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[MATH SL]Statistics Help


Mahuta ♥

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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)=0.15

I 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..". :wtf:

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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.

post-99-1240670835_thumbjpg

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.html

Disclaimer: 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.

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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)=0.15

I was asked to calculate mean and SD.

Alright... I've attached my solution to this question with this post :wtf:

I hope that helps... let me know if you have any more questions...

post-13663-1240689461_thumb.jpg

Edited by 1-2-3
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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. :):wtf:

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Not quite sure I understand why you're finding the prob that 0 are faulty :wtf:

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 by Max
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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?

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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?

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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)

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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?

post-10498-1240742667_thumb.jpg

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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.

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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. :S

*[the factorial thingie, just dont know how to do it vertically on here]

post-10498-1240765995_thumb.jpg

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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)5000-470

X~B(5000,0.1) so.. P(X<470)= post-99-1240793228_thumbgif

The way to do the 'factorial thingie' on a calculator is by inputting 5000 nCr 469

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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,.793

5,10)) OR 1-binomcdf(10,.7935,6)

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