查看完整版本: [-- 16/11更新Stat QuestionsX1[趕急] --]

【好友論壇】- 以足球會好友 -> 【學術交流】 -> 16/11更新Stat QuestionsX1[趕急] [打印本頁] 登錄 -> 注冊 -> 回復主題 -> 發表主題

vincent 2011-09-22 22:49

16/11更新Stat QuestionsX1[趕急]

Let the probability density function of Y be

            cy exp(-2y),     0<=y<無限;
f(y)=
            0,                     elsewhere.

where c is a constant.

(a) Find the value of c that makes f(y) a density function. (not yet confirmed ANS:2)
(b) State the distribution of Y, and give the mean and varience for Y.
(c) Write down the moment generating function for Y.

hohoho2 2011-09-23 14:38
1. (5C2)/(9C6)=10/84=0.1190
2.[(n-r-1)! *r!]/ n!

唔係好記得點systematic地solve lu,但係應該無錯

vincent 2011-09-23 16:50
引用第1樓hohoho22011-09-23 14:38發表的“”:
1. (5C2)/(9C6)=10/84=0.1190
2.[(n-r-1)! *r!]/ n!
唔係好記得點systematic地solve lu,但係應該無錯

可不可以簡單講下第一條個答案點黎

hohoho2 2011-09-23 20:47
probability = no of favourable outcome/total outcome
講左點計total outcome先
呢隻係combination with replacement
有個例子係當你有 3 個波要放落 3 個格,每個格可以放多過一個波 ,問有幾多種排法
呢個時候你可以當係有3個波同3-1 = 2條分隔線排次序
OOO||
OO|O|
OO||O
O|OO|
O|O|O
O||OO
|OOO|
|OO|O
|O|OO
||OOO

10種排法

其實如果全部都理order o既話本身有5!=120咁多種排法,但係你會發現到本身5種排法入面果3粒波之間o既order係無用的,呢到每個排法整多左3!個combination出黎,另外果兩條分隔線個order都係唔使理,呢到每個排法又多左2!個combination
所以total outcome=5!/(2!x3!)=10,即係5C3
或者3(個波)+(3-1)(條分隔線) C 3(波既數量)
當波數係k,格數係n
total outcome就係(n+k-1)Cn <----formula

你呢到有6個波同4個格,即係有9C6個outcome=84

然後計favourable outcome
有邊個係favourable outcome?
每個格最少有1個波,所以有4個波o既位置係fix左,然後剩番兩個亂排出黎果o的結果就係favourable outcome
呢個時候你可以當係2個波4個格咁睇,(2+4-1)C2=5C2=10

大約係咁la

hohoho2 2011-09-25 01:00
BTW今日有同學拎住份assignment問我同樣呢兩條問題

vincent 2011-09-25 01:27
引用第4樓hohoho22011-09-25 01:00發表的“”:
BTW今日有同學拎住份assignment問我同樣呢兩條問題 [表情] [表情] [表情] [表情] [表情] [表情] [表情]

咩同學= =

hohoho2 2011-09-25 01:29
引用第5樓vincent2011-09-25 01:27發表的“”:
咩同學= =

大學同學lor

happykinghk 2011-09-25 15:04
引用第0樓vincent2011-09-22 22:49發表的“Statistics Questions (each 250好友元)”:
1)
Assume that every time you buy an item of the HK Disney series, you receive one of the four types of cards, each with a cartoon character Mickey, Minnie, Donald and Daisy with an equal probability. Over a period of time, you buy 6 items of the series. What is the probability that you will get all four cards?
2)
If n persons, including A and B, are randomly arranged in a straight line, what is the probability that there are exactly r persons in the line between A and B?

我既答案:

1.) (4P4 x 6P2) / (4C1)^6 = 0.176

2.) 同hohoho一樣

hohoho2 2011-09-25 17:09
引用第7樓happykinghk2011-09-25 15:04發表的“”:
我既答案:
1.) (4P4 x 6P2) / (4C1)^6 = 0.176
.......

