# HW3-sol.pdf - Solution to HW3 IA 3.5(5A.1 USU Him-A)7(332-A...

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Unformatted text preview: Solution to HW3 IA 3.5 (5A) (.1 -' USU Him-A! ] . )7: (£332 --A- =1} 3W1. :I‘ [1 (IA): _i/li) 5mm: LAN '2.) AA) + (AA) 1 (AAA run 3.125 [:11 7 hi! the 30ml 1mm:1 ut' the thaw Loim; E'AL1LD and N stand 33;: :1 dime and make] lt‘HpOfllH‘W. hum Wt' Am1 woivefimj Without Arugziaamuvut, thus .1113va qguaurunmuung uiemenhs [GE which A. 2:] 2'3 and :3!) (TINA-A coin-spending AAA 1110 AAA13A1A: [43% A; AA" 2 nickela and 'AE dimes 1 11'1ArlualnmlRAIiAAAAAA and :.Aclmm. 'A' HAJAAIcef‘um. P(7'2‘ 2A})2 #11? : A; 3 m1: 25):: W -' PAT! _ 3A1) ----- %’% A: .' mnl the. 'pmlmh‘llity AlieaArilmLiAm in ['-A"|l')ll.|}'ll‘ rum is w;— 5 A 2A) :25 :AA: P{’l‘"t=t} Us 3/5 115 AAA :1 J'AI‘AJhAlhiliAy hiatngrnm BIS 2}!» US - - I 3.3!.) AA.) quj(3-m.1:' MA AAA (:.AAA T)? .. .f'A. Sn A :31. (h) For "i E '1“ ;; 1.111(3)“: I‘ﬂ Film" HUT” '" (31' ' 'IYHNPJ '1 31'! “A?" " 'Tﬁ' r ' A SU‘ PHQH 2):}:m (.%)£i) W 56(2) 1’ I2; (a) mum A: AAA‘AJ 31,)“ A: AAAAAJ --A PM” :A (1.8] {17-418) AAA AAFUJAAJ “W ‘ \$11" 'ALAUSU' U '” 1"“3‘“ "H181 - 0.1%. if!) 4.3 JA HUI”) ~ {2[)}( (1/5)+|(25}(‘1/5] AA (RAMA/5) 2'3 cents. A7 Expected gain -: MAX) (AAAAAJJAAAAJ |-{ AAAAJAJJUJ 7) —— E1:- on 4,12 AAA) 21323:“ -- {if} diff:- 1/3. 80. fifSN\$5J§IUUj “\$1,663.07. 21.1.“? The 11121111111111“)! 111111111111 15111112221111.) )1), "-3 E} 9 1:71; 112 11:1 2511111 51111. 11(1) ”mix-j 111(1).) 11)?) 122.2)(1111) )- (1111))(112) 112111111112) 2-1111. 4.3-4 1:. :2-2 (~:-2){l).3] ~-|- {3)(02) -l- (SNELS) 2-212 2.5 111111 111.312) -—- (--2)'J(11.3) -1- (3)2102) 1(1))?(1111) 15.5. 50. 112 RT(XZ}W :13 1.21.211 111-111 11 .- 2.11111. 11.111 111.1) ::(1))(11 .11..)1 ()( )(11:17)1 (:1)(11 111) 1 (311111111) 11121111112111.1111 11.1111 131(1):”) .. ('11) ([11211]) 1J(1)'(1137) 1 (211(111121 (:1)J(11.1113)1(-1)9-(11.111):1.112. SL1, Vr1.1'{..-"1'j 2:: [.62 - H.815;5 2: 0.214.311 11ml 17 "\ll 581561101911. 1.511 1-(11)-:2);j1:(1--1).11--(\$2 — 1.21) .1 1111.11 Ll 133(2‘1’2} 2i}: 1125);] 2:) 1111::- 121-“:(431 331:,“ 2-2:: -'~. 1511111311. V1111'(.X‘J :- '1') [-1)]2 :2: «if; 1111:! 1.12111118----1..1235'é 111.1(X)-1-.))(1111 1(1;)(112) 11)(2.;111)m:111;2. 11.1.11 ) (--:1)J(1111)1(11)J(1 1(12)r)]}(11)1111;111:2212. 1111 112112111. 1)):- 1.11111) 1112226)) 1 z ()(-1 )(111-112)1. )(111211-1 :21111. STAT 35000-8630 21325: Homework 3 PROBLEM I Homework problems from the textbook: 3.5, 3.25, 3.30, 4.3, 4.7, 4.12, 4.17, 4.34, 4.49, 4.50, 4.57 Extra problems: Problem 1 Suppose the probability model for the random variable X is given as 3; 1 100 T 200 Xwnmm2 mlﬂ Table 1: P1ol)ability Model £01 random variabie X. a) In the table above, there is a missing value for Pr(X 2 500). So what is it? 0 - 2.. b) What is Pr(100 < X < 300) ? 