数学T paper 2讨论专区

nkrealman

大家帮帮忙，我的paper 2 就是差在 deductive geometry 跟 vector
都不知道怎么办
第9跟15


第2跟3

數學神童

A company has two factories X and Y.Factory X employs 30 workers and factory Y employs 40 workers.The mean pay of the workers in X is RM1250 with standard deviation of RM280.The mean pay of workers in Y is RM1190,with standard deviation of RM290.
i)Calculate the mean pay of all workers in the company
ii)Calculate the standard deviation of the pay of all workers in the company.

peaceboy

回复 463# 數學神童


  A company has two factories X and Y.Factory X employs 30 workers and factory Y employs 40 workers.The mean pay of the workers in X is RM1250 with standard deviation of RM280.The mean pay of workers in Y is RM1190,with standard deviation of RM290.
i)Calculate the mean pay of all workers in the company

mean = [30*1250 + 40*1190]/70
         = 8510/7


ii)Calculate the standard deviation of the pay of all workers in the company




(sum x^2/n - (mean)^2 )^(1/2) = 280
sum x^2/30 - 1250^2 = 280^2
sum x^2 = 49227000


(sum y^2/n - (mean)^2)^(1/2) = 290
sum y^2/40 - 1190^2 = 290^2
sum y^2 = 60008000



sum x^2+sum y^2 = 109235000


var = (sum x^2+sum y^2)/70 - (8510/7)^2 = 109235000/70 - (8510/7)^2
                                                     =1560500 - 72420100/49
                                                     =82538.776
std dev = 287.296

數學神童

A multiple choice quiz has 100 questions each with 4 possible answers of which only one is the correct answer.What is the probability that sheer guesswork yields from 11 to 15 correct answers for 40 of the 100 problems about which the student has no knowledge?
Answer:0.203

peaceboy

A multiple choice quiz has 100 questions each with 4 possible answers of which only one is the corre ...
數學神童 发表于 27-10-2010 11:23 PM

    我算到的答案差很多

(40C11)[(1/4)^(11)][(3/4)^(29)]+(40C12)[(1/4)^(12)][(3/4)^(28)]+(40C13)[(1/4)^(13)][(3/4)^(27)]+(40C14)[(1/4)^(14)][(3/4)^(26)]+(40C15)[(1/4)^(15)][(3/4)^(25)] = 0.389

blazex

The time taken by a man to walk from A to B follows a normal distribution with mean 90 minutes and standard deviation 20 minutes. At B, the man gets on a bus and travel to C. The time taken by the bus to C. The time taken by the bus to travel from B to C is independent of the time taken by the man to walk from A to B and follows a normal distribution with mean 50 minutes and standard deviation 5 minutes. The buses depart from B at 1000 hours and 1100 hours. Given that the man leaves A at 0900 hours and he gets on one of the buses. Determine the probability that he arrives at C before 1200 hours.

peaceboy

回复 467# blazex


    The time taken by a man to walk from A to B follows a normaldistribution with mean 90 minutes and standard deviation 20 minutes.
AtB, the man gets on a bus and travel to C. The time taken by the bus toC. The time taken by the bus to travel from B to C is independent ofthe time taken by the man to walk from A to B and follows a normaldistribution with mean 50 minutes and standard deviation 5 minutes.

Thebuses depart from B at 1000 hours and 1100 hours. Given that the manleaves A at 0900 hours and he gets on one of the buses. Determine theprobability that he arrives at C before 1200 hours.


有难度 orz...

P(get into the 1st bus) = P(t<60)
                                 = p(z< (60-90)/20)
                                = p (z<-1.5)
                               =0.0668
P(reach be4 1200 with 1st bus) = P(t<120)
                                            = p(z<(120-50)/5)
                                            =1

P(get into the 2st bus) = P(60<t<120)
                                 = P(-1.5<z<1.5)
                                 =2(0.5 - P(z>1.5))
                                 = 2 (0.5-0.0668)
                                 =0.8664
P(reach be4 1200 with 2nd bus) = P(t<60)
                                            = P(z<(60-50)/5)
                                             =P(z<2)
                                             = 0.977

so probability reach be4 1200 = 0.0668*1 + 0.8664*0.977
                                          =0.9133

blazex

回复 468# peaceboy


   谢谢。前辈的数学真的太强了。

peaceboy

回复 469# blazex


    客气了，偶不是前辈，答案也不确定对的 *仅供参考

nkrealman

peaceboy又要麻烦你了

peaceboy

回复 471# nkrealman


    好多

1. 10!
a) 6*5*8!
B)10! -8*2!*8!

