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elemprob-page17

# elemprob-page17 - np is the same as E S n and np(1-p is the...

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Example. Find P (1 X 4) if X is N (2 , 25). Answer. Write X = 2 + 5 Z . So P (1 X 4) = P (1 2 + 5 Z 4) = P ( - 1 5 Z 2) = P ( - 0 . 2 Z . 4) = P ( Z . 4) - P ( Z ≤ - 0 . 2) = Φ(0 . 4) - Φ( - 0 . 2) = . 6554 - [1 - Φ(0 . 2)] = . 6554 - [1 - . 5793] . Example. Find c such that P ( | Z | ≥ c ) = . 05. Answer. By symmetry we want c such that P ( Z c ) = . 025 or Φ( c ) = P ( Z c ) = . 975. From the table we see c = 1 . 96 2. This is the origin of the idea that the 95% significance level is ± 2 standard deviations from the mean. Proposition 7.1. We have the following bounds. For x > 0 1 2 π 1 x - 1 x 3 e - x 2 / 2 1 - Φ( x ) 1 2 π 1 x e - x 2 / 2 . Proof. Take the inequalities (1 - 3 y - 4 ) e - y 2 / 2 e - y 2 / 2 (1 + y - 2 ) e - y 2 / 2 and integrate. In particular, for x , large, P ( Z x ) = 1 - Φ( x ) 1 2 π 1 x e - x 2 / 2 e - x 2 / 2 . 8. Normal approximation to the binomial. A special case of the central limit theorem is Theorem 8.1. 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
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Unformatted text preview: 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. Example. Suppose a fair coin is tossed 100 times. What is the probability there will be more than 60 heads? Answer. np = 50 and p np (1-p ) = 5. We have P ( S n ≥ 60) = P (( S n-50) / 5 ≥ 2) ≈ P ( Z ≥ 2) ≈ . 0228 . Example. Suppose a die is rolled 180 times. What is the probability a 3 will be showing more than 50 times? 17...
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