Presentation is loading. Please wait.

Presentation is loading. Please wait.

The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 8. Some Approximations to Probability Distributions: Limit Theorems More Practical Problems Jiaping.

Similar presentations


Presentation on theme: "The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 8. Some Approximations to Probability Distributions: Limit Theorems More Practical Problems Jiaping."— Presentation transcript:

1 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 8. Some Approximations to Probability Distributions: Limit Theorems More Practical Problems Jiaping Wang Department of Mathematics 04/24/2013, Wednesday

2 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(1- p)) 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.

3 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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

4 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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.

5 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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, T n =∑X i, 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.

6 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL 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.

7 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Problem 6 Let the random variable X have the normal distribution with mean μ and variance σ 2. Find the probability density function of Y=e X.


Download ppt "The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL Chapter 8. Some Approximations to Probability Distributions: Limit Theorems More Practical Problems Jiaping."

Similar presentations


Ads by Google