3. a)5C2*4C2
b)9C4 - 4C4

5)6C2*5C3(A only) + 6C3*5C2(B only) + 6C3*5C3 (not A and B)

6) P(A) =8/15 , P(B) = 1/3 , P(AlB) =1/5

a)P(AlB) = 1/5
P(AnB)/P(B) = 1/5
P(AnB)= 1/5*1/3
          =1/15

b) P(AuB)=P(A)+P(B)-P(AnB)
             =8/15+1/3-1/15
             = 4/5
P(only one event occur) = P(AuB) - P(AnB)
                                 = 4/5 -1/15
                                 =11/15

c)1-P(AuB) = 1-4/5
                =1/5

12)P(A)= 1/2*1/2*1/2 [three times get odd num]+ 1/2*1/2*1/2*(3!/2!) [2 time get even number , 1time get odd]
           =1/2


sum = 13 , 1+6+6 , 2+5+6, 3+5+5 , 3+6+4 , 4+5+4
P(B) = 1/216*3!/2! +1/216*3! +1/216*3!/2! + 1/216*3! + 1/216*3!/2! <<<1/216 come from1/6*1/6*1/6
       =3/216+6/216+3/216+6/216+3/216
       =21/216

number appears in the third throw <<<看不懂什么意思...orz

14)P(AnB) = P(A)*P(B)

P(AnB') +P(AnB) = P(A)
P(AnB') + P(A)*P(B) = P(A)
P(AnB') = P(A)- P(A)*P(B)
           =P(A)[1-P(B)]
           =P(A)*P(B')
therefore , A and B' is independent

P(A'nB') = 1-P(AuB)
           =1- [P(A)+P(B) -P(AnB)]
            =1-[P(A)+P(B) -P(A)*P(B)]
            =1-[P(A) + P(B)[1-P(A)]
             =1-P(A) - P(B)[P(A')]
             =P(A') -  P(B)[P(A')]
            =P(A')[1-P(B)]
            =P(A')P(B')
       therefore , A' and B' are independent

nkrealman

问题那么多你还是帮我解完了，感激不尽。

number appears in the third throw 应该是说3三次投出来的号码？probability应该是 1/6 吧

第一题跟dice那题我是不是很清楚你的做法，可以解释吗？

dice 的答案

P(A)=1/2
P(A/C)=1/2 independent

P(B)=1/72
P(/B/C)=1/6 not independent

peaceboy

回复 473# nkrealman


    第一题,10! <<<不用解释吧...

a)6*5*8! <<<既然有四个地方不能放，那么第一个罐子只有6个地方可放，第二个有5个地方可放。剩下8个位子和八个罐子随便防，所以6*5*8!

b)2 particular placed side by side = 8*2!*8!<<<把两个连在一起放的可能性有8个（不包括面对面的），那么那2个罐子之间又可以交换放，剩下8个位子和八个罐子随便防，所以8*2!*8!
10！<<<全部可能性
全部可能性 - 2 particular placed side by side = 10! -8*2!*8!


dice的，如果a+b+c ,要是单数,a,b,c一定要有一个是单数或则3个都是单数(odd+odd+odd = odd , even +even +odd=odd)
dice的数字,1,3,5单数,2,4,6双数，所以单数或者双数的可能性是1/2


所以 odd+odd+odd 的可能性是(1/2)(1/2)(1/2)
odd+even+even的可能性是(1/2)(1/2)(1/2)，不过其排列也有可能是even +odd+even ,  even +even +odd，所以我们算他的排列是3!/2！ (2！是给2个一样even的)

所以总共可能性是 1/2*1/2*1/2 + 1/2*1/2*1/2*(3!/2!)

b的 ，sum=13的可能性有这么多 1+6+6 , 2+5+6, 3+5+5 , 3+6+4 , 4+5+4
我们获得1至6任何一个号码的几率都是1/6
所以我要拿P(1+6+6)的几率是1/6*1/6*1/6 = 1/216
加上我们要算6+1+6，6+6+1等的可能性，所以我们算排列
所以是1/216*3!/2! +1/216*3! +1/216*3!/2! + 1/216*3! + 1/216*3!/2!

3!/2! , 3! <<<<这些都是拿来算排列的...

peaceboy

回复 473# nkrealman


    我还是看不懂event C的逻辑，除非他说number appears in the third throw is 6....

nkrealman

回复 474# peaceboy


   大致上明白了，本来不太明白为什么3!/2!，原来这已经include 号码出现的顺序前后跟  repetition

數學神童

By writing sin x as cos [(pi/2)-x], solve the equation cos 3x=sin x , 0<x<pi. Give your answers as multiples of pi.

whyyie

回复 477# 數學神童

cos (pi/2 - x) = cos 3x
pi/2 - x = 3x
pi/2 = 4x
x = pi/8, 5pi/8

nkrealman

懒惰copy 跟 cut...直接post上来好了
麻烦大家了


第六题



第7题
第8题b是between 75 and 80？
至于12题,<17.5是不是代表 0<x<17.5

peaceboy

回复 479# nkrealman


   6) 10C5*4C3 + 10C4*4C4
i)(8-1)!
接下来很熟悉，见450楼（18页）

7）偶没学

8)n=100 , p=0.7
mean np=70 , variance npq =21
P(x>=75)
P(x>74.5) = P(z >[74.5-70]/(21)^(1/2))
              =P(z>4.5/(21)^(1/2))
             =P(z>0.982)
              =0.1631

b)

P(75<=x<=80)
P(74.5<=x<=80.5) = P(4.5/(21)^(1/2)/<z<10.5/(21)^(1/2))
                            =P(0.982<z<2.291)
                            =P(z>0.982) - P(z>2.291)
                            =0.1521

P(less than 80 l more than 75) = P(75<=x<=80) /P(x>=75)
                                           =0.1521/0.1631
                                            =0.9326

nkrealman

回复 480# peaceboy


   erm...关于那一题可以请你解释吗？不是很明白。我除了懂(n-1)!之外，全部都没有概念