Chapter 8. Some Approximations to
Probability Distributions: Limit Theorems
More Practical Problems
Jiaping Wang
Department of Mathematics
04/24/2013, Wednesday
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Problem 1
Suppose we know in a crab farm, 20% of crabs are male. If
one day the owner catches 400 crabs, what is the chance that
more than 25% of the 400 crabs are male?
Answer: p=0.2, n=400, x=25%(400)=100, np=0.2(400)=80, (np(1p))1/2=(400(0.2)(0.8))1/2=20(0.4)=8
P(X>100)=1-P(X≤100)=1-P[(X-μ)/(np(1-p))1/2≤(100+0.5- 80)/8]=1-P(Z≤2.56)
=0.5-0.4948=0.0052.
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Problem 2
A process yields 10% defective items. If 100 items are
randomly selected from the process, what is the probability
that the number of defectives exceeds 13?
Answer: p=0.1, n= 100, np=10, [np(1-p)]1/2=3,
P(X>13)=1-P(X≤13)=1-P(Z≤(13+0.5-10)/3)=1-P(Z≤1.17)=0.5-0.379=0.121
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Problem 3
In the United States, 1/6 of the people are lefthanded. In a
small town (a random sample) of 612 persons, estimate the
probability that the number of lefthanded persons is strictly
between 90 and 150.
Answer: p=1/6, n=612, np=102, [np(1-p)]1/2= 9.22,
P(90<X<150)=P(X<150)-P(X≤90)=P(X≤149)-P(X≤90)
=P[Z≤(149+0.5-102)/9.22]-P[Z ≤(90+0.5-102)/9.22] =P(Z ≤5.15)-P(Z ≤-1.25)
=1-(0.5-0.3944)=0.8944.
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Problem 4
The weight of an arbitrary airline passenger's baggage has a
mean of 20 pounds and a variance of 9 pounds. Consider an
airplane that carries 200 passengers, and assume every
passenger checks one piece of luggage. Estimate the
probability that the total baggage weight exceeds 4050
pounds.
Answer: μ=20, σ2=9, n=200, Tn=∑Xi,
P(Tn>4050)=P(Tn/n>4050/200)=P(avg>20.25)=P(n1/2(avg-μ)/σ>(200)1/2(20.25-20)/3)
=P(Z>1.18)=0.5-0.381=0.119.
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Problem 5
Let X be exponentially distributed with a mean of θ. Find the
probability density function of the random variable Y=cX with
some positive constant c. Identify the distribution of Y
including the parameters.
Answer: Y=cX is a monotone increasing function as c>0. The inverse function
h(y)=y/c with its derivative h’(y)=1/c and the domain is (0,∞). Also the density function
For X is f(x)=1/θe-x/θ for x>0. We can have
1
𝑦 1
1
𝑦
𝑓𝑌 𝑦 = 𝜃 exp − 𝑐𝜃 𝑐 = 𝑐𝜃 exp − 𝑐𝜃 , 𝑦 > 0
Which is an exponential distribution with mean cθ.
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Problem 6
Let the random variable X have the normal distribution with
mean μ and variance σ2. Find the probability density function
of Y=eX.
Answer: 𝑓(𝑥) =
1
exp
𝜎 2𝜋
−
𝑥−𝜇
2𝜎2
2
. Y=eX is a monotone increasing function from 0 to
∞. The inverse function h(y)=ln(y) with derivative h’(y)=1/y. So
𝑓𝑌(𝑦) =
1
exp
𝜎 2𝜋
−
ln(𝑦)−𝜇
2𝜎2
2
1
,
𝑦
𝑦 > 0.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL