M1

數學小子

唔知呢到有冇人讀M1,
我自己讀pure,不過有d學弟問我m1問到我口啞啞,因為我唔識m1個d statistics
我研究過唔多明,想知int[0,z] 1/sqrt(2pi) e^{(-1/2 x^2)}dx呢舊野既1d詳細既內容,thx

-終場ソ使者-

你唔明,係正常既,我估你對statistics既認識都會CE程度,
我係讀m2,不過有研究過呢個graph,俾d意見啦(我係呢方面冇接受過嚴謹既訓練)
如果你有讀AL chem,應該會見過1個probability distribution of electrons,姐係electrons圍繞住個nucleus出現既prob.(某d位置出現electrons既prob.係特別大,某d位置既prob.特別細)
又舉個例,某某地區既人既平均IQ係107,standard deviation係5,當數據夠多既話,就會plot到1條graph,呢條graph既樣就好似一條intestinal villus咁,不過係左右對稱,比較完整既,而呢條graph既function正確d既寫法係
(呢個graph係叫做normal distribution,F5 core math有教)
=standard deviation
=mean
至於只係for standard normal distribution
(core math有教,呢個interval內所佔有總數既%係68%)
而個area係用黎搵一個interval入面所佔total area既%
呢個graph係點黎,你就要多謝某某數學家/統計學家啦,係佢地好辛苦地plot出黎
上述講到個area in an interval代表該interval既%
即
d constant咁無稜兩可有咩意義呢?就係plot呢條graph既人既聰明之處,因為

如果個個constant 係其它既話,個area in interval 就唔係=1
所以個聰明人就將佢設計成area in interval =1啦
舉個例子,如下圖

紅線既=4,=35
紫線既=2,=37
所以你見到當佢既x value=,就會遇到個max. value,而就係決定佢既"粗幼程度"(大,佢會"粗"d,細,就會"幼"d)
正因為佢既area in interval =1,所以當增加,area不變,"高度"就會下降(因為"粗"左),所以個"高度"並不反映d咩架!
不過個問題係,個f(x)既面積點計呢?
放心,係唔洗要你計,對表就得,個function係比你applied番,冇乜實際用處


PS:以上內容我只係用左幾個小時既研究,有錯莫怪

[H]bunbunbunbun

Yeah that's pretty much all the important points.

(core math有教,呢個interval內所佔有總數既%係68%)
而個area係用黎搵一個interval入面所佔total area既%
呢個graph係點黎,你就要多謝某某數學家/統計學家啦,係佢地好辛苦地plot出黎
上述講到個area in an interval代表該interval既%

The interval shows how many % of the outcome is within the interval. Usually this is used to find the confidence interval, as most statistical models are an estimate rather than an exact model. Confidence interval is the interval of which has n% probability that the interval contains the mean.

Area under graph is 1 because the probability of something happens is 1.

To integrate the function f(x)
consider e^{-x^2}
int [-inf,inf][-inf,inf] e^{-x^2-y^2}dxdy
=int [0,2pi][0,inf] re^{-r^2}drdt
=pi

Thus int[-inf,inf]e^{-x^2}dx = sqrt(pi)
with similar method you get int [-inf,inf]f(x)dx=1

Normal distribution is bare useful as the Central Limit Theorem shows that when the number of samples are large, their distribution tends to a normal distribution (regardless whether it is binomial or Poisson or gamma or anthing) and then you can find the maximum likelihood estimator to find the parameters of the distribution etc...

-終場ソ使者-

回復 3# [H]bunbunbunbun
In fact, I don't know much about statistics, but would you mind telling me detailly on the method you used to solve the integration ?
Why double integration was used?

[H]bunbunbunbun

Using double integral because we can evaluate the integral with polar coordinates (x,y)=(rcos(theta), rsin(theta))


Actually when I think about it, its hard to evaluate directly for the region (a,b), because without infinity involved, the region can't be expressed by polar coordinates easily. (x>0,y>0 would be r>0, 0<theta<pi/2; y>0 would be r>0, 0<theta<pi etc., but a square region is ugly in polar coordinates) so I think for the region (a,b) we'll have to evaluate the function from a to inf and from b to inf then subtract one from the other to get the answer. I'll try that when I have spare time.

-終場ソ使者-

Wow! That's the most cleverest method I have never seen in solving the area under (-inf, inf)!
I have searched much in calculating the area under [a,b], but I just found nothing!
But, can we deduce a general solution by MI or other methods using distribution table?

-終場ソ使者-

琴晚睇番條數,突然俾我諗到1個方法
Let ,
=>, .
Therefore, we have

.
Using integration by parts, we have

.
Therefore, we have
.

firefighter628

我突然覺得自己好眇小