Mahuta ♥ Posted April 25, 2009 Report Share Posted April 25, 2009 Hey,Im in desperate need of help in statistics, here's my problem:I was given certain probabilities:P(X<40)= 0.12 P(X>90)=0.15I was asked to calculate mean and SD.Weird enough, I can only do this manually, i.e using equations and the table of values given, but not using the GDC:(This question is on Paper 2, so the time im expected to spend on it is only the time to do on the GDC. I dont know how to do it, please help.I have seen the text book and it said use normalcdf...but apparently thats when you given the mean...how do i do the opposite?Thanks alot.Oh and only answer if you know this very well otherwise ill get confused more if you say "i think..". Reply Link to post Share on other sites More sharing options...
moneyfaery Posted April 25, 2009 Report Share Posted April 25, 2009 I'm guessing invNorm( ?? I hate stats Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 25, 2009 Author Report Share Posted April 25, 2009 Another question, im putting it here because it would look neater to have 2 questions seperate since this is a big one.So I get (a), but then (b) and © im not sure why the mark scheme says that. Please help. And thank you so much. Reply Link to post Share on other sites More sharing options...
Abu Posted April 25, 2009 Report Share Posted April 25, 2009 First of all. Just because you have something on paper 2 does not mean that you will do the entire thing by GDC. You will have to show some working, and not do the entire calculation on a GDC and spit out the answer. So for that first question, it is on paper 2 because you cannot solve decimalised numbers that quickly by hand, therefore it is on paper 2. Do it the way you know and use your GDC to help you with the calculations.To answer your second question. Only assume that it is a normal distribution when the question says that a distribution is normal. If it doesn't say, it's not a normal distribution.You've worked out the mean from Part (a). So now you have the mean and total sample. Question (b) asks for 3 calculators that are faulty. So you know that the mean is 2, the total sample is 100, and the question is asking for 3 calculators that are faulty. Formula for binomial distribution is attached. The n is 100, the x is 3 and the p is 0.02.Question (c.) asks for the probability that more than 1 calculator is faulty. That means that it can be anything from 1.0000001 to infinity. Lets say 2 since it should be a round number. So you find the probability that 0 calculators are faulty, and 1 calculator is faulty, add them together and subtract from 1. Since 1 is the maximum a probability can go to. If you want to do it via calculator (TI Series), for (b) it'd be binompdf(100,.02,3) and for (c.) either 1-(binompdf(100,.02,0)+binompdf(100,.02,1)) or 1-binomcdf(100,.02,1) . binomcdf is cumulative. Examples here: http://math.elon.edu/statistics112/prob_dist.htmlDisclaimer: It's been more than a year since I've learned this stuff, so any discrepancies are purely memory loss and may be blamed on my failing brain cells. Reply Link to post Share on other sites More sharing options...
moneyfaery Posted April 25, 2009 Report Share Posted April 25, 2009 Do it the way you know and use your GDC to help you with the calculations.And if you're REALLY lazy, you can use rref (reduced row echelon form) to solve the system of equations for the mean and s.d. Reply Link to post Share on other sites More sharing options...
Abu Posted April 25, 2009 Report Share Posted April 25, 2009 And if you're REALLY lazy, you can use rref (reduced row echelon form) to solve the system of equations for the mean and s.d. Don't confuse her more now. Reply Link to post Share on other sites More sharing options...
1-2-3 Posted April 25, 2009 Report Share Posted April 25, 2009 (edited) Im in desperate need of help in statistics, here's my problem:I was given certain probabilities:P(X<40)= 0.12 P(X>90)=0.15I was asked to calculate mean and SD.Alright... I've attached my solution to this question with this post I hope that helps... let me know if you have any more questions... Edited April 25, 2009 by 1-2-3 Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 25, 2009 Author Report Share Posted April 25, 2009 Oy my god! THANKYOU Aboo!I get it allll that except this:So you find the probability that 0 calculators are faulty, and 1 calculator is faulty, add them together and subtract from 1. Since 1 is the maximum a probability can go to.Not quite sure I understand why you're finding the prob that 0 are faulty And, from what you said..if its not normally distributed its always binomially distributed?LOL Irene, yeah im getting this ok! dont give me new terminology. Reply Link to post Share on other sites More sharing options...
Max Posted April 25, 2009 Report Share Posted April 25, 2009 (edited) Not quite sure I understand why you're finding the prob that 0 are faulty Well, there is a possibility that none of those 100 calculators is faulty And, from what you said..if its not normally distributed its always binomially distributed?Nope, if it does not say that the data follows a normal distribution, then it does not follow any 'standard' distribution. Edited April 25, 2009 by Max Reply Link to post Share on other sites More sharing options...
