elemprob-fall2010-page41

elemprob-fall2010-page41 - ≈ P Z ≥ 2 ≈ 0228 An...

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16 Normal approximation to the binomial A special case of the central limit theorem is Proposition 16.1 If S n is a binomial with parameters n and p , then P a S n - np np (1 - p ) b P ( a Z b ) , as n → ∞ , where Z is a N (0 , 1) . This approximation is good if np (1 - p ) 10 and gets better the larger this quantity gets. Note np is the same as E S n and np (1 - p ) is the same as Var S n . So the ratio is also equal to ( S n - E S n ) / Var S n , and this ratio has mean 0 and variance 1, the same as a standard N (0 , 1). Note that here p stays fixed as n → ∞ , unlike the case of the Poisson approximation. An example. Suppose a fair coin is tossed 100 times. What is the proba- bility there will be more than 60 heads? Answer. np = 50 and np (1 - p ) = 5. We have P ( S n 60) = P (( S n - 50)
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Unformatted text preview: ≈ P ( Z ≥ 2) ≈ . 0228 . An example. Suppose a die is rolled 180 times. What is the probability a 3 will be showing more than 50 times? Answer. Here p = 1 6 , so np = 30 and p np (1-p ) = 5. Then P ( S n > 50) ≈ P ( Z > 4) , which is very small. An example. Suppose a drug is supposed to be 75% effective. It is tested on 100 people. What is the probability more than 70 people will be helped? Answer. Here S n is the number of successes, n = 100, and p = . 75. We have P ( S n ≥ 70) = P (( S n-75) / p 300 / 16 ≥ -1 . 154) ≈ P ( Z ≥ -1 . 154) ≈ . 87 . (The last figure came from a table.) 41...
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