gtgt Ta 1 Ta 2 Ta 3 Ta 4 Ta 5 Ta 6 De nearvXwEQ gtXwERby 0 ifwHz39z3916 XM 1 ifwTz39z3916 HW MATH 461561 Lecture Notes 11 4 and de nearvYwEQ gtYwERby Ywz39 ifwcicH orT De ne and if it y E A 0 if 3311 if A Then X Y is a discrete random vector with range space A and density function fy Check the de nition Find the proba bility PX g 0 2 g Y lt fltaygt IPX 3 02 g Y lt 5 IPH 2 H 3 H4 312 Similarly we can de ne a continuous random vector as follows De nition Suppose that X Q gt R and Y Q gt R are two continuous rv s if there is a two variable nonnegative function fy Z 0 such that 1 fcy gt O for any 3611 E A C R2 2 ffltx7w Afltgtygtddy 13 3 PX Y E 11 x cd f2fffyddy for any interval 11 x 0d E R2 Then X Y is called a continuous random vector ag y is called the probability density function of X Y and A is called the range space of X Y Remark For a continuous rv 0r rv s usually a question only HW MATH 461561 Lecture Notes 11 5 describes the density function and range space The sample space is not given ExamplezA continuous rv s X Y has range space A O lt 6 lt 1 O lt y lt 1 and density function as follows 453 if 3631 E A 0 if 3311 65 A fltaygt Find the probability RX 3 12 Y lt 14 12 14 1 PltX 312Y lt14 4yddy a 0 0 HW MATH 461561 Lecture Notes 11 6 ExamplezA continuous rv s XY has range space A O lt 13 0 lt y and density function as follows ye w y if 3611 E A 0 if 23y if A fltaygt Find the probability RX 3 1 Y lt 2 Solution 1 2 RX 3 1 Y lt 2 ye w ydmdy 0 0 2 l 2 eizy exwewyw mwrw 0 0 0 y 2 7y 1 2 k rawyok eeeo 0 1 1 0 6721 2 2 4 2 Tlo e ylo 05 0567 6 Similar to the case of single rv we can de ne the cumulative distribution function of rv s De nition For a given rv s X Y de ne FylPX Y y then Fy is called the joint cumulative distribution function of X and Y According to XY is discrete or continuous we have 1793 Z Z uav ugs Ugy Fy fuvdudv OI HW MATH 461561 Lecture Notes 11 7 The density function 2 F 88y ExampleSuppose that rv s X and Y vary in accordance with the joint pdf 1 y as y fy cy O lt36 lt3 lt1 Find 0 SolutionSince 1Plt9gt ycltxygtdscdy clt2ya1dy we get 0 2 HW MATH 461561 Lecture Notes 13 1 ExampleSuppose that X Y has density function 35 if 3311 6 01 x 01 fltygt 0 if 3631 E R2 01 x 01 Find 1PltX lt 21 Solution 1 1 1 2 1PltX lt 21 c ydyld Sty gilt261 0 z 0 2 2 2 2 6 3 3 5 1 fl 12 d 11 0 2 8l 22 6 240 24 De nition Suppose that XY has range space A and density function fcy If XY is a discrete rv s de ne fXltgt Z fltgtygtgt Z MuKM veinFA fXa is called the marginal density function of rv X and fyy is called the marginal density function of rv Y If X Y is a continuous rv s de ne fXltasgt fltscygtdy fyltygt fltscygtda yinFA 96196710619 fXa is called the marginal density function of rv X and fyy is called the marginal density function of rv Y Example In a box there are 4 white balls 3 black balls 2 red balls We randomly select 2 balls without replacement De ne X number of white balls Y number of black balls HW MATH 461561 Lecture Notes 13 2 Find density function fy and marginal densities fXa and fYltygt39 Solution We can nd 022 1 PXOYO PXOYi1 lt 7 gt 367 lt 7 gt C3 3 PXOY2 2 PX1Y lt gt029 36 lt C403 12 PXilYi111 PX2YO lt 7 gt 367 lt 7 gt Dist and marginal dists table YX 0 1 2 My 1 8 615 0 1 0 3 3 2 O O fxltgt a 0fo 029 a 0fo 029 a CS 7729 HW MATH 461561 Lecture Notes 13 3 De nition Two random variables X and Y are said to be inde pendent if for any intervals A and B we have IPltX e AY e B IPltX e APY e B Thereom Two random variables X and Y are independent if and only if ag y fxltgtfyltygt where fy is the joint density function and fXC fyy are the marginal density functions of rv X and Y respectively De nition 71 random variables X1 X2 Xn are said to be inde pendent if for any intervals A1 and An we have IPX1 A1 Xn e An IPX1 A1PX2 6 A2 IPXn e An Thereom 71 random variables X1X2 Xn are independent if and only if flt1a2am n fX11gt 39anngt where f12 ajn is the joint density function and inc is the marginal density functions of rv X 1 g 239 g 71 De nition 71 random variables X1 X2 Xn are said to be inde pendent if for any intervals A1 and An we have IPX1 A1 Xn e An IPX1 A1PX2 6 A2 IPXn e An Thereom 71 random variables X1X2 Xn are independent if and only if flt1a2am n fX11gt 39anngt where f12 ajn is the joint density function