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## Elements of Mathematical Statistics and Probability Theory

by: Ms. Lonie Herzog

35

0

5

# Elements of Mathematical Statistics and Probability Theory STAT 345

Marketplace > University of New Mexico > Statistics > STAT 345 > Elements of Mathematical Statistics and Probability Theory
Ms. Lonie Herzog
UNM
GPA 3.56

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

## Popular in Statistics

This 5 page Class Notes was uploaded by Ms. Lonie Herzog on Wednesday September 23, 2015. The Class Notes belongs to STAT 345 at University of New Mexico taught by Staff in Fall. Since its upload, it has received 35 views. For similar materials see /class/212226/stat-345-university-of-new-mexico in Statistics at University of New Mexico.

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Date Created: 09/23/15
Stat 345 Solutions Section 51 3Td ed Problem 51 Check de nition 5 17 pg 143 1 nyzy 2 0 for all Ly yes 2 Zmzyfmw i 1 3 nyzy PX LY y yes Problem 52 a PXlt25Ylt3 PX1Y1PX15Y2 b PX lt 25 PX 1Y 1PX 15Y 2PX 15Y 3 c PYlt3PX1Y1PX15Y2i d PX gt18Y gt 47 PX 3Y 5 Problem 53 We rst need to nd the prnfs Lets start with X fX96 2y fXY957l Then7 Problem 54 a See Problem 53 above b To nd the conditional pmf7 we nd PY le 15 W for all values of PX15 Y Thus we have7 for example7 PX15Y2 18 1 PY2X15 7 l PX 15 38 3 The pmf is then y 1 2 3 4 5 leX15y 0 g g 0 0 and we can Check that this sums to 1 C We have PX 15Y 2 pX15lY2 pY2 18 and the rest of the probabilities are all 0 dEYlX 15 ZnyY le 15 3 using the distribution from part b Problem 55 We are given that nyzy Cz y for x 17273 and y 17273 and we need to nd C We know that the 2m 2y nyzy 1 23 022ml c2z1z2z3 c111213212223313233 360 Then setting 36c 17 we have c Problem 56 Ylt4PX1Y1PX1Y2PX1Y3 7 b PX 1 is the same as a7 since Y must be less than 4 C PY2 PX 1Y 2PX 2Y 2PX 3Y d PXlt2Ylt2PX1Y1 Stat 345 Solutions Section 55 3rd edition Problem 5 67 First7 we nd the marginal distributions The marginal distribution for X is The marginal distribution for Y is 1 s col M ml col The covariance is given by corX Y EXY 4 EXEY First7 we nd EXY EXY ZZmny Ly lt1gtlt3gt lt1gtlt4gt lt2gtlt5gt M6 9375 Now we nd EX and EY Thus COUltX7 Y 9375 4 18754625 07031 The correlation is given by COTTltX7 M VarxVarY So we need to nd the variances lt E l 5 H E gtlt Vw l E 5 7 H C 3 E 16 718752 08594 VarY EY2 4 EY2 9 16 25 36 4 46252 07344 Thus the correlation is 07031 corquot XY 08850 x0859407344

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