ReCom.org
Portal Page Forum Wiki Social Groups Scholarship Holders Infobase Site Map About
Go Back   ReCom.org > Forum > ReCom Cafe > Education > SPM & STPM

SPM & STPM Post questions on SPM and STPM topics ONLY.

[STPM] Maths T needs help!!

Reply
 
Thread Tools
markwongsk Male
Developer
 
markwongsk's Avatar
 
Join Date: Nov 2008
Posts: 717
  #11 Old 15-03-2010 Default Re: [STPM] Maths T needs help!!

Quote:
Originally Posted by smallkid View Post
I found this question in fajar bakti ace ahead (volume 2)
Exam practise chapter 8
Q21.

Suppose the time T taken for a car wash is normally distributed with mean 6minutes and standard deviation 1.5 minutes.

c) If 1000 cars are washed, what is the probability that not more than one car takes more than 10.62 minutes.

Answer : 0.3399

Anyone knows how to do it?

I've tried poisson approximation, but i'm still no where near the solution
My answer does not match the answer given but I'll show you my method and see if you can find what's wrong:

First find the probability of a car wash taking more than 10.62 minutes:

P(X>10.62) = P(Z > (10.62-6)/1.5) = P(Z>3.0 = 0.001

Now use Binomial distribution since all outcomes are independent:

P(X<=1) = P(0) + P(1) = 0.999^1000 + 1000C1*0.999^999*0.001
= 0.7358

I think this answer makes sense. Let's do some logical check:

We know standard deviation is 1.5 and mean is 6. 10.62 lies 3 SD out of the mean and the probability will be low.

If we take the mean of 1000 tries, np = 1000(0.001) = 1. The mean is 1 car taking more than 10.62 minutes. So an answer less than 0.5 isn't that logical.

Can someone confirm this? I haven't been doing stats for 2 years ...
markwongsk is offline   Reply With Quote
Annabelle88 Female
Member
 
Annabelle88's Avatar
 
Join Date: Dec 2009
Posts: 151
  #12 Old 15-03-2010 Default Re: [STPM] Maths T needs help!!

Quote:
Originally Posted by markwongsk View Post
My answer does not match the answer given but I'll show you my method and see if you can find what's wrong:

First find the probability of a car wash taking more than 10.62 minutes:

P(X>10.62) = P(Z > (10.62-6)/1.5) = P(Z>3.0 = 0.001

Now use Binomial distribution since all outcomes are independent:

P(X<=1) = P(0) + P(1) = 0.999^1000 + 1000C1*0.999^999*0.001
= 0.7358

I think this answer makes sense. Let's do some logical check:

We know standard deviation is 1.5 and mean is 6. 10.62 lies 3 SD out of the mean and the probability will be low.

If we take the mean of 1000 tries, np = 1000(0.001) = 1. The mean is 1 car taking more than 10.62 minutes. So an answer less than 0.5 isn't that logical.

Can someone confirm this? I haven't been doing stats for 2 years ...
Since among the 1000 tries only 1 car more than 10.62 min.Why the probabilities sholud more than 0.5? not lower?If more than 0.5 then there will be half of the tries more than 10.62,rite?
Correct me if i m wrong.Thanks

Last edited by Annabelle88; 15-03-2010 at 11:41 AM.
Annabelle88 is offline   Reply With Quote
markwongsk Male
Developer
 
markwongsk's Avatar
 
Join Date: Nov 2008
Posts: 717
  #13 Old 15-03-2010 Default Re: [STPM] Maths T needs help!!

Quote:
Originally Posted by Annabelle88 View Post
Since among the 1000 tries only 1 car more than 10.62 min.Why the probabilities sholud more than 0.5? not lower?If more than 0.5 then there will be half of the tries more than 10.62,rite?
Correct me if i m wrong.Thanks
Firstly, remember we want probability not more than 1, which means it's P(0) or P(1). Second remember that the mean is around 1. So the distributively it's all around 1. Since P(X > 1) should balance P (X < 1) It only makes sense that the probability of P(X = 0) is going to be larger than any P (N) for N > 1.

Also, remember when we calculated np = 1, we are modelling the binomial as a normal. In reality we would need continuity correction, and take P(X > 1.5) instead of P(X > 1) so...

I'm not sure whether this is correct, but it makes sense to me. Hope someone points out my flaw...
markwongsk is offline   Reply With Quote
Holyboy27
Less Junior Member
 
Join Date: Aug 2007
Posts: 68
  #14 Old 04-05-2010 Default Re: [STPM] Maths T needs help!!

Quote:
Originally Posted by smallkid View Post
I found this question in fajar bakti ace ahead (volume 2)
Exam practise chapter 8
Q21.

Suppose the time T taken for a car wash is normally distributed with mean 6minutes and standard deviation 1.5 minutes.

c) If 1000 cars are washed, what is the probability that not more than one car takes more than 10.62 minutes.

Answer : 0.3399

Anyone knows how to do it?

I've tried poisson approximation, but i'm still no where near the solution
For stpm, most of the questions which have a few parts, the earlier parts usually have something to do with the last question. I noticed this is part c, what is part a and b. Bcos the question mentioned that it's originally a normal distrb. In part c, I think they want you to change from normal distr to binomial distrb. And the answers in earlier parts are used for this conversion.
__________________
Don't take life too seriously, your not gonna survive it anyway
Holyboy27 is offline   Reply With Quote
dkpit
Super Junior Member
 
Join Date: Dec 2009
Posts: 32
  #15 Old 30-06-2010 Default Re: [STPM] Maths T needs help!!

