ch_04 - 4—1 4-2 4-3 CHAPTER 4 From the evenness of f(x):...

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Unformatted text preview: 4—1 4-2 4-3 CHAPTER 4 From the evenness of f(x): 1 - F(x) = F(-x). From the definition of X“: u = F(xu), l - u = F(x1-u). Hence 1 ’ u = l ' F(xu) = F(-xu) = F(x1—u) From the symmetry of f(x): 1 ¢ F(n+a) = F(n—a). Hence [see (4—8)] P{n-a < 5 < n+a} = F(n+a) — F(n—a) = 2F(r)+a) — 1 This yields l-a = 2F(n+a) - l F(r)+a) = l — 02/2 11+a = x1_a/2 F(a-n) = 01/2 2H) = xa/Z (a) In a linear interpolation: x -x Xu 2 x3 + b ’ (u-ua) for X3 < xu < xb ub'ua From Table 4-1 page 106 0.00565 0.00885 x 0.05 = l.2819 20.9 L: 1.25 + Proceeding simiplarly, we obtain an“ mm (b) If 2 is such that x = n + 02 then 2 is N(0,l) and 6(2) = Fx(n+az). Hence, 15 4—4 pk — 2G(k) = l = 2 erfk (a) From Table 4-1 I- E (b) From Table 3—1 with linear interpolation: (c) P{n-zua < x < n + zua} = 2G(zu) — l = '1 Hence, G(zu) = (1+1)/2 u = (l+’y)/2 4-5 (a) F(x) = x for 0 s x s 1; hence, u = F(xu) =’ xu (b) F(x) = l—e'2x for x z 0; hence, u = l-e’zxu xu = — ~5- £n(l—u) 4-6 Percentage of units between 96 and 104 ohms equals 100p where p = P{96 < R < 104} = F(lO4) — F(96) (a) F(R) = 0.l(R-95) for 95 s R s 105. Hence, p = 021004-95) - 0.1(96-95) = 0.8 (b) p = G(2.5) - G(-2.5) = 0.9876 4-7 From (4—34), with 0c= 2 and [3: 1m we get {(x) = c2 xe""U(x) X F(x) = c2 I ye‘cydy = l — e'c" - cxe‘cx o 4-8 {(g - 10)2 < 4} I {8 < 3 < 12} p{(§ - 10)2 < 4} a c(12 - 10) - c(3 - 10) = 0.954 _ (x-lO)2 fx 1 2 P{8 < g < 12} g e f(xl(§ - 10)2 < 4} = 0.954/EF for 8 < x < 12 and zero otherwise 4-9 F(x) = (1 - e-ux)u(x-c) f(x) =‘(1 - e'“c)a(x-c) + e‘°xu(x—c) ______.,_.____,___w__,_9,__a,____9Fw_____________________________.__ 4—10 (a) P{1 :_x 5_2} = G(%) - cc%) = 0.1499 3 6(1) — cgo.5) = 0.1499 = (b) P{1 5_§ 5_2|§ 3_1} 1 _ C(o.5) o 3085 0.4857 because {1 :_§ :_2, g 3_1} = {1 §_§ §_2} 4-11 1 _ _ _ _ y " "' If §(tl) ix l I then ! 0 x ti : Y = 80:) x = G-1(y) = H(y) Hence, 'l7 1024 - 1000 4-12 (a) P{§ < 1024} = G( ) = G(1.2) = 0.8849 20 p{951 < ¥ < 1024} (b) Pb} < 1024|=5 > 961} = ——Pl{x > 951} = 21$;£l_:_§£i;2§) . o 8819 1 — 961.95) (c) P{31 < J; < 32} a P{961 < :5 i 1024} - 0.8593 ~— 1 3 3 4-13 P{§ = 0} = E P{§ = l} = — P{§ - 2} a - P{§ - 3} = — 8 8 900 4—14 (a) 1. f(x)=-1—- Z (900)5(x-k) x 2900 k=0 k 900 2 2. fx(x) s 1 X e-(k-aso) /450 6(x_k) 15/2? k=0 10 15 (b) P{435 §_x :_460} a C(ig) — G(- IE) 0.5888 4-15 If x > b then {x < x} = S F(x) = 1 If x < a then {x < x} = {w} F(x) = 0 MM“ 4-16 If y(ci) :_w, then x(ci) §_w because x(;i) 5'y(:i); Hence, {y<v/}C{x<w} P{y <w} <P{x<w} — ~._ - __ ~— ~ 4. Therefore Fy(w) §_Fx(W) WWW‘ 4-17 4-18 4-19 4-20 Writing a similar equation for P(BI§ _<_ x0) we conclude that, if P(AI}5 = x) = P(B;)5 From (4-80) x -fktdt 0 f(x) = kx e = kx e -kx2/2 M It follows from (2-41) with A1={§:x} AZ-{§>x} It follows from Pb} _<_ x,A} P{:5 _<_ x,A} Fx(X|A) 3 P(A) P{Al?