咁應該你o岩了,我好似貪快照sub formula無理到o的outcome係with unequal probability

happykinghk 2011-09-25 17:41
引用第8樓hohoho22011-09-25 17:09發表的“”:
咁應該你o岩了,我好似貪快照sub formula無理到o的outcome係with unequal probability [表情]

做咩咁客氣
其實我都唔肯定,諗住一齊研究下咁解

fatman 2011-09-25 18:05
呢兩條係elementary 既 prob題目黎

joN3 2011-09-25 19:34
引用第10樓fatman2011-09-25 18:05發表的“”:
呢兩條係elementary 既 prob題目黎 [表情]


vincent 2011-09-26 01:09
happykinghk
hohoho2

謝謝

vincent 2011-10-17 23:45
題目更新了
希望大家幫幫手
好急但唔識做

plmplm 2011-10-18 01:09
好明顯係功課黎

hohoho2 2011-10-18 10:45
引用第0樓vincent2011-09-22 22:49發表的“(each重賞500好友元)Stat QuestionsX6[趕急]”:
1)
A student is getting ready to take an important oral exam and is concerned about her possibility of having an 'on' day or an 'off' day. She figures that if she has an 'on' day, then each of her examinars will pass her, independent of each other, with probability 0.8; whereas if she has an 'off' day, this probability will be reduced to 0.4. Supposethat the student will pass the examination if a majority of examinars pass her. If the student feels that she is twice as likely to have an 'off' day as she is to have an 'on' day, should she request an examination with 3 examinars or with 5 examinars?
2)
.......

1.你諗下係邊個distribution
2.俾左moment generating function,問你點搵expectation同variance,應該notes有大大條formula
3.prove independent要P(A)=P(A|B)就得,你check下係唔係一樣
4.probability加埋=1,咁就搵到k,expectation 同 variance用番原本條式就做到
5.Bayes' theorem
6.P(6)=P(7)=P(8)=0
P(5)=1/70
P(4)=(4!/3!)/70
P(3)=[5!/(3!2!)]/70
P(2)=[6!/(3!3!)]/70
P(1)=[7!/(4!3!)]/70
E(X)同Var(X)都係照做

自己試下先,今次其實straightforward過上次,不過係一要學左方法先識咁lor

fatman 2011-10-18 11:37
照睇似跟formula就得

vincent 2011-10-18 12:04
引用第14樓plmplm2011-10-18 01:09發表的“”:
好明顯係功課黎 [表情] [表情] [表情]

yes但其他科有midterm搞到溫唔切stat

vincent 2011-10-18 12:07
引用第15樓hohoho22011-10-18 10:45發表的“”:
1.你諗下係邊個distribution [表情]
2.俾左moment generating function,問你點搵expectation同variance,應該notes有大大條formula
3.prove independent要P(A)=P(A|B)就得,你check下係唔係一樣
4.probability加埋=1,咁就搵到k,expectation 同 variance用番原本條式就做到
.......

ok i try it first@@
唔識再問你!

vincent 2011-11-03 00:59
再更新了

hohoho2 2011-11-03 01:19
http://www.stat.washington.edu/~hoytak/teaching/current/stat394/

googled

joN3 2011-11-03 02:19
更新後 TREE DIAGRAM +BEYES .

joN3 2011-11-03 02:23
第二條係 少少數學層面 sequence

vincent 2011-11-03 14:27
引用第20樓hohoho22011-11-03 01:19發表的“”:
http://www.stat.washington.edu/~hoytak/teaching/current/stat394/
googled [表情] [表情] [表情]

咁都比你搵到

hohoho2 2011-11-03 17:49
引用第23樓vincent2011-11-03 14:27發表的“”:
咁都比你搵到 [表情] [表情]

I always google if I don't know the answer for assignment
About 60% of the time it can be solved

vincent 2011-11-03 19:35
引用第24樓hohoho22011-11-03 17:49發表的“”:
I always google if I don't know the answer for assignment [表情]
About 60% of the time it can be solved [表情]

可否解釋下第二條個ans
唔太明佢做緊咩:(

hohoho2 2011-11-03 22:41
引用第25樓vincent2011-11-03 19:35發表的“”:
可否解釋下第二條個ans
唔太明佢做緊咩:(

