Homework_3_solutions - Solutions to Section 2 2.59(a Let X be the number of-occupiecl lanes then X is binomial with n = 1| and p = 03m Than 10 PLX 5

Homework_3_solutions - Solutions to Section 2 2.59(a Let X...

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Unformatted text preview: Solutions to Section. 2.? 2.59 (a) Let X be the number of-occupiecl lanes, then X is binomial with n = 1|] and p = 03m Than . 10 PLX 5 9121. pg: = 10) = 1 — (10) [0.75)1“(1 — 0.75)” = 0.944, (b) ELY] = up = 10 )l 0.75 2 7.5 _ and ox = npﬂ — J: 1/10 x 0.75 21:1 — 035) = 1.369. 2.61] [3.) Let X be the number of tirnes a. participant will be selected, than X is binomial with n = l2 and p = 0.1. Then mm a up = 12 x 0.1 2 1.2. (b) 12 P(X%2)= (2 ) (0.132(1 — 0.1)m = 0.230. {0) Now p r“— U.2 , so that Put a 2) '= (122) [0.232(1 h 0.2310 2' 0.233. 2.61 (a) (10] 130 (10) 3|] ) [1:30) 30 J pp; 5 g) = U (“J-)0&quot; + inf); + T334- = 0.300 +0444 +0214 2 0.050. 4 d. 4 ' (b) For 5 2} a (i) (0.25)&quot;(1 ~ 0.2534 + (‘14) {0.25)1(1 — 0.25)a + G) (025)2(1 — 0.25)”- = 0.310 + 0.422 + 0.211 = 01040. The approximation is fairly actura’ae. 32.62 (a) X has a. hypergeometrio distribution with N = 50, M = 2, and n = 3. Then . fc Ju(:)(3‘f§3 f ““012 __ 3 ﬁnE-ETEEJ— DI'E— 11‘ 03] P(X = 0) = (3(ng = 0.332. 3 [c] Using p = MIN = 2f50 = 0.04., P(X = D) e (g) (0.04.)°(1 — 11.111)a = 0.835. The approximation' 15 very secuxate. 2.134 (a) X has a. Binomial distribution with n— .‘ QUE] and 33: 1/20: D. [15. Then POL. &gt; 5) — 1 - Pix- &lt; 4) = 1— 2 (20&quot;) (c as) (1 s ﬁ.05}3&quot;”‘i i=0 {h} The Poisson distribution ﬁts because an error on a. page is a. rare event and the number of pages in the msnuscript' 1s large. Using A_ _ D 05 x 20!]— — ll], and the table of cumulative Poisson prohahﬂities, \$341101. P[X&gt;5)=1- P(X&lt;1)=1— EB =1uuuse=cﬂs71 2. 7|] (e) Let E denote the event that the Eastern Conference team wins the series Then P(E) = PIE, 1 games) + P{E, 5 games) +P(E, 6 games) + P(E, 7 games). For the Eastern Conference team to win in 51' games, it must have won 3 out of the ﬁrst 3' —- 1 games, as well as the last one. So _ HE. jeame5)=( g1: ) pie-4 PIE} = (3)1149&quot; + (ﬂee 4!— @11in + (2)134?- [h)- Let W denote the event that the Western Conference team wins the series. Then the formula. for the probability that .the Western Conference team wins in 3' games is similar to that of the Eastern Conference, but with p and q switched. Then land P{Series ends in 55 games] = IP[E,I 3' games) +P(W 3' games) _ (3' ; ﬂee-4 + (9*)1 p:— (.73 — 1) [13495—4 + [31135-4] fl Solutions to Section 2.3 2.71 (a) '» 100 — 90 a &lt; x &lt; = ————— = . . PjEl _ _ 1013) 100 _ 70 I] 333 [b] To ﬁnd the 90th percentileI set 100 — b &lt; &lt; = = . . P(b_X,_lﬂﬂ) 100‘?“ U]. - Solving for 6 gives b = 97. 2.72 (a) If city A is at mile CI and city 0 is at mile T5, then city B is at mile 25. Then the probability that the car is towed more than 10 miles is P(1u5X515)+P(355X565)=%+%=0.467. (b) The probability that the ear is towed more than 20 miles is PMS g x 5 55 = g = [1.133. (c) P(Towed &gt; 10 miles HX &gt; 20) _ P[X &gt; 20) P(35 g X s 65) 13(20 5 X g 75) 30x75 3575— P(Towed &gt; 10 milele 3- 20} i] = 0.545. 2.73 (a) To ﬁnci the pth quentile, set F(m] = 1 — 3““: =1: and solve for x, which yields 3 = -1{}].1:L[1 —p). - .