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L Let YY Be a Bernoulli Random Variable with Success Probability Pr(Y=1)=p\operatorname { Pr } ( Y = 1 ) = p

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L Let YY be a Bernoulli random variable with success probability Pr(Y=1)=p\operatorname { Pr } ( Y = 1 ) = p , and let Y1,,YnY _ { 1 } , \ldots , Y _ { n } be i.i.d. draws from this distribution. Let p^\hat { p } be the fraction of successes (1s) in this sample. In large samples, the distribution of p^\hat { p } will be approximately normal, i.e., p^\hat { p } is approximately distributed N(p,p(1p)n)N \left( p , \frac { p ( 1 - p ) } { n } \right) . Now let XX be the number of successes and nn the sample size. In a sample of 10 voters (n=10)( n = 10 ) , if there are six who vote for candidate A\mathrm { A } , then X=6X = 6 . Relate XX , the number of success, to p^\hat { p } , the success proportion, or fraction of successes. Next, using your knowledge of linear transformations, derive the distribution of XX . 24

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