75% of the participants in a mths quiz are male. Marks(x) for 30 female participants are given by [sigma](x-72)=66 and marks (y) for male participants are [sigma](y-80)=-135.Find the mean marks for (i) female participants ans: 74.2(ii) all participants.ans:77.43
Can someone help me out with this question? I cant get the ans for (ii).
dkpit is offline   Reply With Quote
ling910520 Male
Less Junior Member
 
Join Date: Sep 2009
Posts: 86
  #16 Old 01-07-2010 Default Re: [STPM] Maths T needs help!!

Mean marks for all participants
= sum of the mark of all participants / number of all participants

number of all participants = 120 ,(there is total 120 participants since 25 % of participants accounts for 30 participants)

sum of the mark of all participants = sum of the mark of all female participants + sum of the mark of all male participants

sum of the mark of all female participants = 2226 , (i assume you can get this because you already solve question (i) )

sum of the mark of all male participants
(use same formula as question (i) to separate the constant and y)
[sigma]y - [90x80]=-135
[sigma]y=7065

substitute in the value that you calculated and you will get 77.425. round off to become 77.43
ling910520 is offline   Reply With Quote
The Following User Says Thank You to ling910520 For This Useful Post:
dkpit
Super Junior Member
 
Join Date: Dec 2009
Posts: 32
  #17 Old 01-07-2010 Default Re: [STPM] Maths T needs help!!

i am wondering whether this question refers to 30 female participants out of a certain no. or the 25% are these 30 ppl. are there any ways to distinguish which one the question refers to?
dkpit is offline   Reply With Quote
HarryZai Male
Super Junior Member
 
Join Date: Jul 2010
Posts: 1
  #18 Old 04-07-2010 Default Re: [STPM] Maths T needs help!!

Differentiation: reply me ASAP yah! I need help urgently.

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius, r.

[A: 32/81 pi r^3]
HarryZai is offline   Reply With Quote
Sillyboy Male
Slightly Senior Member
 
Sillyboy's Avatar
 
Join Date: Apr 2006
Posts: 915
  #19 Old 05-07-2010 Default Re: [STPM] Maths T needs help!!

Quote:
Originally Posted by HarryZai View Post
Differentiation: reply me ASAP yah! I need help urgently.

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius, r.

[A: 32/81 pi r^3]
http://www.themathpage.com/acalc/applied.htm

Have a look at problem 4!
Sillyboy is offline   Reply With Quote
The Following User Says Thank You to Sillyboy For This Useful Post:
kaiyi Female
Member
 
kaiyi's Avatar
 
Join Date: May 2009
Posts: 335
  #20 Old 08-08-2010 Default Probability

Please help me regarding this question, am i doing the correct thing?

calls on Sun 1 2 3 4 5 6
Probability 0.05 0.10 0.30 0.25 0.20 0.10

Let X - calls
E(X) = 3.75
Var(X)= 1.6875

(a) Imagine the number of emergency service calls is recorded for 6
randomly chosen Sundays throughout a year. Find the expected value and
standard deviation of the total service calls over the 6 Sundays.

W = X1 +X2 + X3 ...+ X6
E(W) = E(X1 +X2 + X3 ...+ X6)
= 3.75^6
= 2780.9

Var(W) = Var(X1 +X2 + X3 ...+ X6)
= 6(1.6875)

(b)The person in charge move move to a different area, and wants to
determine the true mean number of emergency service calls on Sunday in the
new area. Assuming that the standard deviation of calls in the new area is the
same, how many Sundays should he include in his sample to ensure that he
estimates the true mean to within 0.5 with 98% probability?

Q~N(mean, 1.2990)

P(mean-0.5<Q<mean+0.5) = 0.98
P(Z<2.33) = 0.9901
mean +0.5 = 2.33
mean = 1.83

(X1-mean)/ s.deviation = mean -0.5
(X1-1.83)/ 1.299= 1.83-0.5
X1 = 3.5577

(X2-mean)/ s.deviation= mean +0.5
(X2-1.83)/ 1.299= 1.83+0.5
X2 = 4.8567

3.5577 <X< 4.8567
4 sundays are needed.
__________________
The real voyage of discovery consists not in seeking new landscapes, but in having NEW EYES.
kaiyi is offline   Reply With Quote
Reply

Bookmarks

Tags
mathematics, stpm

Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Similar Threads
Thread Thread Starter Forum Replies Last Post
Differences between the 3 maths subjects in STPM? infested_ysy Education 5 16-06-2009 08:48 PM
I need stpm 06,05 further maths answer.. Help me pls.... UlquiorraSchiffer SPM & STPM 0 09-09-2007 09:46 AM
Where to get latest STPM further maths book? qfeng Education 9 20-01-2007 09:21 AM
Further Maths lolilo Education 37 05-06-2005 04:45 AM
need help for maths project joebf86 Education 4 15-09-2004 07:04 AM


All times are GMT +8. The time now is 12:46 PM.


Powered by vBulletin® Version 3.7.6
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

ReCom stands for Reborn Community. It has no affiliation with other organizations that may share the same name. The views expressed in this website solely represent the authors point of view and do not necessarily reflect the views of ReCom Anchors and other ReCom users.


 

Page generated in 0.10956 seconds with 17 queries