5 P<' (X3: 2.00) -m C)- E)... rig-66’ f) What is E(X ) . wgzo I. Cg)Whatis SD(X X)? 5: VQJL(X) 156606 W“ lZl” Cl h) What 13 VaT(X). m {Elkzﬁ‘m Cﬁﬂ)2:“ w ”8 00C) ““" 39.61 ml SgégOQ i) What is E(:2X + 3),.5'D(—2X + 3} and Varr(:2X -1- 3). [592x333 : ._2 (EGO +3 :— m2(32c>)-i—3 :2: .. \$37 533 (“9.343) “ml-Ql S’SDQ‘Q 4: £838 me(—~e>ca~3):: (2') WLQ): 62, £406 £17000 32,000 30000 Page 2 of 5 STAT 35000-83: 21325 : Homework 3 PROBLEM 2 Problem 2 You draw a card from EL standard deck of 52 cards. If you get a red card, you win nothing. If you got a spade, you win \$5. For any club, you win \$10 plus an extra \$20 for the ace of clubs. a) Create a probability model for the amount you win at this game. 1)) Find the expected amount you’ll win in 1 game. And also the standard deviation. (3} Find the expected amount you’ll win in 100 independent games. And also the standard deviation. (”9 .. Clemé AC3?“ #CGNCXS ‘ ()3) l3 (w) Rec, 0 l/2 ”“ 96 State 5’ \3 5 mtg-~23“- O'tk—e‘VQMM: f0 [‘2’ i0 3/13: 4.3077 Adz, 0? emu 30 \ 50 3/59, :2. norm; Q5) [EM 5:: oy— %+%%+§%:§2 24;” 4M3 '3 0 ‘700 12952.5” ._[ Etvﬂ ~— opegiqo fig tw 99.? 3,; - it QBLFQ, UMQQ g; nugget—3, ~Q.ig““.-: 2.719777 gegm :: J25? 5777 m 3'44- CC) E New” '“*+'V\1L0b‘3 :: [00 E820] .._.. 7413 LOG; 3» Cw‘er-l— m~~+ Wm) :NllOO SEW) :2 sirﬂﬁ Page 3 of 5 STAT BBODO—Sec 21325: Homework 3 PROBLEM 3 Problem 3 The amount of time, in hours, that a computer functions before breaking down is a continuous random variable with probability density function given by Ac"”"/100 :1: Z 0 3: = ﬂ ) {0 :1: < 0 1;) Calculate A. b) What is the probability that a. computer will function between 50 and 150 hours before breaking down? c) What is the probabiiity that a computer wiil function less than 100 hours before breaking down? €30 «WK/1'00 Q we 6 CLVL =2: {00% 863 \$3 TWIOO'A 0 I?) woom m :1 00! ~63 ctr. \ﬂ) ‘9 (co¢:_<tso) SO \“ \~ 5 93A W “ ““fir ﬂoggﬁi 2} I. e Chg ‘06 werow‘i a) R, (x000)? loci Ci Am I «~\ ‘ .43» :2: *5 v: LO ﬁzgg %’ ‘ ,tgcmWH ,ﬁ ocB‘szKQ m...— Page 4 of 5 STAT 35000~Sec 21325: Homework 3 PROBLEIVI 4 Problem 4 The density function of a. continuous random variable X is given by a. + (751:2, O S a: g 1 MS?) : _ O, 0therw1se I 1. ifE{X) 33/5, ﬁndaandb. _ i .1: agoskioxq‘) AAL *2 0dr" 4%,. l S‘BiUt’W LE: 3; SxCQﬁ_QDXL>Q\ﬁ 2 1%: "\— 2%" L) 2. Lot g(r.r:): l 2. Calculate E[g(X)]. @ “gs-F: :L GK”? TI); Q S— C) ,3): J i+2m 1 £3 FL ‘ : ESL. 0&5”: S g . “Pm—4"“ “IX 5 5’ \$x. .—— 13., a: 13-— E-EQO] ‘3 30 1 + 2x1 “P“ S i; A” g 3. What is the probability X is greater than 0.5, La Pr(X > 0.5). i ‘3 g; "A RQ>K>OWE§B ‘1“: S L“??? gx 3AM Chg ,3 igwomS‘ .._ 21.; .1. Lyn—7»? “’ 53 c2. S 3 3 ‘2; ‘1 , l r», _.i.._._ t: 2"": ”ﬂ“ ”V "g: 8’ '"" .20 Q COS Page 5 of 5 ...
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