Lecture 8 - TA Office Hours: Vadiraj Hombal: Thursdays...

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TA Office Hours: Vadiraj Hombal: Thursdays 4:30-6:30 PM JEC 6027 Max Henderson: Sundays 4:00-7:00 PM McNeil Room Scott Alderman: Tue/Thur: 5:00-7:00 PM MRC 318 Li Huijiang: TBA Exam 1 Results: Average: 79.79 Standard Deviation: 21.45 For Questions About Your Individual Grade Contact: Ajay Malaviya: malava@rpi.edu
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x 1 234 p(x) 0.4 0.3 0.2 0.1 5-41 2 x based on a sample of 2 compute the probability distribution of x 4 16 possible outcomes =
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11 . 522 . 533 . 54 0.16 0.24 0.25 0.20 0.10 0.04 0.01 x ) ( x p () 2 0 . 5 vs. 2 V ( x ) 1 Ex V x = = == ) ( 2.5) 0.16 0.24 0.25 0.20 0.85 bp x ≤= +++ =
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d) n 4 p(x 1.5) prob(1,1,1,1) 4prob(2,1,1,1) 4 (2,2,1,1) 2 4prob(3,1,1,1) 0.240 prob = ≤= + + ⎛⎞ + ⎜⎟ ⎝⎠ = c) Generate the sampling distribution of the range for a sample size of 2 by examining the probabilities of the 16 sampling outcomes for n=2.
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Let X 1 , X 2 ,…X n be a random sample from any distribution with mean value μ and standard deviation σ . Then E(Xbar) = μ V(Xbar) = σ 2 /n and σ Xbar = σ / n (If T 0 is the sample total, E(T 0 )=n μ , V(T 0 )=n σ 2 ), i.e., the variance of the sum is the sum of the variances) (Recall from problem 5-41 that V(X)=nV(Xbar))
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If X 1 , X 2 ,…X n is a random sample from a
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Lecture 8 - TA Office Hours: Vadiraj Hombal: Thursdays...

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