# 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

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

Theorem 8.4 The practical importance of the Central Limit Theorem is that for large n, the sampling distribution of 𝑋 can be closely approximated by a normal distribution: 𝑃 𝑋≤𝑏 =𝑃

𝑛 𝑋−𝜇 𝜎

𝑛 𝑏−𝜇 𝜎

= P(Z ≤

𝑛 𝑏−𝜇 𝜎

)

Where Z is the standard normal random variable. Central Limit Theorem: Let X1, X2, …, Xn be independent and identically distributed random variables with E(Xi)=μ and 𝑋 −𝜇

1

𝑛 V(Xi)=σ2 296.8 ≈ 𝑃 𝑍 > = 𝑃 𝑍 > −0.34 𝑛 24 2.04 = 0.5 + 0.1331 = 0.6331.

2. We have seen that 𝜎 𝜎 𝑃[𝜇 − 1.96( 𝑛) ≤ 𝑋 ≤ 𝜇 − 1.96( 𝑛)]=0.95  𝜎

𝜇 − 1.96( 𝑛)=297.5-1.96(10/(24)1/2)=293.5 and 𝜎

𝜇 + 1.96( 𝑛)=297.5+1.96(10/(24)1/2)=301.5

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

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 σ2=1 ounce. The mean ounces of fill μ, 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. 𝑃 𝑋 − 𝜇 ≤ 0.3 = 𝑃 −0.3 ≤ 𝑋 − 𝜇 ≤ 0.3 = 𝑃 −0.3 6 ≤ 𝑍 ≤ 0.3 6 =P(-1.8≤Z≤1.8)=2(0.4641)=0.9282.

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Approximate Binomial by Normal Distribution Let 𝑋 = 𝑛𝑖=1 𝑌𝑖 with Bernoulli trial Yi having probability p for success. 𝑋 1 𝑛 So = 𝑌𝑖 = 𝑌, then we have E(X/n)=p, V(X/n)=p(1-p)/n, the normality 𝑛 𝑛 𝑖=1 follows from the CLT. As 𝑋 = 𝑛𝑌, X has approximately a normal distribution with a mean of np and a variance of np(1-p). 𝑃(𝑋 ≤ 𝑎) ≈ 𝑃(𝑍 ≤

𝑎 + 0.5 − 𝑛𝑝

)

𝑛𝑝 1 − 𝑝 If we can make sure 𝑝 ± 2 𝑝(1 − 𝑝)/𝑛 will lie within the interval (0,1) where 0.5 is the correct factor.

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

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 𝑝 ± 2 𝑝(1 − 𝑝)/𝑛 = 0.06 ± 0.024 which is entirely within (0,1). Thus The normal approximation should work. 𝑃 𝑋 ≥4 =1−𝑃 𝑋 ≤3 ≈1−𝑃 𝑍 ≤

3 + 0.5 − 6

100 0.06 0.94

= 1 − 𝑃 𝑍 < −1.05

=1-(0.5-0.3531)=0.8531.

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL

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 precinct’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 𝑋 𝑃 ≥ 0.55 = 1 − 𝑃 𝑋 < 0.55𝑛 = 1 − 𝑃 𝑋 ≤ 54 𝑛 54 + 0.5 − 50 ≈1−𝑃 𝑍 ≤ = 1 − 𝑃 𝑍 ≤ 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+…+X20≤ 50], where X1, …, X20 are independent random variables having a common mean μ= 2 and a common standard deviation σ= 10. Answer: P[X1+X2+…+X20≤ 50] = P[(X1+X2+…+X20)/20≤ 50/20 =𝑃 𝑋20 ≤ 2.5 = 𝑃 20(𝑋 20 − 𝜇)/𝜎 ≤ 20(2.5 − 2)/10 ≈ 𝑃(𝑍 ≤ 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

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