2008 First Semester STATS课程官方问答贴

luvwei

哈哈

我晕了

是第四题 拉～～～

z-score

原帖由 luvwei 于 2008-4-10 11:03 发表
哈哈

我晕了

是第四题 拉～～～

(a) 我们不能算这个probability, 因为我们不知道passenger luggages的distribution.

(b) 4000 / 40 = 100.

(c) Yes you can. X bar ~ Normal(95, 11/sqrt(40))
IP(X bar > 100) use either R or SPSS to calculate the probability.


[ 本帖最后由 z-score 于 2008-4-10 13:02 编辑 ]

luvwei

thank you~~~~

chocolover

感谢~ 会好好做练习题的

chocolover

chapter 7 好难理解 ...

z-score

原帖由 chocolover 于 2008-4-10 23:01 发表
chapter 7 好难理解 ...

我说的可能比较简练, 但是我感觉大家现在还不用担心. 在Test之前把我上载的所有past test都作一遍, 我认为Chapter 7和probability的问题会变得容易一些.

天煞孤星

lz精通surveycraft, converso这类的软件么？

t-multiplier

原帖由 天煞孤星 于 2008-4-12 08:20 发表
lz精通surveycraft, converso这类的软件么？

不好意思，我从来没用过这些软件。

mayigo

麻烦atm能解释一下108 第四题如何解题，我怎么看不明白，用什么方法计算呢

c) Yes you can. X bar ~ Normal(95, 11/sqrt(40))
IP(X bar > 100) use either R or SPSS to calculate the probability.

mayigo

(c)        Let P be the weight of passers and their luggage.
Pr(x>100)=1-pr(x<100)=1-0.6753= 0.3247
The probability of the plane being overloaded is 32.47%
我这样做对吗？

z-score

原帖由 mayigo 于 2008-4-12 22:53 发表
(c)        Let P be the weight of passers and their luggage.
Pr(x>100)=1-pr(x

你能告诉我你怎么计算的吗?
我用R计算得出
> 1-pnorm(100,95,11/sqrt(40))
[1] 0.002021431

mayigo

原帖由 z-score 于 2008-4-14 08:38 发表


你能告诉我你怎么计算的吗?
我用R计算得出
> 1-pnorm(100,95,11/sqrt(40))
[1] 0.002021431

我就是用excel算出100的百分比，
输入x=100
        mean=95
        standard deviation =11
        true
得出结果，有什么问题吗？

[ 本帖最后由 mayigo 于 2008-4-14 13:50 编辑 ]

z-score

原帖由 mayigo 于 2008-4-14 13:32 发表
我就是用excel算出100的百分比，
输入x=100
        mean=95
        standard deviation =11
        true
得出结果，有什么问题吗？

哦，你在Excel要用NORMDIST的function算．

1-NORMDIST(100,95,11/xqrt(40),TRUE)

~Kurosagi~

妳們做的好快啊.....

我stat 108 asssi 2 的question 3 還沒有搞定..

3d  好煩...

rabby

108第四题不就只有A&B吗 为什麼会有C呢?

~Kurosagi~

有C 啊.....

妳在仔細看看

z-score

嗯，虽然我还没看，但是肯定有C.

rabby

喔原来我做错 做成07年的assignment 2
幸好有这帖子 不然我就完了

~Kurosagi~

對暸


question 3d 那裏

號混亂...

痲煩教一下

謝謝

z-score

原帖由 ~Kurosagi~ 于 2008-4-14 15:36 发表
對暸


question 3d 那裏

號混亂...

痲煩教一下

謝謝

你是说309和310楼的解答吗？

~Kurosagi~

已經有暸???

奇怪為什么我往前看沒有找到那

再去看看

謝謝

~Kurosagi~

謝謝謝謝誒!!!!!!

z-score

原帖由 qian.qian 于 2008-4-15 14:31 发表
4. The distribution of a random variable X is given by the following probability function.(12 mark(s))
P(X = x) =
1/2 , if x = 0
1/4 , if x = 1
1/8 , if x = 2 or 3
0, otherwise.

(a) Calcu ...

哦，这个不能用2^IE(X)因为2^X不是一个linear transformation. Linear transformation指得是IE(aX + b) = aIE(X) + b.

所以这个我们要用sum(2^X * IP(X = x)) from negative infinity to positive infinity.

水鱼皇后

2008 Assignment 2

Download data set 2008Assig2.csv
It contains the same variables as before and some additional ones:
CMR Child Mortality rate (number of deaths per thousand children)
Pop1000: Population size in thousands
TotFert Fertility rate (children per woman)
NatInc Gross National Income per person.

I want to analyse the number of children that die from different diseases per thousand
children. So I can tell countries that have high death rates from ones that don’t.
Prepare the data and then perform a PCA on the number of deaths per thousand
children for the 8 different sources of mortality.


我想问一下在做PCA之前 要怎么prepare the data...
perform a PCA on the number of deaths per thousand children for the 8 different sources of mortality
8 different sources of mortality是proportion...怎么跟CMR Child Mortality rate(number of deaths per thousand children)用在一起..
需要怎么样的standardisation？

ps. 麻烦阿童木把data set放上来...

z-score

原帖由 qian.qian 于 2008-4-15 16:07 发表
谢谢,阿童木...
那么IE(2^x)=2^0*1/2+2^1*1/4+2^2*1/8+2^3*1/8?

是的。

kirstens

P(d)=0.52
(a) What is the distribution of X? the probability distribution of the number Y = X − 1 of failures before the first success, should be Geometric distribution

(b) What is the probability that the couple will have at least 5 children?
cdf for Geometric distribution = 1-(1-p)^k
Note P(at least 5)=1-P(k<=4)
therefore P(K>=5)=1-P(K<=4)=1-[1-(1-p)^k]=(1-p)^k=...

(c) Find E[X].
E(X)=1/p

(d) A different childless couple (with the same birth probabilities) decide to keep having children
until they have both a son and a daughter. Let Y be the number of children that they will have. Give an expression for P(Y = n) (hint: it is very helpful here to first condition on the
gender of the first child!)
to first condition on the gender:
-if first is a boy, E(X)=1/p
-if first is a girl, E(X')=1/q
the probability for get a boy is q and for girl is p therefore E(Y)=qE(X)+pE(X')

z-score, would you check if I'm right? haven't been playing with this for a long time.

z-score

原帖由 kirstens 于 2008-4-16 13:09 发表
P(d)=0.52
(a) What is the distribution of X? the probability distribution of the number Y = X − 1 of failures before the first success, should be Geometric distribution

(b) What is the  ...

kirstens你很强，你的逻辑很清楚。佩服一个。