12.07.2015 Views

Course Notes - Department of Mathematics and Statistics

Course Notes - Department of Mathematics and Statistics

Course Notes - Department of Mathematics and Statistics

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BSummary <strong>of</strong> Formulae1. Normal DistributionIf X is a normal r<strong>and</strong>om variable with parameters µ X (mean) <strong>and</strong> σ 2 X(variance)• Mean: µ X• St<strong>and</strong>ard deviation: σ X =√σ 2 XA st<strong>and</strong>ard normal r<strong>and</strong>om variable Z has mean µ Z = 0 <strong>and</strong> σ 2 Z = 1.To transform a normal r<strong>and</strong>om variable X into a st<strong>and</strong>ard normal(<strong>and</strong> vice versa):2. Binomial DistributionZ = X − µ Xσ X<strong>and</strong> X = Zσ X + µ X .If X is a binomial r<strong>and</strong>om variable with n trials <strong>and</strong> probability πthen• Mean: µ X = nπ• St<strong>and</strong>ard deviation: σ X = √ nπ(1 − π)• If nπ <strong>and</strong> n(1 − π) are both greater than 5, then X is approximatelynormally distributed with mean µ X <strong>and</strong> variance σ 2 X .3. Distributions <strong>of</strong> <strong>Statistics</strong>• The mean ¯X <strong>of</strong> a r<strong>and</strong>om sample <strong>of</strong> size n has mean µ ¯X = µ X<strong>and</strong> st<strong>and</strong>ard deviation σ ¯X = √ σ Xn.• The sample proportion P computed from a binomial distributionwith parameters √n <strong>and</strong> π has a mean <strong>of</strong> µ P = π <strong>and</strong> st<strong>and</strong>ardπ(1−π)deviation σ P =n. If nπ <strong>and</strong> n(1 − π) are both greaterthan 5, then P will be approximately normally distributed.• The distribution <strong>of</strong> the difference between two sample means ¯X 1 −¯X 2 has a mean <strong>of</strong> µ ¯X1 − ¯X 2= µ 1 − µ 2 <strong>and</strong> a st<strong>and</strong>ard deviation <strong>of</strong>√σ ¯X1 − ¯Xσ12=2 n 1+ σ2 2n 2.271

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