佢應該漏左o的union符號
我其實唔識,下面係睇完答案即場理解估出黎,可能錯
將E1 U E2 U... Ex-1當1個event
P(AUB)=P(A)P(B|A)
p(E1 U E2 U....Ex-1 U Exc)
=P(E1 U E2 U... Ex-1))P(Exc|E1 U E2 U... Ex-1)
去到最後就係果個答案
後半part自己研究下la,我呢幾日比較忙

vincent 2011-11-03 23:13
引用第26樓hohoho22011-11-03 22:41發表的“”:
佢應該漏左o的union符號
我其實唔識,下面係睇完答案即場理解估出黎,可能錯 [表情]
將E1 U E2 U... Ex-1當1個event
P(AUB)=P(A)P(B|A)
.......

okok thx

vincent 2011-11-16 15:26
Let the probability density function of Y be

        cy exp(-2y),   0<=y<無限;
f(y)=
        0,               elsewhere.

where c is a constant.

(a) Find the value of c that makes f(y) a density function. (not yet confirmed ANS:2)
(b) State the distribution of Y, and give the mean and varience for Y.
(c) Write down the moment generating function for Y.

又又再更新了

MonsieurGarcon 2011-11-16 23:49
引用第28樓vincent2011-11-16 15:26發表的“”:
Let the probability density function of Y be
        cy exp(-2y),   0<=y<無限;
f(y)=
        0,               elsewhere.
.......


唔知岩唔岩...

用gamma, k = 2 , theta = 1/2

只要搞掂b果條搵到個distribution其他只係put數落公式
當然,我估佢係想你先做a果題,in番個c出黎先
btw, c 應該係4

vincent 2011-11-17 00:01
引用第29樓MonsieurGarcon2011-11-16 23:49發表的“”:
唔知岩唔岩...
用gamma, k = 2 , theta = 1/2
.......

c=4點搵嫁?

MonsieurGarcon 2011-11-17 00:03
你代下k=2,theta=1/2落gamma度就會迫到c=4

MonsieurGarcon 2011-11-17 00:05
再唔係洗唔洗慢慢寫番in個c出黎d step?

vincent 2011-11-17 00:11
引用第32樓MonsieurGarcon2011-11-17 00:05發表的“”:
再唔係洗唔洗慢慢寫番in個c出黎d step? [表情]

我諗要
thx stat神

hohoho2 2011-11-17 00:11
點解我計到4/3o既,可能我計錯
in個pdf -infinity to infinity應該會=1
因為係probability黎

vincent 2011-11-17 00:13
引用第34樓hohoho22011-11-17 00:11發表的“”:
點解我計到4/3o既,可能我計錯[表情]
in個pdf -infinity to infinity應該會=1
因為係probability黎

又多個幫手

MonsieurGarcon 2011-11-17 18:54
引用第34樓hohoho22011-11-17 00:11發表的“”:
點解我計到4/3o既,可能我計錯[表情]
in個pdf -infinity to infinity應該會=1
因為係probability黎


如果你真係由負無限in到正無限,咁梗係計錯
個pdf係由0開始

MonsieurGarcon 2011-11-17 19:12
引用第33樓vincent2011-11-17 00:11發表的“”:
我諗要 [表情]
thx stat神 [表情]


希望你睇得明

S(infinity, 0) cy exp(-2y) dy = 1...............S(infinity, 0) 即係將舊野由0 in 到無限

c S(infinity, 0) y exp(-2y) dy = 1.......將個c 抽出黎

[y exp(-2y)/-2](infinity,0) - S(infinity, 0) exp(-2y)/-2 dy = 1/c.......呢度用左integration by parts黎in 番 y exp(-2y) 呢舊野, 我選擇左之後抽左個 -1/2 出黎先, 跟住d steps都係普通野而且好難打出黎所以我跳

......

做完d in野最後會變成咁...

1/2 (0-1) = -2/c............以上integration by parts既前面最後會in係0, 而後面會in到係-1, 所以有舊"(0-1)"出左黎, 而等號後既 -2/c 就係因為抽左個-1/2然後掟埋過1/c果度變成既

所以最後c = 4

vincent 2011-11-17 19:26
引用第37樓MonsieurGarcon2011-11-17 19:12發表的“”:
希望你睇得明 [表情]
S(infinity, 0) cy exp(-2y) dy = 1...............S(infinity, 0) 即係將舊野由0 in 到無限
.......

