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The Edexcel S2 (24/05/12 - AM) Revision Thread

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Reply 240
Avatar for Kaiser7
Kaiser7
6
Can someone explain to me number 4(e) in the January 2009 exam?

What graph is that? And why is it more suitable than a pdf? :redface:
Original post by Legendary
I will do FP3 if I can be asked otherwise ill just upload the paper. :colone:


Please send me the FP3 paper when you get it.

Though the paper is a long time away, when the time will come then we will see who does the solutions. Gibbo might be doing it.
(edited 11 years ago)
Reply 242
Avatar for H.J.P
H.J.P
7
HELP AGAINQUESTION: The number of cars passing an observation point in a 10 minute interval is modelled by a Poisson distribution with mean 1.

3) (b) Find the probability that in a randomly chosen 60 minute period there will be

(i) exactly 4 cars passing the observation point,

(ii) at least 5 cars passing the observation point. (5)

The number of other vehicles, other than cars, passing the observation point in a 60 minute interval is modelled by a Poisson distribution with mean 12.

(c) Find the probability that exactly 1 vehicle, of any type, passes the observation point in a 10 minute period. (4)


for PART C:
I did P(X=1) for each poisson distribution for 10 mins.

The mark scheme did this:
P ( 0 car and 1 others) + P (1 cars and 0 other )
= e-1 x 2e-2 + 1e-1 x e-2
= 0.3679 x 0.2707 + 0.3674 x 0.1353
= 0.0996 + 0.0498
= 0.149


Can anyone explain this? :smile:
Reply 243
Avatar for Gibbo81
Gibbo81
1
Original post by raheem94
Please send me the FP3 paper when you get it.

Though the paper is a long time away, when the time will come then we will see who does the solutions. Gibbo might be doing it.


Hi there,

I've created a thread for the post exam discussion, it's locked for now (they seem to be trying to keep the place tidy this year).

I'll post the solutions asap the day after the exam.
Original post by Gibbo81
Hi there,

I've created a thread for the post exam discussion, it's locked for now (they seem to be trying to keep the place tidy this year).

I'll post the solutions asap the day after the exam.


:ta:
Can someone explain 7c on Solomon Paper c?

The random variable X follows a continuous uniform distribution over the interval [2, 11].
c) Find P( |X − 5| < 2).
(edited 11 years ago)
Original post by Gibbo81
Hi there,

I've created a thread for the post exam discussion, it's locked for now (they seem to be trying to keep the place tidy this year).

I'll post the solutions asap the day after the exam.


Thank you very much Gibbo :smile:

Can you do FP2 and FP3 too? Pretty please with sugar on top? :please:
Original post by john_bishop1
Can someone explain 7c on Solomon Paper c?

The random variable X follows a continuous uniform distribution over the interval [2, 11].
c) Find P( |X − 5| < 2).


Some posted a similar question but with different values. It's exactly the same principal:

P(∣X−16∣<3)P(\left |X-16 \right |<3)

P((X−16)<3)=P((X<16+3))=P(X<19)P((X-16)<3)=P((X<16+3))=P(X<19)

P(−(X−16)<3)=P(−X<3−16)=P(−X<−13)=P(X>13)P(-(X-16)<3)=P(-X<3-16)=P(-X<-13)=P(X>13)

P(13<X<19)=(19−13).124=14P(13<X<19) = (19-13).\frac{1}{24}=\frac{1}{4}
Reply 248
Avatar for Gibbo81
Gibbo81
1
Original post by Dreamweaver
Thank you very much Gibbo :smile:

Can you do FP2 and FP3 too? Pretty please with sugar on top? :please:


Will do FP2 for sure. Not sure about FP3 as I'm not 100% that anyone is doing it at my school this year. If they are I might give it a go, or at worse, post the paper.
Original post by Gibbo81
Will do FP2 for sure. Not sure about FP3 as I'm not 100% that anyone is doing it at my school this year. If they are I might give it a go, or at worse, post the paper.


Legend <3

There's a FP2/FP3 thread in full swing atm. Please feel free to make a post exam thread. Alternatively, you could post it on the current thread, after the correct amount of time after the exam has elapsed, and i'll stick it in the OP (with full credits :smile:).
Reply 250
Avatar for Groat
Groat
13
Let's keep the discussion on topic please.
Reply 251
Avatar for number23
number23
OP
15
The main thing im finding difficult is when they say "why is f(x) not a suitable model" and then you have to draw a better one :s

what could be wrong with pdf distributions?
Original post by Kaiser7
Can someone explain to me number 4(e) in the January 2009 exam?

What graph is that? And why is it more suitable than a pdf? :redface:


Yeah, I also didn't understand this :s-smilie:
Original post by number23
The main thing im finding difficult is when they say "why is f(x) not a suitable model" and then you have to draw a better one :s

what could be wrong with pdf distributions?



Original post by Phenylethylamine_
Yeah, I also didn't understand this :s-smilie:


I think that they're just trying to get you to understand that the length of a phone call with be distributed away from the mean, like a graph of a normal distribution, as opposed to a straight line model which doesn't seem to make much sense
Original post by Phenylethylamine_
Yeah, I also didn't understand this :s-smilie:


I can't remember quite what the question is, but it's about phone calls, right?

How likely is it that ALL phone calls will fit that model (the pdf line)? Not very. It's more likely that most phone calls will last around 3-5 minutes, and then slowly trail off; however, you also have some phone calls which last 60+ minutes, so the curve you should draw should tail off like that to account for such calls. They're trying to get you to point out how the model they gave you was inaccurate.
Original post by knowledgecorruptz
I can't remember quite what the question is, but it's about phone calls, right?

How likely is it that ALL phone calls will fit that model (the pdf line)? Not very. It's more likely that most phone calls will last around 3-5 minutes, and then slowly trail off; however, you also have some phone calls which last 60+ minutes, so the curve you should draw should tail off like that to account for such calls. They're trying to get you to point out how the model they gave you was inaccurate.


Thanks for that :biggrin:
Original post by Refrigerator
I think that they're just trying to get you to understand that the length of a phone call with be distributed away from the mean, like a graph of a normal distribution, as opposed to a straight line model which doesn't seem to make much sense


Thanks for that :biggrin:
Reply 257
Avatar for number23
number23
OP
15
Original post by Refrigerator
I think that they're just trying to get you to understand that the length of a phone call with be distributed away from the mean, like a graph of a normal distribution, as opposed to a straight line model which doesn't seem to make much sense


so unifrom pdf's dont account for variance?
A very common question i've noticed that most people can't answer is why, for example, don't you test all of something? The answer is that testing destroys said something and renders them all useless. So instead of testing everything and having no product, you would test a few and make a prediction based on that. Free 2 marks :smile:
Original post by number23
so unifrom pdf's dont account for variance?


I don't think it was a uniform pdf in the question, it was a line with gradient 1/10 or something I think

I think a uniform pdf has a variance, but mean and variance aren't the only 2 things to take into account with a distribution. I don't really know any more than we've been taught but I can't imagine the only difference between a normal and uniform distribution being their variance/mean

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