Sections 8.4: The Central Limit Theorem
Jiaping Wang
Department of Mathematics
04/22/2013, Monday
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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: 𝑛 𝑏−𝜇
≤ = P(Z ≤ 𝜎 𝜎
Where Z is the standard normal random variable.
𝑛 𝑏−𝜇 𝜎
)
Central Limit Theorem : Let X
1
, X
2
, …, X n be independent and identically distributed random variables with E(X i
V(X i
)= σ 2 <∞. Define 𝑌 𝑛
= 𝑛(
𝑋 𝑛
−𝜇 𝜎
) where 𝑋 𝑛
=
1 𝑛
)= μ and 𝑛 𝑖=1
𝑋 𝑖
.
Then Y n converges in distribution toward a standard normal random variable.
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Suppose that we wish to find an interval, (a, b), such that
𝑃 𝑎 ≤ which is equivalent to 𝑛 𝑎−𝜇 𝑛 𝑏−𝜇
𝑃( ≤ ≤ )=0.95, approximately, it is 𝜎 𝜎 𝜎 𝑛 𝑎−𝜇
𝑃( 𝜎 𝑛 𝑎−𝜇
≤ 𝑍 ≤
=-1.96 and 𝜎 𝑛 𝑏−𝜇
)=0.95, so we can have 𝜎 𝑛 𝑏−𝜇
=1.96  𝑎 = 𝜇 − 𝜎
1.96𝜎 𝑛
1.96𝜎
And b = 𝜇 + .
𝑛
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From 1976 to 2002, a mechanical golfer, Iron Byron, whose swing was modeled after that of Byron Nelson (a leading golfer in the 1940s), was used to determine whether golf balls met the Overall Distance Standard. Specifically,
Iron Byron would be used to hit the golf balls. If the average distance of 24 golf balls tested exceeded 296.8 yards, then that brand would be considered nonconforming. Under these rules, suppose a manufacturer produces a new golf ball that travels an average distance of 297.5 yards with a standard deviation of 10 yards.
1. What is the probability that the ball will be determined to be nonconforming when tested?
2. Find an interval that includes the average overall distance of 24 golf balls with probability of 0.95.
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Answer:
1. Assume n=24 is large enough for the approximation. The μ=297.5 and 𝜎
=
10
= 2.04.
Thus, 𝑃
296.8−297.5
= 𝑃 𝑍 > −0.34
𝑛 24 2.04
= 0.5 + 0.1331 = 0.6331.
2. We have seen that 𝜎
𝑃[𝜇 − 1.96( 𝜇 − 1.96( 𝜇 + 1.96( 𝑛
𝑋 ≤ 𝜇 − 1.96( 𝜎 𝑛
) ]=0.95  𝜎 𝑛 𝜎
) =297.5-1.96(10/(24) 1/2 )=293.5 and 𝑛
) =297.5+1.96(10/(24) 1/2 )=301.5
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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 σ 2 =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(1.8≤Z≤1.8)=2(0.4641)=0.9282.
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Let
𝑋 = 𝑛 𝑖=1
So 𝑋 𝑛
=
1 𝑛 𝑛 𝑖=1
𝑌𝑖 with Bernoulli trial Yi having probability p for success.
𝑌𝑖 =
, then we have E(X/n)=p, V(X/n)=p(1-p)/n, the normality follows from the CLT. As
𝑌,
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.
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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 ≤
=1-(0.5-0.3531)=0.8531.
3 + 0.5 − 6
100 0.06 0.94
= 1 − 𝑃 𝑍 < −1.05
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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.
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Apply the central limit theorem to approximate P[X
X
1
, …, X
20 and a common standard deviation σ= 10.
1
+X
2
+…+X
20
≤ 50], where are independent random variables having a common mean μ= 2
Answer: P[X
1
+X
2
+…+X
20
≤ 50] = P[(X
1
+X
2
+…+X
20
)/20≤ 50/20
= 𝑃
20
≤ 2.5 = 𝑃 20(
20
− 𝜇)/𝜎 ≤ 20(2.5 − 2)/10 ≈ 𝑃(𝑍 ≤ 0.224) =0.5+0.0871
=0.5871.
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Let X have a binomial distribution Bin(200,0.15). Find the normal approximation to P[25 <X < 35].
Answer: P(25<X<35)=P(X<35)P(X≤25)=P(X≤34)-P(X≤25)
34 + 0.5 − 200 0.15
≈ 𝑃 𝑍 ≤ − 𝑃 𝑍 ≤
200 0.15 0.85
=P(Z≤0.89)-P(Z≤-0.89)=0.3133(2)=0.6266.
25 + 0.5 − 200 0.15
200 0.15 0.85
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Roll a fair coin 100 times, use CLT to find the approximate probability that more than 60 tails shows.
Answer: n=100, p=0.5, X=number of tails.
P(X>60)=1P(X≤60) ≈ 1 − 𝑃 𝑍 ≤
60+0.5−100 0.5
100 0.5 0.5
=1P(Z≤2.1)=0.5-0.4772=0.0822.
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Page 405: 8.2, 8.4, 8.6 (a);
Page 417: 8.12, 8.16, 8.18, 8.28.
Due next Monday, 04/29/2013.
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