stats week 5

# stats week 5 - Sarah Anderson Statistics Week 5 6...

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Sarah Anderson Statistics Week 5 14.) This is a binomial distribution with parameters n = 6 and p = 0.95 A. P(X = 6) = (6 choose 6)(0.95)^6 (1 - 0.95)^(6 - 6) = 0.95^6 ≈ 0.7351 B. P(X = 5) = (6 choose 5)(0.95)^5(0.05)^1 ≈ 0.2321 C. The mean μ of a binomial distribution is given by μ = np = 5.7 D. The variance σ² is given by σ² = np(1-p) = 0.285 The standard deviation σ equals the square root of the variance. 16.) n=6 p=0.3 6.) Admissions Probability X * P(x) (x-u)sq. (x-u)sq.P(x) 1 000 0.6 600 12100 7260 1200 0.3 360 8100 2430 1500 0.1 150 152100 15210 total 1 1110 24900 a. u= 1110 b.osq.=24900 O=157.79 7.) Amount P(x) xP(x) (x-u)sq.P(x) 10 0.5 5 60.5 25 0.4 10 6.4 50 0.08 4 67.28 100 0.02 2 124.82 total 21 259 a. u= 21 b. 259 O=16.093

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q=0.7 use this P(x=K) = ( n nCr k ) * p^k * q^(n-k) a) k=4 . ...p(x=4) = 0.0595 b) k=0 . ...p(x=0) = 0.1176 c) k=2 . ...p(x=2) = 0.3241 d) E(x) = n*p = 1.8 , answer = 2*1.8 = 3.6 24. a) All six would be .8^6 = 26.2% c) The probablity none have the transponder is .2^6 = 0.0064% b) For the second one, the fastest way to the answer is to find (a) the probability of none, which you
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## This note was uploaded on 04/12/2011 for the course STAT 2610 taught by Professor Sanjeev during the Spring '11 term at Bemidji State.

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stats week 5 - Sarah Anderson Statistics Week 5 6...

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