7. A study of the home pages of Fortune 500 companies reports that the mean number of bad links per home page is 0.4 and the mean number of spelling errors per home page is 0.16. Find the probability that a randomly selected home page will contain
(1) exactly 0 bad links.
(2) 5 or more bad links.
(3) exactly 0 spelling errors.
(4) 10 or more spelling errors.
8. Suppose that the number of travelers per day at the WinnipegInternationalAirport has a mean of 500 and a variance of 40000. The random variable is thus discrete. Further assume that the airport opens 365 days every year. Let be the mean number of travelers per day during a year.
(1) What is the sampling distribution of ? What theorem are you using to support your claim?
(2) What is the approximate probability that is less than 480?
(3) What is the approximate probability that there are more than 200,750 travelers in a randomly selected year?
好深既statictics 3 --- probability
2008-05-07 8:44 am
回答 (3)
2008-05-07 8:04 pm
✔ 最佳答案
7(1)
bad links follows Poisson (0.4)
Pr(exactly 0 bad links)=e^(-0.4)=0.6703
(2)
Pr(5 or more bad links)
=1-Pr(1)-Pr(2)-Pr(3)-Pr(4)
=1-0.6703-0.2681-0.0536-0.00715-0.000715
=0.000135
(3)
spelling errors follows Poisson (0.16)
Pr(exactly 0 spelling errors)
=e^(-0.16)
=0.8521
(4)
Pr(10 or more spelling errors)
=1-Pr(0)-Pr(1)-...-Pr(9)
8
(1) The sampling distribution is a normal distribution by central limit theorem
(2)
Pr (X<480)
=Pr(Z<(480-500)/200)
=Pr(Z<-0.1)
=0.4602
(3)
S~(182500,14600000)
Pr(S>200750)
=Pr(Z>(200750-182500)/sqrt(14600000))
=Pr(Z>4.7762)
=0.000001
2008-05-07 12:04:35 補充：
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2008-05-07 16:07:03 補充：
係bad question﹐但是一來這類error題是用poisson﹐二來它比得一個參數﹐不用poisson無甚麼好用
2008-05-10 11:06 pm
7
(1)
bad links follows Poisson (0.4)
Pr(exactly 0 bad links)=e^(-0.4)=0.6703
(2)
Pr(5 or more bad links)
=1-Pr(1)-Pr(2)-Pr(3)-Pr(4)
=1-0.6703-0.2681-0.0536-0.00715-0.000715
=0.000135
(3)
spelling errors follows Poisson (0.16)
Pr(exactly 0 spelling errors)
=e^(-0.16)
=0.8521
(4)
Pr(10 or more spelling errors)
=1-Pr(0)-Pr(1)-...-Pr(9)
8
(1) The sampling distribution is a normal distribution by central limit theorem
(2)
Pr (X<480)
=Pr(Z<(480-500)/200)
=Pr(Z<-0.1)
=0.4602
(1)
bad links follows Poisson (0.4)
Pr(exactly 0 bad links)=e^(-0.4)=0.6703
(2)
Pr(5 or more bad links)
=1-Pr(1)-Pr(2)-Pr(3)-Pr(4)
=1-0.6703-0.2681-0.0536-0.00715-0.000715
=0.000135
(3)
spelling errors follows Poisson (0.16)
Pr(exactly 0 spelling errors)
=e^(-0.16)
=0.8521
(4)
Pr(10 or more spelling errors)
=1-Pr(0)-Pr(1)-...-Pr(9)
8
(1) The sampling distribution is a normal distribution by central limit theorem
(2)
Pr (X<480)
=Pr(Z<(480-500)/200)
=Pr(Z<-0.1)
=0.4602
2008-05-07 9:58 pm
This is really a bad question. How could we calculate the probability without knowing the distribution before?
What happen if there is exactly one home page containing 200 bad links and 80 spelling errors ?
2008-05-07 13:59:06 補充：
We should make an assumption on the distribution.
Since the sample size is large enough (500), I would prefer to adopt the Normal Distribution rather than Poisson Distribution.
I believe that this is the assumption behind the question. Am I correct?
What happen if there is exactly one home page containing 200 bad links and 80 spelling errors ?
2008-05-07 13:59:06 補充：
We should make an assumption on the distribution.
Since the sample size is large enough (500), I would prefer to adopt the Normal Distribution rather than Poisson Distribution.
I believe that this is the assumption behind the question. Am I correct?
收錄日期: 2021-04-16 16:47:19
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