— For 13-: 0.5, the median is 6.931. For 31 = [1.751 the Tﬁth percentile is 13.863. (1:) There are no jobs in 15 minutes if the arrival time of the ﬁrst job is above 15 minutes, 50 P(No jobs in 15 minutes] = P(X &gt; 15) z l—F{15] = e-&quot;Alilﬁi = 0.223. 2.74 (a) lithe mean time to failure is 10,000 hours, then A é l/lODDU. To ﬁnd the median1 set ‘ Fm = 1 -— 3—5/1”” '4'.“ and solve for m, which yields m 2 400001141 — 0.5) = 5931.4?2. (13) PLX 2 1000) = 1 — F(10EID] = e—lmﬂm = 0.905. (c) Because of the memoryless property of the monential distribution, Pg 2' 2000p; _&gt;__ 1000) = Put 2 1000) = 0.905. 2'75 Let X,- be the failure time of the itb bulb. Then Xi is exponential with failure rate A = 1f10000. Since T = X1 +Xg+.. .+X5 is the sum of 5 exponential distributions with failure rate A = 1,110000, T has a. gamma distribution with the some failure rate A = 1/1000!) and r = 5. Then __ _ r 5 : Em = X = menus = 50’ one and r I 5 V T = — = —.———— = i all J A: [1 [100000 500, one, use 2.76 Let X be the bets. random variable under study. For ELK) to be 3/4, a 3 n+3: E or o. = 3b. For Vsr(X] to be ME, __L_ .7 __1b2_ _ 1 (n+ Win + 5+ 1) — {4b)2(4b+ 1) ‘ 32 or . . 33:32:33,163: [40+1) or b 5 1/4. Then solving for u yields a = 3/4. Solutions to Section 2.9 ' . ' ' 2.78 (at) Hz 5 1.68) = 0.9535, PﬁZ &gt; 0.75) = 1 -— 0.7734 = 0.2266, 19(2- 5 4.42} = 0.0078. Hz &gt; —1_)= 1 - 0.1537 = 3.3413. [b] -. _ _ Pa 5 z 5 3) :.o.9732 _— 3.33.13 = 3.1353, ' F(—2 g z 5 —1)= 3.1353, - . P(—1.5 5 z s 1.3) = 3.3332 - 3.3633 = 3.3634, ' 33(41 5 z 5 2) = 0.9772 — 3.1337 = 3.3133. (t) PLZ g z.1} = 1 - 0.1 : n.9, P(Z &gt; —z.u5] = 1 - 0.35 = 0.35, ' H325 5 Z &lt;_: 2.31} = 0.25 — 0.01 = 0.24., Pf—‘zga 5 E g 3.31) = 3.35 — [1.31 = 0.74. 2.39 (a) 2,3 50.525, 315 =_ 1.34, and 2.315 = 1-44: 3,3 = :1 + (0.323) x 3 = 5.513. :15 = 4+ (1.04] x 3 = 2’12, . 3.575 = 4-i- (1.44) X 3 = 8.32. 2.80 (a) Let X be the weight of coffee in a. can. Then . 1313: &lt; 13} = 1: (z = 13;... &lt; 356%) = Hz &lt; -3.2) = 3.4237. (b) .. . . ' _ 13—131 T X—p. 13.33131 P[16&lt;X&lt;16.5) - P( 3.5 &lt;2- a .5 0'5 ) = P(——B.2 &lt; Z &lt; 0.8) = 0.7'331 — 0.420? = 13.3574. (c) To ﬁnd the 10th percentile, set 'P(zg”'9'“)=u.1 0' 0r . -. _ . 319 = ,u + £3.17 = 16.1 + {—1.28} x 0.5 = 15.46. 2.81 [a] W has a normal distribution. {‘3} EﬁW} = E(U) —- E(V] = lﬁﬂ — 120 = 40 and _ . VariW) = Varcx} + VarﬁY) = 3112 + 332 = 1335. ' . W—p. 50—40) U—V EU --'= -P 50 =P = P[ &gt; J [10&gt; J (z a &gt; 1525 PM &gt; 0.250} = 1 — 0.0020 = 0.303. |[ 2.82 {a} X —-Y has a normal distribution with Eu: — Y] = 0.520 — 0.525 = 0.001 and 5x4 = «(3)2 +10)2 = 5 a: lﬂ‘f. (13} Xl—Y— — . 0 PLXFY &gt; 0) n P (2*: &quot;Ti &gt; %?[i—§—_}) = PtZ &gt; —2) = 1—0022? = 0.0773. {c} Sinoe'the number of pairs that'ﬁt together has a. binomial distribution with n = 10 and p s ELQWZ, this probability i5 . (1§)(0.0772}9(0.022311 + GE) (0.077211“{0.0223)'5 .'—_ 0105 + 0.794 = 0.979. 2.33 [a] I? has a nomal distribution with mean 0 = 90 and SD = (Th/E: 2mm 2 4. 03} Put &gt; 100) = P (z = X ‘ “if &gt; 1°” ' 9&quot;) = 1312 2- 2.5) = 0.0002. ' ﬁx 4 - (c) To ﬁnd the 90th percentile, set. P (2: g 51-1133) = 0.90 “x 01' ﬁllll] S [.0 + 30.190 '3 9D ‘1‘ (1.23} X 4 = 95.12. ...
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