_{ : X} 3 W We replace in (4-80) all probabilities with conditional probabilities assuming {15 1 x0}. This yields an JP(A|§ = x, :5 _<_ xo)f(xl§ : xo)dx = P(A|§ _<_ X0) on: But f(x|§ : x0) = 0 for x > x0 and {)5 = x, x _<_ x0} = {)5 = x} for x 1x0. Hence, x o IP(AI§ = x)f(xl§ :xo)dx = P(AI§ _<_ x0) < < a: < for x __ x0, then P(Al)~c _ x0) P0311: _ x0) MW 19 x) 4-21 (a) Clearly, f(p) = l for 0 s p 5 l and 0 otherwise; hence 0.7 P {0.3 s p S 0.7} = dp = 0.4 ~ 0.3 (b) We wish to find the conditional probability P{0.3 s p s 0.7IA) where A = {6 heads in 10 tosses}. Clearly P{A|p=p} = p°(l-p)4. Hence, [see (4-81)] 6 _ 4 6 _ 4 fop (1-p)dp This yields 0.7 107 0.7 No.3 5 g s 0.7IA} = I” Wde = 4329 0.3 p6(l —p)4dp = 0.768 4—22 (3) In this problem, f(p) = 5 for 0.4 s p s 0.6 and zero otherwise; hence [see(4-82)] 0.6 P(H) = 5 J pdp = 0.5 0.4 (b) With A = {60 heads in 100 tosses) it follows from (4-82) that ' 0.6 f(plA) = 06°(l-p)“° / I“ p6°(1—p>4°dp for 0.4 s p s 0.6 and 0 otherwise. Replacing f(p) by f(p|A) in (4—82), we obtain 0.6 P(H|A) = I pf(p|A)dp = 0.56 0.4 20 4—23 n = 900 p = q = 0.5 up = 450 Vnpq = 15 k -np k ~np k1 = 420 k2 = 465 2 :1 1 = -2 Vnpq Vnpq P(420 _<_ k 1 465} = 0(1) - [1 - G(-2)] = (3(1) + 0(2) - 1 = 0.819 4-24 For a fair coin Vnpq = /;/2. If k1 = 0.49n and k2 = 0.52n then k — np k - np 2— g 0.52:1 - n/2 = 0.04/5 1 ~ g _ 0.02/5 /an v/rT/z /npq P{k1 §_k §_k2} = c(o.oa/E) + c(o.oz/E ) - 1 :_o.9 From Table 4-1 (page lO6)it follows that 0.02/5 > 1.3 n > 652 21 4-25 (a) Assume n = 1,000 (Note correction to the problem) P(A) = 0.6 up = 600 npq = 240 k = 650 k I 550 2 1 k - np k - up 2 =- 50 - 3.23 3—— = - 3.23 Vnpq #240 lnpq P{550 jhk §_650} - 26(3.23) - 1 ‘ 0.999 (b) P{0.59n :1: : 0.5m} - zc( 0‘01“)- 1 V0.24n n ' ZG(/m) - 1 'I 0.475 Hence, (Table 3-1) n = 9220 4-26 With a 3 0, b - T/h it follows that -1/4 p 8 l-e = 0.22 up = 220 npq = 171.6 k2 = 100 k2 - np —-—-—— = - 9.16 and (4-100) yields '“Pq Mo 5 k i 100} = G(-9.16) = 0. M 4-27 The event A = {k heads show at the first n tossings but not earlier} occurs iff the following two events occur B = {k-l heads show at the first n-l tossing} C = {heads show at the nth tossing} And since these two events are independent and -1 k—l -1- k-l 9(a) = (§_1)p q“ ‘ ’ P<c> - p we conclude that PcA) = P<B>P(c) = (2:1)pkq“’k £22 4-28 4-29 4-30 4-31 4-32 2 d 1 -x /2 1 X -x2/2 > e-x2/2 Multiplying by l/JE; and integrating from x to m, we obtain °° 2 Je-C /2 dc I l - C(x) X 2 l e-x /2 >‘_j; J27 2/ x 2____’o x40 l — e x The first inequality follows similarly because 2 2 2 _ 1L [Ci _ jgae-x /2] ‘ (l _ 4%qe-x /2 < e-x /2 X X —.