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## [STPM] Maths T needs help!! Developer
#11 15-03-2010 Re: [STPM] Maths T needs help!!

Quote:
 Originally Posted by smallkid 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 ...  Member
#12 15-03-2010 Re: [STPM] Maths T needs help!!

Quote:
 Originally Posted by markwongsk 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.  Developer
#13 15-03-2010 Re: [STPM] Maths T needs help!!

Quote:
 Originally Posted by Annabelle88 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...  Holyboy27
Less Junior Member
#14 04-05-2010 Re: [STPM] Maths T needs help!!

Quote:
 Originally Posted by smallkid 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.
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Don't take life too seriously, your not gonna survive it anyway  dkpit Super Junior Member   Join Date: Dec 2009 Posts: 32 #15 30-06-2010 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).  ling910520 Less Junior Member   Join Date: Sep 2009 Posts: 86 #16 01-07-2010 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  The Following User Says Thank You to ling910520 For This Useful Post:
 dkpit Super Junior Member   Join Date: Dec 2009 Posts: 32 #17 01-07-2010 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?  HarryZai Super Junior Member   Join Date: Jul 2010 Posts: 1 #18 04-07-2010 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]  Slightly Senior Member
#19 05-07-2010 Re: [STPM] Maths T needs help!!

Quote:
 Originally Posted by HarryZai 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!  The Following User Says Thank You to Sillyboy For This Useful Post:
 kaiyi Member   Join Date: May 2009 Posts: 335 #20 08-08-2010 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 Tags mathematics, stpm Thread Tools Show Printable Version Email this Page 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 infested_ysy Education 5 16-06-2009 08:48 PM UlquiorraSchiffer SPM & STPM 0 09-09-2007 09:46 AM qfeng Education 9 20-01-2007 09:21 AM lolilo Education 37 05-06-2005 04:45 AM joebf86 Education 4 15-09-2004 07:04 AM

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