ECON 214 - The Normal Distribution.pdf

Suppose we wish to determine the probability of 20 or

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Suppose we wish to determine the probability of 20 or more heads in 30 tosses of a coin.
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Slide 50 Normal approximation of the binomial distribution By the binomial we have P(X ≥ 20 / n = 30, P = .5) = .0494 Using normal approximation means Slide 30(.5) 15 np (1 ) 30(.5)(.5) 2.74 np p
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Slide 51 Normal approximation of the binomial distribution To determine the appropriate normal approximation, we need to interpret “20 or more” as if values on a continuous scale. Thus we must subtract .5 from 20 to get 19.5 since the lower bound of 20 starts from 19.5 Hence P(X ≥ 19.5) = P(Z ≥ 1.64) = .0505 Slide ( 20/ 30, .5) ( 19.5/ 15, 2.74) B N P X n P P X
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Slide 52 Normal approximation of the binomial distribution When the Normal is used to approximate the Binomial, correction for continuity will always involve either adding .5 to or subtracting .5 from the number of successes specified. Generally the continuity correction is as follows: P(X < b ) => subtract .5 from b ( b is exclusive) P(X > a ) => add .5 to a ( a is exclusive) P(X ≤ b ) => add .5 to b ( b is inclusive) P(X ≥ a ) => subtract .5 from a ( a is inclusive)
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Slide 53 Normal approximation of the Poisson distribution When the mean, λ , of the Poisson distribution is large, we can approximate with the normal. Approximation is appropriate when λ ≥ 10 Then The correction for continuity similarly applies. Slide
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Slide 54 Normal approximation of the Poisson distribution Suppose an average of 10 calls per day come through a telephone switchboard. What is the probability that 15 or more calls will come through on a randomly selected day. Using Poisson we obtain P(X ≥ 15 / λ = 10) = .0835
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