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Unformatted text preview: 1c The mean of the population data is μ = ∑x ∕ N The variance of the population data is Item x A 33 5.76 B 45 92.16 C 54 345.96 D 27 70.56 E 18 302.76 177 817.2 μ = 35.4 163.44 2c The mean of the sample data is The variance of the sample data Item x A 33 1 B 45 169 E 18 196 96 ∑(x − x‾)² = 366 32 183 3c The mean of the means equals the mean: 35.4 N = 5, n = 3 The variance of the means is Since the population is finite, you must include the Finite Population Correction Factor in the formula FPCF = (N − n) ∕ (N − 1 N =5 n =3 FPCF =0.5 163.44 27.24 4b C(N, n) = C(36, 6) =1,947,792 5b 200 The mean of the means equals the mean: 200 6d 50 5 7d μ =80 σ =30 Here we are dealing with the distribution of the population data. P(x < 77) z = (x  μ) ∕ σ z =0.1 P(z < 0.1) =0.4602 8a P(74 < x < 86) Again, we are dealing with the population data. x =86 x =74 z =0.2 z =0.2 P(0.2 < z < 0.2) =0.1585 9c 30.8 129.2 10b μ =80 σ =30 n =25 6 z =0.5 P(z < 0.5) =0.3085 11d μ =80 σ =30 n =64 P(72.5 < x‾ < 87.5) 3.75 72.5 z =2 87.5 z =2 P(2 < z < 2) =0.9545 12c μ =80 σ =30 n =64 Find the 90% MOE....
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This document was uploaded on 10/27/2011 for the course ECONE 270 at Indiana.
 Fall '10
 ToFu

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