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Edexcel S2 - 27th June 2016 AM

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48 hours = 2 days

Therefore X~Po(1.6) and since the next car is caught within the next 48 hours then we know that there is at least 1 car that speeds within the next 48 hours

therefore P(X=>1) = 1 - P(X=0)
gnight everybody I'm going to bed

good luck
Original post by apzoe
It's just that there have been several exercises already in past papers with questions like "given that.. find that..".

I know you have to apply the formula P(B|A)=P(A intersection B)/P(A). It's just about using it correctly, right?


In the question above, just work out prob that between 5 and 8 mins she isn't served.

You don't always have to use the conditional probabilities formula (in fact you never have to, but in many cases it wouldn't make it easier).
(edited 7 years ago)
Original post by apzoe
It's just that there have been several exercises already in past papers with questions like "given that.. find that..".

I know you have to apply the formula P(B|A)=P(A intersection B)/P(A). It's just about using it correctly, right?


Yes, that is correct. Sometimes you have to 'interpret' the intersect though. Like in this case:

P(t>15t>10)=P(t>15t>10)P(t>10)=P(t>15)P(t>10) P(t> 15 | t>10) = \frac{P(t>15 \cap t>10)}{P(t>10)} = \frac{P(t>15)}{P(t>10)} as (t>15t>10)=t>15 (t>15 \cap t>10) = t>15 as t> 15 implies that t>10, whereas t>10 doesn't imply that t>15
(edited 7 years ago)
Is cumulative distribution function the same as cumulative probability density function?
https://www.youtube.com/watch?v=BJPcDV0rx88

June 2013, Question 3b.

Can anyone explain to me why P(X>n)? I thought it would be X<n..

Thanks


For Part c:

We're given that 4 were caught

1 on the first day and 2 on the second so calculate these using X~Po(0.8) as each occurrence happened within a 1 day period.

Since we're given that 4 happened we can use P(AnB)/P(B)
which is simply the probabilities of X=1 on the first day multiplied by the probability X=3 on the second day all over the prob of 4 accidents within the 2 day period, where X~(1.6)

Hope this makes sense
Original post by M3WIZARD
Any thoughts on tomorrow's paper? - Hopefully it's going to be a 6 Question paper ending on a nice easy mean or median sampling distribution :smile:

boundaries would be 74/75 for an A then
Reply 608
Original post by SeanFM
Yes, that is correct. Sometimes you have to 'interpret' the intersect though. Like in this case:

P(t>15t>10)=P(t>15t>10)P(t>10)=P(t>15)P(t>10) P(t> 15 | t>10) = \frac{P(t>15 \cap t>10)}{P(t>10)} = \frac{P(t>15)}{P(t>10)} as (t>15t>10)=t>15 (t>15 \cap t>10) = t>15 as t> 15 implies that t>10, whereas t>10 doesn't imply that t>15


Yes I see now. Thank you for the example btw. I solved it and got the same result as you did.
Sadly there are only 2 examples of such questions in regular past papers I could find to practice on, but now it makes more sense.

Thank you ^^
Original post by Moral010
Is cumulative distribution function the same as cumulative probability density function?

It sounds like it might be. I've never heard of it though, and a quick google doesn't seem to turn up many relevant results for 'cumulative probability density function'. But if you've been taught S2 in a class then it's probably the same thing.
Reply 610
Original post by Moral010
Is cumulative distribution function the same as cumulative probability density function?


Never heard of "cumulative probability density function".

At A2 we only studied the probability density function and the cumulative distribution function. Have you been taught something more or are you combining them and confusing yourself? :tongue:
You know when you have lots of different functions on a cumulative distribution function and they ask you to find median or quartile and you can only pick one function to be equal to the median or quartiles.
How do you know which one to pick!?
Thank you!


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Original post by Maths help pls
https://www.youtube.com/watch?v=BJPcDV0rx88

June 2013, Question 3b.

Can anyone explain to me why P(X>n)? I thought it would be X<n..

Thanks


You could go down the (x=<n) route, but you would have to use 0.95 (1-0.05).

Basically, the shop owner is trying to make it unlikely that he will run out of what he's selling (i.e P(X=<n) =<0.95) 9r P(X>n) (selling more than n copies) has probability less than 0.05
Original post by Bloom77
You know when you have lots of different functions on a cumulative distribution function and they ask you to find median or quartile and you can only pick one function to be equal to the median or quartiles.
How do you know which one to pick!?
Thank you!


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Trial and error. Only one of the 'sections' will have the right value.

I hope this post explains it.

Say the CDF is defined with 3 functions -
f(x)ax<b[br]g(x)bx<c[br]h(x)cx<d f(x) a \leq x < b[br]g(x) b \leq x < c[br]h(x) c \leq x < d

Where f, g and h are different functions and a, b and c are used for different ranges.

Say you're trying to find where F(x) = 0.5, you know it's going to be either from f, g, or h. Firstly, test f(b) (the upper limit of the domain of f(x)). If it is less than 0.5, then you know that it's not going to be in that function. If f(b) is greater than 0.5, then x is going to be somewhere between a and b.

Say it isn't between a and b, the next one is to test g(c) and apply the same logic.

Then you find the right function. Similar logic can be applied to any value of F(x).
Original post by apzoe
Never heard of "cumulative probability density function".

At A2 we only studied the probability density function and the cumulative distribution function. Have you been taught something more or are you combining them and confusing yourself? :tongue:


Probably confusing myself, they were just named differently in this revision website... Thanks :smile:
Original post by SeanFM
Trial and error. Only one of the 'sections' will have the right value.

I hope this post explains it.

Say the CDF is defined with 3 functions -
f(x)ax<b[br]g(x)bx<c[br]h(x)cx<d f(x) a \leq x < b[br]g(x) b \leq x < c[br]h(x) c \leq x < d

Where f, g and h are different functions and a, b and c are used for different ranges.

Say you're trying to find where F(x) = 0.5, you know it's going to be either from f, g, or h. Firstly, test f(b) (the upper limit of the domain of f(x)). If it is less than 0.5, then you know that it's not going to be in that function. If f(b) is greater than 0.5, then x is going to be somewhere between a and b.

Say it isn't between a and b, the next one is to test g(c) and apply the same logic.

Then you find the right function. Similar logic can be applied to any value of F(x).


Thank you! That makes sense :smile:


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Screenshot_2016-06-26-23-41-30.pngguys, can you help me with q7b June 2013 R?
Original post by Supermanxxxxxx
Time to do an all nightier then


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You need to be able to think in the morning, so go to bed!
Original post by kosvengali
Screenshot_2016-06-26-23-41-30.pngguys, can you help me with q7b June 2013 R?


We know X represents the correct answer, so by logic, the amount of wrong answers is 20-X

S represents the final score, which will obviously be the amount he gets - amount he loses

So, each correct is 4 points, so it's 4X - 1(20-X)

Which = 5X-20

Because 1 point's lost for a wrong answer
Original post by tiny hobbit
You need to be able to think in the morning, so go to bed!


I'm literally get 54/75 on the Jan 2014 ial I need to cram


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