Understanding Normal and Poisson Approximations in Probability
Explore the concepts of Normal and Poisson approximations in probability theory through examples like finding broken biscuits in boxes and calculating flu percentages in samples. Discover how these approximations can simplify complex calculations in binomial distributions.
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Presentation Transcript
Hypothesis ristu.saptono@staff.uns.ac.id
Outline Normal and Poisson Approximation for Binomial Distribution Confidence Interval
Poisson Approximation Reminder for probability of Poisson ? ? = ? =??? ? ?! The value of mean ? = ?? On easy way For large values of n and small values of p ? > 20,?? < 5 ?? ?? < 5 Test ? = ?? New parameters
Example Question A factory puts biscuits into boxes of 100. The probability that a biscuit is broken is 0.03. Find the probability that a box contains 2 broken biscuits
Example Solution This is a binomial distribution with n = 100 and p = 0.03. These values are outside the range of the tables and involve lengthy calculations. Using the Poisson approximation (test: np = 100 x 0.3 = 3 5) Let X be the random variable of the number of broken biscuits The mean = np = 100 0.3 = 3 P(X = 2) = 0.224 (from tables) OR =POISSON.DIST(2,3,FALSE) The probability that a box contains two broken biscuits is 0.224.
Normal Approximation Binomial distribution (discrete) Normal distribution (continuous) Requirement if ?~???(?,?) with ? large enough and ?? 10 ? 1 ? = ?? 10 Normal Approximation ?~?(? = ??,? = ??? )
Example Suppose in a sample of ? = 10000 people, 9% have the flu. Notice that ? = 10000(.09) = 900 and ? = 10000(.09)(.91) = 819. Find ?(? 950) with the normal approximation.
Solution : Test First, check the number of expected successes and failures. ?? = 10000(0.09) = 900, ?? = 10000(.91) = 9100. Both are much larger than 10. So ? ?(? = 900,? = 819).
Solution I will turn ? = 950 into a ? ?????. ? =? ?? ? 819 ?(? 950) = ?(? 1.75) = 0.9599 Since we only have negative Z score table then ? ? 950 = ? ? 1.75 = 1 ?(? < 1.75) =950 900 = 1.75
Discussion 13.1: A factory puts biscuits into boxes of 1000. The probability that a biscuit is broken is 0.03. Find the probability that a box contains at least 10 broken biscuits Suppose in a sample of ? = 100 people, 4% have the flu. Find the probability at worst 3 people get a flu.
Illustration Population Sample
Illustration ?2 ?2 Population Sample ? ?
?2 ?2 Glossary Popula tion Sam ple ? is average of sample ?2 is variance of sample ? is standard deviation of sample ? ? ? is average of population ?2 is variance of population ? is standard deviation of sample
Formula Sample Average (mean) Population Average (mean) ? = ??? ? = ??? ? ? Variance Variance ?2= ??? ?2 ? 1 ?2= ??? ?2 ?
Confidence Interval (1 ?)100% To make sure that the mean of population is in the range of sample with certain confidence level Confidence level is equal to ? where: ? = ? ? ?? + ?(? ?? 2) 2
Confidence Interval 1 ? 100% - Step Take a Confidence Level (?) Find the ? = ??/2 Make a range ? ? ??? ? ? ? + ? ??? ?
Discussion 13.2 Discussion 13.2 Perhatikan nilai ujian statistika untuk 80 orang mahasiswa berikut: 79 49 48 74 81 98 87 80 80 84 90 70 91 93 82 78 70 71 92 38 56 81 74 73 68 72 85 51 65 93 83 86 90 35 83 73 74 43 86 88 92 93 76 71 90 72 67 75 80 91 61 72 97 91 88 81 70 74 99 95 80 59 71 77 63 60 83 82 60 67 89 63 76 63 88 70 66 88 79 75 Make a Confidence Interval 95%
Confidence Interval for proportion (?) Take a Confidence Level (?) Find the ? score for ?/2 Make a range ? ??/2 ??? ? ? ? + ??/2 ??? ?
Proportion and Estimation ? =?? ?=?1+ ?2+ + ?? ? Where ?,?? are number of trial and number of succeed, respectively. ?? ~?(?) distributed in Bernoulli Please proof that ? ? = ? ? =?(1 ?) ??? ?
Discussion 13.3 Suppose we have taken a random sample (i.i.d. or simple) of n=500 voters, where X=220 support Richard Guy, a candidate for political office. Find interval estimate for 95% confidence is
