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UIUC STAT 400 - 400Ex5_7ans

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STAT 400 Lecture AL1 Answers for 5.7 Spring 2015 Dalpiaz Normal Approximation to Binomial Distribution: Normal Binomial mean n  p standard deviation   1 ppn  1. Binomial distribution, n = 25, p = 0.50. Normal approximation: mean = n  p = 25  0.50 = 12.5. n  p  ( 1 – p ) = 25  0.50  0.50 = 6.25. SD = 25.6 = 2.5. a) P(X = 17) = PMF @ 17 = 0.0322. b) P(X = 17) = P(16.5  X  17.5)  5.25.125.17Z5.25.125.16P = P(1.60  Z  2.00) = 0.9772 – 0.9452 = 0.0320.c) P(X  11) = 1 – CDF @ 10 = 1 – 0.2122 = 0.7878. d) P(X  11) = P(X  10.5)  5.25.125.10ZP = P(Z  – 0.80) = 1 – 0.2119 = 0.7881. e) P(10  X  14) = CDF @ 14 – CDF @ 9 = 0.7878 – 0.1148 = 0.6730. f) P(10  X  14) = P(9.5  X  14.5)  5.25.125.14Z5.25.125.9P = P(– 1.20  Z  0.80) = 0.7881 – 0.1151 = 0.6730.2. Let X = number of passengers who do not cancel their reservations. Then X has Binomial distribution, n = 100, p = 0.85. Normal approximation:  = 100  0.85 = 85,  2 = 100  0.85  0.15 = 12.75.  = 3.57. P ( X  92 ) = P ( X  92.5 )  57.3855.92ZP = P ( Z  2.10 ) = 0.9821. Binomial: P ( X  92 ) = 0.9878. 2.5. A fair 6-sided die is rolled 180 times. The sum of the outcomes is likely to be around __________, give or take __________ or so. The average of the outcomes is likely to be around __________, give or take __________ or so.  = 6216654321 = 3.5.  =             65.365.355.345.335.325.31 222222 = 65.17 625.625.225.025.025.225.6   1.708. E ( Sum ) = n   = 180  3.5 = 630. SD ( Sum ) = n    180  1.708  22.9. The sum of the outcomes is likely to be around 630 , give or take 23 or so. E ( Average ) =  = 3.5. SD ( Average ) = σn  180 708.1  0.1273. The average of the outcomes is likely to be around 3.5 , give or take 0.13 or so.A fair 6-sided die is rolled 180 times. The number of 6’s is likely to be around __________, give or take __________ or so. Let X denote the number of 6’s. Binomial distribution, n = 180, p = 1/6 . E ( X ) = n  p = 180  1/6 = 30. SD ( X ) =   1 ppn  = 6561180  = 5. The number of 6’s is likely to be around 30 , give or take 5 or so. 3. Binomial distribution, n = 180, p = 1/6. a) P(X = 35) = 14535 65 61 35180 C = 0.0464. Normal approximation: mean = n  p = 180  1/6 = 30. n  p  ( 1 – p ) = 180  1/6  5/6 = 25. SD = 25 = 5. b) P(X = 35) = P(34.5 < X < 35.5)  P(0.90 < Z < 1.10) = 0.8643 – 0.8159 = 0.0484. c) P(X  35) = P(X > 34.5)  P(Z > 0.90) = 1 – 0.8159 = 0.1841. Binomial: P(X  35) = 18035180 65 61 180 Ckkkk  0.18283. d) P(20  X  40) = P(19.5 < X < 40.5)  P(– 2.10 < Z < 2.10) = 0.9642. Binomial: P(20  X  40) = 4020180 65 61 180 Ckkkk  0.965.e)* Use Normal approximation to find the probability that the sum of the results is between 600 and 640 (both inclusive)? [ Recall:  = 3.5,  2 = 17.5/6 = 35/12 . ] P(600  Sum  640) = P(599.5 < Sum < 640.5)  1235 1805.31805.640Z1235 1805.31805.599P = P(– 1.33 < Z < 0.46) = 0.5854. 4. Poisson distribution,  = 1.4  52 = 72.8. a) P( X = 68 ) = ! 688.728.7268 e = 0.0411. Normal approximation:  =  = 72.8.  = λ = 8.72 = 8.5323. b) P( X = 68 ) = P( 67.5 < X < 68.5 )  P( – 0.62 < Z < – 0.50 ) = 0.3085 – 0.2676 = 0.0409. c) P( X  70 ) = P( X < 70.5 )  P( Z < – 0.27 ) = 0.3936. Poisson: P( X  70 ) = 700 8.72.8.72 40078.0! kkke d) P( 65  X  80 ) = P( 64.5 < X < 80.5 )  P( – 0.97 < Z < 0.90 ) = 0.8159 – 0.1660 = 0.6499. Poisson: P( 65  X  80 ) = 8065 8.72.8.72 65218.0! kkke“Hype” for 0.5 continuity correction: Let X be a Binomial ( n = 400, p = 0.80 ) random variable. P ( X  312 ) = 0.1738. A B A B 1 =BINOMDIST(312,400,0.8,1)  1 0.173821 2 2 Without 0.5 continuity correction: With 0.5 continuity correction: P ( X  312 ) P ( X  312 ) = P ( X  312.5 ) z = 20.080.0400 80.0400312 = – 1.00 z = 20.080.0400 80.04005.312 = – 0.9375 P ( Z < – 1.00 ) = 0.1587. P ( Z < – 0.94 ) = 0.1736. P ( Z < – 0.9375 ) = 0.17425. 


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UIUC STAT 400 - 400Ex5_7ans

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