我睇得明
moment generating function for Y又點搞

MonsieurGarcon 2011-11-17 19:36
引用第38樓vincent2011-11-17 19:26發表的“”:
我睇得明 [表情]
moment generating function for Y又點搞 [表情]


你做到b果題,又搵到個distribution
個mgf真係代數入去就得架喇喎

vincent 2011-11-17 19:40
引用第39樓MonsieurGarcon2011-11-17 19:36發表的“”:
你做到b果題,又搵到個distribution
個mgf真係代數入去就得架喇喎 [表情]

佢話state the distribution...
係咪姐係就咁講出黎就得?
係咪gamma dis. ??

hohoho2 2011-11-17 19:41
引用第36樓MonsieurGarcon2011-11-17 18:54發表的“”:
如果你真係由負無限in到正無限,咁梗係計錯 [表情]
個pdf係由0開始 [表情]

都係負無限開始,只係個f(x)細過0果陣=0所以in完都係0姐
我做做下抄漏左個y

MonsieurGarcon 2011-11-17 19:48
引用第40樓vincent2011-11-17 19:40發表的“”:
佢話state the distribution...
係咪姐係就咁講出黎就得?
係咪gamma dis. ?? [表情]


真係就咁講gamma 就得
不過提提你, gamma 可能唔夠準確
其實咩beta, exponential 等等都係gamma 家族
但係以我暫時諗到既除左gamma應該冇其他更準確既distribution

MonsieurGarcon 2011-11-17 19:50
引用第41樓hohoho22011-11-17 19:41發表的“”:
都係負無限開始,只係個f(x)細過0果陣=0所以in完都係0姐
我做做下抄漏左個y [表情]


如果個boundary 一直都只係黐住 cy exp(-2y) 呢舊野當然由負開始in 都冇所謂
不過假如之後用左integration by parts 得番 y exp(-2y) 或者 exp(-2y) 咁就唔知仲work 唔work

vincent 2011-11-17 19:55
引用第42樓MonsieurGarcon2011-11-17 19:48發表的“”:
真係就咁講gamma 就得
不過提提你, gamma 可能唔夠準確
其實咩beta, exponential 等等都係gamma 家族
.......

ok
咁mean = 1/入
varience= 1/入^2

???

hohoho2 2011-11-17 20:35
引用第43樓MonsieurGarcon2011-11-17 19:50發表的“”:
如果個boundary 一直都只係黐住 cy exp(-2y) 呢舊野當然由負開始in 都冇所謂
不過假如之後用左integration by parts 得番 y exp(-2y) 或者 exp(-2y) 咁就唔知仲work 唔work [表情]

我o既意思係1 = Sf(y)dy(-infinity to infinity)= S0dy(-infinity to 0) + Scyexp(-2y) (0 to infinity)
=0+Scyexp(-2y) (0 to infinity)

concept上咁好似完整o的,我見佢連in左佢=1都唔知所以講清楚o的

MonsieurGarcon 2011-11-18 00:26
引用第44樓vincent2011-11-17 19:55發表的“”:
ok
咁mean = 1/入
varience= 1/入^2
.......


你學左gamma dis. 未?
mean = 1/入
var = 1/入^2
呢d只係exponential ge mean & var

gamma ge mean = k乘theta
var = k乘theta^2

當gamma ge k=1, 就會變成exponential dis.

MonsieurGarcon 2011-11-18 00:29
引用第45樓hohoho22011-11-17 20:35發表的“”:
我o既意思係1 = Sf(y)dy(-infinity to infinity)= S0dy(-infinity to 0) + Scyexp(-2y) (0 to infinity)
=0+Scyexp(-2y) (0 to infinity)
concept上咁好似完整o的,我見佢連in左佢=1都唔知所以講清楚o的 [表情]


我諗佢明卦


查看完整版本: [-- 16/11更新Stat QuestionsX1[趕急] --] [-- top --]


Powered by phpwind v8.7 Code ©2003-2011 phpwind
Time 0.272513 second(s),query:2 Gzip disabled