and inc is the marginal density functions of rv X251 3 239 g n ExampleSuppose that rv s X and Y vary in accordance with the HW MATH 461561 Lecture Notes 13 joint pdf fy2y 0ltltylt1 Find fXc and fyy Are X and Y independent SolutionSince fX 2ydy 20511 y22gti 21 32 0ltlt1 fyltygt y2ltx was 2031 22gt 312 0 lt y lt 1 It is obvious that flt7 1 7e fxgtfyygt 9311 6 071 X 01 Therefore X and Y are not independent ExampleSupp0se that X Y has density function ase W if 3611 E 0 00 x 0 00 fltygt 0 if 3331 E R2 0 00 x 0 00 Find fXc and fyy Are X and Y independent Solution Since HW MATH 461561 Lecture Notes 13 5 fX 6 zydy 0 336 0 g 6 lt 00 My Wd OO e me y lgo 640 0 eyOO e fdsc e yl e xl e y O yltoo It is obvious that M311 fxltgtfyltygt as y E 0 00 X 0 00 Therefore X and Y are independent Transformation of Random variables Given two random variables X and Y X Y and XY are two new random variables If we de ne gs y 36 and 956 y my then X Y fXY and XY gXY Therefore X Y and XY are transformed from rv s XY The question is how to nd the density function of the transformed rv s Theorem 1 Let X be a rv having density function with range space AX and Y aX b where a y O and b are two HW MATH 461561 Lecture Notes 13 6 constants then 1a If X is a discrete rv Y has density m My y e Ay where Ay y y ab E AX 1b If X is a continuous rv Y has density fYltygt 1 11 1 i m a where Ay y y ab E AX 2 Let X be a continuous rv having density function with range space AX and Y X2 then 2a If AX C 0 00 Y has density m m y e Ay where Ay y y 2 E AX 2b If AX C oo0 Y has density fyltygt m y e Ay where Ay y y 2 E AX 2c If AX is symmetric interval Y has density fyltygt mm fem y e Ay where Ay y y 2 E AX 3 Let X and Y be two independent rv s having density function fXa with range space AX and fyy with range space Ay then 3a If X and Y are discrete Z X Y has density fZltZgt Z fXltgtfyltZ as 2 E AZ all 6 gt7 yEAY HW MATH 461561 Lecture Notes 13 7 where AZ z zy E AXy E Ay 3b If X and Y are continuous Z X Y has density fzz fXcfyz d z E AZ where AZ z zy E AXy E Ay 4 Let X and Y be two independent continuous rv s having density function fXc with range space AX and fyy with range space Ay LetZ XY Then Z has density fzltzgt 0 fXltyzgtfyltygtydy z e AZ where AZ z z 33316 E AXy E Ay HW MATH 461561 Lecture Notes 16 1 Consider a random experiment in which X the number of customers visiting a service center in a given unit of time is a random variable and visitings are independent of one another Then X has Poisson distribution k 7A IPXkll k012 where A is the average number of customers visiting a service center in a given unit of time Remark Above experiment can be generalized as an abstract model for a class of experiments as long as we can use different ways to interpret the terms quotcustomerquot and quotunit of time For example 1 The number of calls received by a switchboard during a given period of time 2 The number of bacteria per small volume of uid 3 The number of machine breakdowns during a given day 4 The number of traffic accidents at a given intersection during a given time period Example The average number of traffic accidents on a certain section of highway is two per week Assume that number of accidents follows a Poisson distribution with A 2 1 Find the probability of no accidents on this section of highway during a 1week period 2 Find the probability of at most three accidents on this section of highway during a 2week period Solution Let X be the number of accidents on this section of highway during a week Since the average number of accidents per week is A 2 we have 20672 039 6 2 0135335 IPX 0 HW MATH 461561 Lecture Notes 16 2 2During a 2week period the average number of accidents on this section of highway would be 22 4 Let Y be the number of accidents on this section of highway during a 2week period Since the average number of accidents per 2week is 4 we have 1W s 3 230 239 0018316 0073264 0146528 0195367 0433475 Remark IfX has a Poisson distribution p Then EX and VarX A De nition A continuous rv X has a normal distribution with mean u and variance 02 denoted by NW 0 if it has the following density function fa e 2a ooltltoo In the case that u 0 and a 1 N01 is called the standard normal distribution

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