# ppt - Department of Mathematics

January 31, 2018 | Author: Anonymous | Category: Math, Statistics And Probability, Normal Distribution

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Chapter 8. Some Approximations to Probability Distributions: Limit Theorems Sections 8.4: The Central Limit Theorem

Jiaping Wang Department of Mathematics 04/22/2013, Monday

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Theorem 8.4 The practical importance of the Central Limit Theorem is that for large n, the sampling distribution of π can be closely approximated by a normal distribution: π πβ€π =π

π πβπ π

β€

π πβπ π

= P(Z β€

π πβπ π

)

Where Z is the standard normal random variable. Central Limit Theorem: Let X1, X2, β¦, Xn be independent and identically distributed random variables with E(Xi)=ΞΌ and π βπ

1

π V(Xi)=Ο2 296.8 β π π > = π π > β0.34 π 24 2.04 = 0.5 + 0.1331 = 0.6331.

2. We have seen that π π π[π β 1.96( π) β€ π β€ π β 1.96( π)]=0.95 ο  π

π β 1.96( π)=297.5-1.96(10/(24)1/2)=293.5 and π

π + 1.96( π)=297.5+1.96(10/(24)1/2)=301.5

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Example 8.7 A certain machine that is used to fill bottles with liquid has been observed over a long period, and the variance in the amounts of fill has been found to be approximately Ο2=1 ounce. The mean ounces of fill ΞΌ, however, depends on an adjustment that may change from day to day or from operator. If n= 36 observations on ounces of fill dispensed are to be taken on a given day (all with the same machine setting), find the probability that the sample mean will be within 0.3 ounce of the true population mean for the setting. Answer: n=36 is large enough for the approximation. π π β π β€ 0.3 = π β0.3 β€ π β π β€ 0.3 = π β0.3 6 β€ π β€ 0.3 6 =P(-1.8β€Zβ€1.8)=2(0.4641)=0.9282.

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Approximate Binomial by Normal Distribution Let π = ππ=1 ππ with Bernoulli trial Yi having probability p for success. π 1 π So = ππ = π, then we have E(X/n)=p, V(X/n)=p(1-p)/n, the normality π π π=1 follows from the CLT. As π = ππ, X has approximately a normal distribution with a mean of np and a variance of np(1-p). π(π β€ π) β π(π β€

π + 0.5 β ππ

)

ππ 1 β π If we can make sure π Β± 2 π(1 β π)/π will lie within the interval (0,1) where 0.5 is the correct factor.

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Example 8.8 Six percent of the apples in a large shipment are damaged. Before accepting each shipment, the quality control manager of a large store randomly selects 100 apples. If four or more are damaged, the shipment is rejected. What is the probability that this shipment is rejected? Answer: Check π Β± 2 π(1 β π)/π = 0.06 Β± 0.024 which is entirely within (0,1). Thus The normal approximation should work. π π β₯4 =1βπ π β€3 β1βπ π β€

3 + 0.5 β 6

100 0.06 0.94

= 1 β π π < β1.05

=1-(0.5-0.3531)=0.8531.

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Example 8.9 Candidate A believe that she can win a city election if she receives at least 55% of the votes from precinct I. Unknown to the candidate, 50% of the registered voters in the precinct favor her. If n=100 voters show up to vote at precinct I, what is the probability that candidate A will receive at least 55% of that precinctβs votes? Answer: Let X denote the number of voters in precinct I who vote for candidate A. The probability p that a randomly selected voter favors A is 0.5, then X can be Considered as a binomial distribution with p=0.5 and n=100. Approximately by normal Distribution, we need find π π β₯ 0.55 = 1 β π π < 0.55π = 1 β π π β€ 54 π 54 + 0.5 β 50 β1βπ π β€ = 1 β π π β€ 0.9 100 0.5 0.5 =0.5-0.3159=0.1841. The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Additional Example 1 Apply the central limit theorem to approximate P[X1+X2+β¦+X20β€ 50], where X1, β¦, X20 are independent random variables having a common mean ΞΌ= 2 and a common standard deviation Ο= 10. Answer: P[X1+X2+β¦+X20β€ 50] = P[(X1+X2+β¦+X20)/20β€ 50/20 =π π20 β€ 2.5 = π 20(π 20 β π)/π β€ 20(2.5 β 2)/10 β π(π β€ 0.224)=0.5+0.0871 =0.5871.

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Additional Example 2 Let X have a binomial distribution Bin(200,0.15). Find the normal approximation to P[25