——..W—. *- If P(A) = p then P6) = 1-p. that A does not occur at all. Clearly Pl - l-Ql where Q1 equals the probability If pn << 1, then Ql =8 (l-p)n = 1 — np P = p 1 With p = 0.02, n = 100, k = 3, it follows from (4-107) that the unknown probability equals 33-2 axe 100 -2 ( 3 )(o.02)3(o.9a)97 = :- g— e With n = 3, r = 3, kl = 2, k2 = 2, k3 = 1, pl 8 p2 = p3 = 1/6, it follows from (4-102) that the unknown probability equals 5! 1 M Mr With t = 2, k1 8 k, k2 = n-k, pi = 9, p2 = l-p = q, we obtain k - np1 = k - up 1 n-k—nq = np - k k2 - np2 = Hence, the bracket in (4-103) equals 2 2 (k -np ) (k ~np ) 2 2 1 1 + 2 2 _ (k-np) (l + l) s (k-np) np1 np2 n P q an asin fl-9Q. 23 4-33 4-34 (a) (b)‘ (c) mutations of n objects). ‘..... P(M) =- 2/36 ME) = 34/35. hence, [see (2-41)] The events M and H form a partition, P(A) = P(A|M)P(M) + P(A|fi)P(fi) (i) Clearly, P(A1M) = 1 because, if M occurs at first try, X wins. The probability that X wins after the first try equals P(Alfi). But in the eXperiment that starts at the second rolling, the first player is Y and the probability that he wins equals P(Zb = l-p. Hence, P(A|fib - P(FD = l—p. And since P(M) = 1/18 mi) =- 17/18 (1) yields l7 18 _~ p =.__ 1 p + (l-p) 18 18 Each of the n particles can be placed in any one of the m boxes. There are n particles, hence, the number of possibilities equals N - mn. In the m preselected boxes, the particles can be placed in N = n! ways (all per— A Hence p = nI/mn. n particles m—l interior walls All possibilities are obtained by permuting thevnfim-l objects consisting of the m-l interior walls with and n particles. The (m—l)! permutations of the walls and the n! permutations of the particles must count as one. Hence __ V agrim l). N g 1 N ,3 (111-1)! A Suppose that S is a set consisting of the m boxes. Each placing of the particles specifies a subset of S consisting of n elements (box). The number of such subsets equals (2) (see Prob» 2-26). Hence, ' 1 N = (2) NA 24 4-35 If k1 + k3 << n, then k3, = n and k3(1’1 + P2) ‘ [’"(1‘1 + k2)](pl + 92) 3 “(pl + P2) I 1 - + p ) = p z e'n(P1+P2) 133 1 2 3 n. ‘ n(n-1) (n-k3+1) nk1+k2 —r—r—r = -——-——-r—r—-— = —1—-r klikz‘k3‘ k1. 2. kllkzl Hence, k1 k n! pkl pkz :3 ~ e'm’l (“91) '“P2 (“92) I t l ' I ‘ k1.k2.k3. l 2 3 k1. k2. 4-36 The probability p that a particular point is in the interval (0,2) equals 2/100. (3) From (3-13) it follows that the probability p1 that only one out of the 200 points is in the interval (0,2) equals p1 = [210°] x 0.02 x 0.09199 (b) With np = 200 x 0.02 = 4 and k = 1, (3—41) yields p1 2 e‘4 x 4 = 0.073 25 ...
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ch_04 - 4—1 4-2 4-3 CHAPTER 4 From the evenness of f(x):...

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