Abu Posted April 26, 2009 Report Share Posted April 26, 2009 The SL syllabus only has 2 distributions, binomial and normal, which is different from the HL syllabus.The point to remember here is: If it gives you a mean and a population, and asks you to find the probability of an occurrence within that population, then it's a binomial. From www.ibmaths.com (explains it better than I do): The binomial distribution can be used when there are a known number of trials and only 2 outcomes to each trial. It is written as X~B(n, p), where n is the number of trials and p is the probability of a success. As for normal distributions, the question will have to state that the scores or whatever is normally distributed therefore then and only then can you use the normal distribution to solve the question.Clear? Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 26, 2009 Author Report Share Posted April 26, 2009 YES! Definatly! I get this, thank you so much. Pretty lucky we only have two.And thanks Max about question (c ). Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 26, 2009 Author Report Share Posted April 26, 2009 Event (E) is that only one 4 occurs when a die is rolled.It says calculate that event E occurs three times in the five throws, in which case you use the binomial thingie because it didnt say its normal distribution.But, it then says find the probability that event E occurs ATLEAST 3 times in the five throws. They did that substrating from one thingie, but I dont get why.Why isnt it like probability that it occurs 3 times + P.4time + P.5times? Reply Link to post Share on other sites More sharing options...
Abu Posted April 26, 2009 Report Share Posted April 26, 2009 I'm not understanding what you're saying here.At least 3 times would mean that you work out the probability of it happening thrice, 4 times and 5 times, add it together and subtract from 1.1 - (binompdf(5,1/6,3)+binompdf(5,1/6,4)+binompdf(5,1/6,5)) or 1-binomcdf(5,1/6,2) Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 26, 2009 Author Report Share Posted April 26, 2009 The mark scheme says that.Now, I understand what you said, but not the part where you said substract from 1 (I know I keep asking about this one, but I dont understand the '1' thingie). How come in the Method 2, they're not substracting from one, where as in Method 1 they are? Reply Link to post Share on other sites More sharing options...
Abu Posted April 26, 2009 Report Share Posted April 26, 2009 Oops I made a mistake. You either add the probabilities of 3, 4 and 5. Or add up the probabilities of 0, 1 and 2 and take it away from 1 since that is the maximum a probability can go to. The question asks for 'at least 3' which means that out of 5, you only need the probabilities of 3 and above i.e. 3, 4 and 5. Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 26, 2009 Author Report Share Posted April 26, 2009 I tried doing this manually, but I got stuck. I did: X~B(5000,0.1) so.. P(X<470)=(5000,0.1)*(0.1)470(1-0.1)5000-470 But the calculator refused to do it. Im assuming thats what I should do, but the mark scheme says something else. *[the factorial thingie, just dont know how to do it vertically on here] Reply Link to post Share on other sites More sharing options...
Abu Posted April 27, 2009 Report Share Posted April 27, 2009 Well you've got 2 mistakes in there. When it says fewer than 470, it'd usually be up to and including 469. And you're not even looking at the formula!X~B(5000,0.1) so.. P(X<470)=(5000,0.1)*(0.1)470(1-0.1)5000-470X~B(5000,0.1) so.. P(X<470)= The way to do the 'factorial thingie' on a calculator is by inputting 5000 nCr 469 Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 27, 2009 Author Report Share Posted April 27, 2009 Yeah! I realized that but later, but anyhow...Im totally satisfied with STATS now. D The way to do the 'factorial thingie' on a calculator is by inputting 5000 nCr 469Lol, sorry didnt think about that. Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted April 29, 2009 Author Report Share Posted April 29, 2009 Ok this kind of messed up my rules...Isnt it when you're told its normal distribution you dont use the X~B thing?This question they said its normally distributed but then in (b), they expected me to use X~B(n,p)...how is that? Reply Link to post Share on other sites More sharing options...
Abu Posted April 29, 2009 Report Share Posted April 29, 2009 Not sure how to explain it to you. The normal distribution question usually asks for one item, or an item at random.Question (b) is asking for the probability with regards to 10 items. Besides, tell yourself this. In the SL syllabus, you only do the normal and the binomial distribution. How would you do that question using the normal distribution? You can only do that question using the binomial.(i) X ~ B (10,.7935) = binompdf(10,.7935,10)(ii) X ~ B (10,.7935) = (binompdf(10,.7935,7)+binompdf(10,.7935,8)+binompdf(10,.7935,9)+binompdf(10,.7935,10)) OR 1-binomcdf(10,.7935,6) Reply Link to post Share on other sites More sharing options...
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