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Unformatted text preview: 379. The probability is 0.651 that at least one sample from the next five will contain more than one defective Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1. a) P( X ≥ 5) = 1 − P( X ≤ 4) ⎡⎛125 ⎞ 0 ⎛125 ⎞ 1 ⎛125 ⎞ 2 125 124 123 ⎤ ⎢⎜ ⎜ 0 ⎟0.1 (0.9 ) + ⎜ 1 ⎟0.1 (0.9 ) + ⎜ 2 ⎟0.1 (0.9 ) ⎥ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥ = 1− ⎢ ⎢ ⎛125 ⎞ ⎥ ⎛125 ⎞ 4 ⎢+ ⎜ ⎥ ⎟0.13 (0.9 )122 + ⎜ ⎟0.1 (0.9 )121 ⎜ ⎟ ⎜4⎟ ⎢ ⎝3⎠ ⎥ ⎝ ⎠ ⎣ ⎦ = 0.9961 b) P( X > 5) = 1 − P( X ≤ 5) = 0.9886
380. Let X denote the number of defective components among those stocked. a) b) c)
Section 37 381 . P( X = 0) = (100 )0.020 0.98100 = 0.133 0 P( X ≤ 2) = (102 )0.02 0 0.98102 + (102 )0.0210.98101 + (102 )0.02 2 0.98100 = 0.666 0 1 2 P( X ≤ 5) = 0.981 = 1) = (1 − 0.5) 0 0.5 = 0.5 b) = 4) = (1 − 0.5) 3 0.5 = 0.5 4 = 0.0625 c) = 8) = (1 − 0.5) 7 0.5 = 0.58 = 0.0039 d) ≤ 2) = P( X = 1) + P( X = 2) = (1 − 0.5) 0 0.5 + (1 − 0.5)1 0.5 = 0.5 + 0.5 2 = 0.75 e.) P ( X > 2) = 1 − P ( X ≤ 2) = 1 − 0.75 = 0.25
a) P( X P( X P( X P( X 382. E(X) = 2.5 = 1/p giving p = 0.4 a) P( X = 1) = (1 − 0.4) 0 0.4 = 0.4 3 b) P ( X = 4) = (1 − 0.4) 0.4 = 0.0864 4 c) P ( X = 5) = (1 − 0.5) 0.5 = 0.05184 d) P ( X ≤ 3) = P ( X = 1) + P ( X = 2) + P ( X = 3) = (1 − 0.4) 0 0.4 + (1 − 0.4)1 0.4 + (1 − 0.4) 2 0.4 = 0.7840 e) P ( X > 3) = 1 − P ( X ≤ 3) = 1 − 0.7840 = 0.2160
Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. a) E(X) = 4/0.2 = 20 b) P(X=20) = 383. 384. ⎛19 ⎞ ⎜ ⎟(0.80)16 0.24 = 0.0436 ⎜3⎟ ⎝⎠ 315 ⎛18 ⎞ ⎜ ⎟(0.80)15 0.2 4 = 0.0459 ⎜3⎟ ⎝⎠ ⎛ 20 ⎞ 17 4 d) P(X=21) = ⎜ ⎟ ⎜ 3 ⎟(0.80) 0.2 = 0.0411 ⎝⎠
c) P(X=19) = e) The most likely value for X should be near μX. By trying several cases, the most likely value is x = 19. 385. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random variable with p = 0.8 a) b) c) P ( X = 4) = (1 − 0.8) 3 0.8 = 0.2 3 0.8 = 0.0064 P ( X ≤ 4) = P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4) = (1 − 0.8) 0 0.8 + (1 − 0.8)1 0.8 + (1 − 0.8) 2 0.8 + (1 − 0.8) 3 0.8 = 0.8 + 0.2(0.8) + 0.2 2 (0.8) + 0.2 3 0.8 = 0.9984 P ( X ≥ 4) = 1 − P ( X ≤ 3) = 1 − [ P ( X = 1) + P ( X = 2) + P ( X = 3)] = 1 − [(1 − 0.8) 0 0.8 + (1 − 0.8)1 0.8 + (1 − 0.8) 2 0.8] = 1 − [0.8 + 0.2(0.8) + 0.2 2 (0.8)] = 1 − 0.992 = 0.008 386. Let X denote the number of people who carry the gene. Then X is a negative binomial random variable with r=2 and p = 0.1 a) P ( X ≥ 4) = 1 − P ( X < 4) = 1 − [ P ( X = 2) + P ( X = 3)] b) 387. ⎡⎛1⎞ ⎤ ⎛ 2⎞ = 1 − ⎢⎜ ⎟(1 − 0.1) 0 0.12 + ⎜ ⎟(1 − 0.1)1 0.12 ⎥ = 1 − (0.01 + 0.018) = 0.972 ⎜⎟ ⎜1⎟ ⎝⎠ ⎣⎝1⎠ ⎦ E ( X ) = r / p = 2 / 0.1 = 20 Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable with p = 0.02. a) P ( X = 10) = (1 − 0.02) 9 0.02 = 0.98 9 0.02 = 0.0167 b) P ( X > 5) = 1 − P ( X ≤ 4) = 1 − [ P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4) + P ( X = 5)] = 1 − [0.02 + 0.98(0.02) + 0.982 (0.02) + 0.983 (0.02) + 0.983 (0.02) + 0.984 (0.02)]
= 1 − 0.0961 = 0.9039
May also use the fact that P(X > 5) is the probability of no connections in 5 trials. That is, ⎛ 5⎞ P( X > 5) = ⎜ ⎟0.020 0.985 = 0.9039 ⎜ 0⎟ ⎝⎠ c) E(X) = 1/0.02 = 50
388. X: # of opponents until the player is defeated. p=0.8, the probability of the opponent defeating the player. (a) f(x) = (1 – p)x – 1p = 0.8(x – 1)*0.2 (b) P(X>2) = 1 – P(X=1) – P(X=2) = 0.64 (c) µ = E(X) = 1/p = 5 (d) P(X≥4) = 1P(X=1)P(X=2)P(X=3) = 0.512 (e) The probability that a player contests four or more opponents is obtained in part (d), which is po = 0.512. Let Y represent the number of game plays until a player contests four or more opponents. Then, f(y) = (1po)y1po. 316 µY = E(Y) = 1/po = 1.95 389. p=0.13 (a) P(X=1) = (10.13)11*0.13=0.13. (b) P(X=3)=(10.13)31*0.13 =0.098 (c) µ=E(X)= 1/p=7.69≈8 X: # number of attempts before the hacker selects a user password. (a) p=9900/366=0.0000045 µ=E(X) = 1/p= 219877 V(X)= (1p)/p2 = 4.938*1010 σ= 390. V ( X ) =222222 (b) p=100/363=0.00214 µ=E(X) = 1/p= 467 V(X)= (1p)/p2 = 217892.39 σ= V ( X ) =466.78 Based on the answers to (a) and (b) above, it is clearly more secure to use a 6 character password.
