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Part II – Free Response 1a) Mean = 0.8 Standard deviation = √ (0.8)(0.2) 120 = 0.0365 1b) np = (0.8)(120) = 96 n(1 – p) = (0.2)(120) = 24 *Both are more than 10 so the sampling distribution is normal 1c) Shape – Not normal np = (0.8)(30) = 24 n(1 – p) = (0.2)(30) = 6 *Since n(1 – p) is less than 10, then the sampling distribution is not normal Center - Same Mean = 0.8 Spread – Larger Standard deviation = √ (0.8)(0.2) 30 = 0.073 1d) The largest possible sample that could be drawn is 50 because that is 10% of the population size (500). 2a) Mean = 0.22 Standard deviation = √ 2b) z = 0.3−0.22 0.0586 (0.22)(0.78) 50 = 0.0586 = 1.36 P(z > 1.36) = normalcdf(1.36, e99,0,1) = 0.0869