Chapter 8. Some Approximations to Probability Distributions: Limit Theorems - PowerPoint Presentation

Slide1

Chapter 8. Some Approximations to Probability Distributions: Limit Theorems

More Practical Problems

Jiaping WangDepartment of Mathematics04/24/2013, WednesdaySlide2

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(1-p))

1/2=(400(0.2)(0.8))1/2

=20(0.4)=8P(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.Slide3

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=0.121Slide4

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.Slide5

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, σ

2=9, n=200, Tn=∑Xi

,

P(T

n

>4050)=P(Tn/n>4050/200)=P(

avg

>20.25)=P(n

1/2(

avg

-

μ

)/

σ

>(200)1/2(20.25-20)/3)

=P(Z>1.18)=0.5-0.381=0.119.Slide6

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 i

ts derivative h’(y)=1/c and the domain is (0,∞). Also the density function

For X is f(x)=1/

θ

e

-x/

θ

for x>0. We can have

Which is an exponential distribution with mean c

θ

.

 Slide7

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:

Y=e

X

is a monotone increasing function from 0 to

∞. The inverse function h(y)=

ln

(y) with derivative h’(y)=1/y. So