391 . p = 0.005 , r = 8 a.) b). c) Mean number of days until all 8 computers fail. Now we use p=3.91x1019 P( X = 8) = 0.0058 = 3.91x10 −19 1 μ = E( X ) = = 200 days 0.005 μ = E (Y ) =
392. 1 = 2.56 x1018 days 3.91x10 −91 or 7.01 x1015 years Let Y denote the number of samples needed to exceed 1 in Exercise 366. Then Y has a geometric distribution with p = 0.0169. a) P(Y = 10) = (1 − 0.0169)9(0.0169) = 0.0145 b) Y is a geometric random variable with p = 0.1897 from Exercise 366. P(Y = 10) = (1 − 0.1897)9(0.1897) = 0.0286 c) E(Y) = 1/0.1897 = 5.27 Let X denote the number of transactions until all computers have failed. Then, X is negative binomial random variable with p = 108 and r = 3. 393. a) E(X) = 3 x 108 b) V(X) = [3(1−1080]/(1016) = 3.0 x 1016 394.
395 . (a) p6=0.6, p=0.918 (b) 0.6*p2=0.4, p=0.816
Negative binomial random variable: f(x; p, r) = ⎜ ⎜ When r = 1, this reduces to f(x; p, r) = (1−p)x1p, which is the pdf of a geometric random variable. Also, E(X) = r/p and V(X) = [r(1−p)]/p2 reduce to E(X) = 1/p and V(X) = (1−p)/p2, respectively. ⎛ x − 1⎞ ⎟(1 − p) x − r p r . ⎟ ⎝ r − 1⎠ 317 396. Let X denote a geometric random variable with parameter p. Let q = 1p.
∞ ∞ ⎛1⎞ 1 E ( X ) = ∑ x (1 − p ) x −1 p = p ∑ xq x −1 = p ⎜ 2 ⎟ = x =1 x =1 ⎝p ⎠ p V ( X ) = ∑ ( x − 1 ) 2 (1 − p) x −1 p = ∑ px 2 − 2 x + p
x =1 x =1
1 p ∞ ∞ ( 1 p )(1 − p) x −1 = p ∑ x 2 q x −1 − 2∑ xq x −1 +
x =1
∞ ∞ ∞ x =1 ∑q
x =1 ∞ x −1 = p ∑ x 2 q x −1 −
x =1
∞ 2 p2 + 1 p2 = p ∑ x 2 q x −1 −
x =1 1 p2 1 p2 1 p2 1 p2 d = p dq ⎡ q + 2q 2 + 3q 3 + ...⎤ − ⎣ ⎦ d = p dq ⎡ q(1 + 2q + 3q 2 + ...) ⎤ − ⎣ ⎦ d = p dq ⎡ (1−qq )2 ⎤ − p12 = 2 pq(1 − q) −3 + p(1 − q) −2 − ⎣ ⎦ [ 2(1 − p) + p − 1] = (1 − p) = q = p2 p2 p2 Section 38 397. X has a hypergeometric distribution N=100, n=4, K=20
20 80 1 3 100 4 ( )( ) = 20(82160) = 0.4191 a.) P ( X = 1) = ( ) 3921225
b.) ( )( ) = 4845(1) = 0.001236 c.) P ( X = 4) = ( ) 3921225
20 80 4 0 100 4 P ( X = 6) = 0 , the sample size is only 4 d.) E ( X ) K ⎛ 20 ⎞ = 4⎜ ⎟ = 0 .8 N ⎝ 100 ⎠ ⎛ 96 ⎞ ⎛ N −n⎞ V ( X ) = np (1 − p )⎜ ⎟ = 4(0.2)(0.8)⎜ ⎟ = 0.6206 ⎝ 99 ⎠ ⎝ N −1 ⎠ = np = n 318 ...
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.
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