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#11
15032010
Re: [STPM] Maths T needs help!!
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First find the probability of a car wash taking more than 10.62 minutes: P(X>10.62) = P(Z > (10.626)/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 ... 

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#12
15032010
Re: [STPM] Maths T needs help!!
Quote:
Correct me if i m wrong.Thanks Last edited by Annabelle88; 15032010 at 11:41 AM. 

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#13
15032010
Re: [STPM] Maths T needs help!!
Quote:
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... 

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#14
04052010
Re: [STPM] Maths T needs help!!
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#15
30062010
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](x72)=66 and marks (y) for male participants are [sigma](y80)=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). 
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#16
01072010
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 
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Super Junior Member

#17
01072010
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?

Super Junior Member

#18
04072010
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] 
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#19
05072010
Re: [STPM] Maths T needs help!!
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Have a look at problem 4! 

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#20
08082010
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(mean0.5<Q<mean+0.5) = 0.98 P(Z<2.33) = 0.9901 mean +0.5 = 2.33 mean = 1.83 (X1mean)/ s.deviation = mean 0.5 (X11.83)/ 1.299= 1.830.5 X1 = 3.5577 (X2mean)/ s.deviation= mean +0.5 (X21.83)/ 1.299= 1.83+0.5 X2 = 4.8567 3.5577 <X< 4.8567 4 sundays are needed.
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