数学T paper 2讨论专区

數學神童

X1,X2, ... , X200 are uniform independent random variables with probability density functions
f(x)=1, for 0<=x<=1 and
f(x)=0 otherwise.
If Y=X1,X2, ... , X200, find the mean of Y and show that the variance of Y is 50/3.

Lov瑜瑜4ever

要用tree diagram 比较容易看

1st condition ,  take 2R from bag A
P= 3/5 * 2/4 = 3/10

...
peaceboy 发表于 22-8-2010 03:27 PM

没错没错
就是10/21了
thxq

atnub_fuji

[quote]刚刚读大学罢了。。。
哇。。数学系。。好羡慕你哦。。现在又申请研究院。。祝你成功^^
walrein_lim88 发表于 2-8-2010 01:44 AM [/size%

不错。。。。还有回来帮小弟小妹们哦？

白羊座aries

fx = (1/100 ) e^(-x/100) , x>0
           0                        ,ow

calculate the mean

peaceboy

fx = (1/100 ) e^(-x/100) , x>0
           0                        ,ow

calculate the mean
白羊座aries 发表于 26-8-2010 01:14 AM

    够力，看不懂.....

Log

回复 362# 數學神童
mean X1,E(X1)=∫ x.f(x) dx
                    =∫x(1) dx  ( upper limit =1 , lower limit=0)
                    =1/2
E(X&#178;1）=∫ x&#178; f(x)
           = ∫ x&#178; (1) dx   ( up=1, low.=0)
           = 1/3
var(x1) = E(X&#178;1）- {E(X1)}&#178;
           = 1/12
so, E(Y)=E（X1+X2+...+X200)
           = (200)(1/2)=100                 (as they r unifrom)
Var(Y)=Var(X1+X2+..+X200)
         =(200)(1/12)=50/3

Log

回复 365# 白羊座aries
mean X
=∫ x f(x) dx
=∫ x. (1/100) (e^-x/100) dx
use integration by parts,
∫ x. (1/100) (e^-x/100) dx
= x ∫ (1/100) (e^-x/100)  dx   - ∫ [ ∫ (1/100) (e^-x/100)  dx ] .d(x)/ dx  .dx
=do it just like usual get
= -x.e^(-x/100) - 100e^(-x/100) +c
then, 才来insert  upper limit=∞ and lower limit=0 to the equation above
mean x = 100

Winterblade

A school offers Japanese and Arabic as 2 foreign languages. The probability that a student who studi ...
Lov瑜瑜4ever 发表于 22-8-2010 01:31 PM

Anyone manage to solve this?

peaceboy

回复 369# Winterblade


A school offers Japanese and Arabic as 2 foreign languages. Theprobability that a student who studies Japanese also studies Arabic is1/4 and 1/5 of the students who study Arabic also studies Japanese too.Among the students, 1/5 of them do not study any of the 2 languages.What is the probability that a student in the school studies bothlanguages?

What is the mean of that red color sentence?

P(A l J) = P(A n J) / P (J) = 1/4
P(A n J) = P(J)/4

p(J l A) =  P(A n J) / P (A) = 1/5
P(A n J) = P(A)/5

P(J)/4 =P(A)/5
P(J) = (4/5) P(A)

P(J U A) = 4/5
P(A) + P(J) - P(J n A) = 4/5
P(A) + (4/5) P(A) -  P(A)/5 = 4/5
8/5 P(A) = 4/5
p(A) = 4/8 = 1/2

P(A n J) = P(A)/5
            = (1/2) /5
            =1/10

blazex

A continuous variable X is such that P(X≤x) =mx+c for 1≤x≤5. Calculate the values of m and c.

Allmaths

A continuous variable X is such that P(X≤x) =mx+c for 1≤x≤5. Calculate the values of m and c.
blazex 发表于 28-8-2010 06:13 PM

是m=1/4, c=-1/4 吗?


注:小弟不是很会的...

blazex

是m=1/4, c=-1/4 吗?


注:小弟不是很会的...
Allmaths 发表于 28-8-2010 06:33 PM

是的，要如何计算？

Allmaths

是的，要如何计算？
blazex 发表于 28-8-2010 06:41 PM

Winterblade

回复 370# peaceboy


    Ooo thx thx

blazex

X cm  is the diameter of cylindrical rods under production. Assume that X has the probability density function
    f(x)= k(x-0.9)(1.1-x), 0.9<x<1.1   (< means greater or equal)
         =0, otherwise
State the mean. A rod is regarded as defective if its diameter deviates from 1cm by more than 0.09 cm. What percentage of defective rods should be expected?

Log

state the mean
so, mean = (0.9+1.1)/2= 1.0 cm
bcoz if u notice that f(x)= k(x-0.9)(1.1-x), 0.9<x<1.1 , this is quadratic equation. u sketch it then u will figure out the curve is symmetrical at x=1.0 that is the midpoint value btween 0.9 n 1.1.
find k first, k=750
P( defective) = 1 - ∫* 750(x-0.9)(1.1-x) dx= 0.0145
*Upper.L=1.09
  Lower.L=0.91
% defective=1.45%

nkrealman

第六题不会做，拜托啦各位

最后那个部分不会，找displacement 那个

whyyie

peaceboy

whyyie 发表于 12-9-2010 12:22 PM

    dx/dt = (3x+5-3x-6)/(3x^2+11x+10)
             = -1/(3x^2+11x+10)

(3x^2+11x+10) dx = -1 dt
integrate both side
x^3 + 11x^2 / 2 + 10x = -t+c
when x=1,t=2
c=37/2
x^3 + 11x^2 / 2 + 10x = -t+37/2
2x^3 +11x^2 +20x +2t -37 =0



dy/dx = (2y^2 +3)/4y

4y/(2y^2 +3) dy = dx
integrate both side
ln(2y^2+3) = x+c
y=2 , x=-1
c=(ln 11) +1
ln(2y^2+3) = x+(ln 11) +1
2y^2+3 = e^(x+ln11 +1)
            = 11e^(x+1)
y=+-[(11e^(x+1) -3)/2]^(1/2)

一点不一样

Allmaths

whyyie 发表于 12-9-2010 12:22 PM

刚学differential equation不久...有错请纠正...