# S1 questions

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#1
Hi,

I have a few S1 questions that I am stuck on.
The first is here.
I thought that to find a for y on x, it is a = (mean of y) - b(mean of x) and so x on y would just be a' = (mean of x) - b'(mean of y) so am not sure why I got it wrong?

Thanks
0
5 years ago
#2
You don't need to find 'a' here in the manner you have, you have correctly found mean of Y, so just sub this back into the equation with mean of X and solve for C.

4.5=0.40(14.5)-C
4.5=5.8-C
C=1.3

Hopefully that helps a bit
1
#3
(Original post by nathanfield)
You don't need to find 'a' here in the manner you have, you have correctly found mean of Y, so just sub this back into the equation with mean of X and solve for C.

4.5=0.40(14.5)-C
4.5=5.8-C
C=1.3

Hopefully that helps a bit
Thank you so much - that makes much more sense.
Just wondering why my method did not work?
0
5 years ago
#4
(Original post by musicangel)
Thank you so much - that makes much more sense.
Just wondering why my method did not work?
Your method was actually okay, you just calculated it wrong; 4.5-(0.40x14.5)= -1.3, use -1.3 as '-c' then you get C as 1.3.
So your method was fine just a minor mistake! So still well done.

Just much easier to remember that the common point on XonY and YonX is (XBar, YBar) so by subbing those values in to the equation its much easier to find the unknown.
0
#5
(Original post by nathanfield)
Your method was actually okay, you just calculated it wrong; 4.5-(0.40x14.5)= -1.3, use -1.3 as '-c' then you get C as 1.3.
So your method was fine just a minor mistake! So still well done.

Just much easier to remember that the common point on XonY and YonX is (XBar, YBar) so by subbing those values in to the equation its much easier to find the unknown.
Thank you so much for this!
I was wondering why my original method did not work in the question attached and why you have to use binomial? In some cases, (e.g. 2 interviews, one on the 5th and one before that so (SFFF)x4 x P(S) ) my method works so was wondeirng which way I should work out the questions?

Thanks! 0
5 years ago
#6
where do we get the markscheme of jan 2017 paper? any idea?
0
5 years ago
#7
(Original post by sayema1)
where do we get the markscheme of jan 2017 paper? any idea?
Hi Sayema - are you talking about Jan 2017 IAL for S1? I can find them for you.. I have done the paper already

thank you for the rep 👀🤗
0
5 years ago
#8
Hi Sayema - are you talking about Jan 2017 IAL for S1? I can find them for you.. I have done the paper already
yeeehhhh thanks a loot though  0
5 years ago
#9
(Original post by sayema1)
yeeehhhh thanks a loot though  Thank you for the rep 😇🙃

Haha no worries 😉.... Found it- hope it helps.

0
#10
Also, just wondering why this was wrong?
Thanks 0
5 years ago
#11
Is this edexcel? What paper is this from ?

Thank you 0
#12
(Original post by wantodothebest)
Is this edexcel? What paper is this from ?

Thank you Hey, no its OCR. Can't remember what paper it is from but will try to find out
0
5 years ago
#13
(Original post by musicangel)
Hey, no its OCR. Can't remember what paper it is from but will try to find out
Dw about it, I do Edexcel and I freaked because the format was different haha
0
5 years ago
#14
(Original post by musicangel)
Thank you so much for this!
I was wondering why my original method did not work in the question attached and why you have to use binomial? In some cases, (e.g. 2 interviews, one on the 5th and one before that so (SFFF)x4 x P(S) ) my method works so was wondeirng which way I should work out the questions?

Thanks! It looks like the first few questions are geometric which appears to be the way you have done this which is good!
For the last question, however, it states that there are TWO successes.
This is imperative as Geometric probabilities must only obtain ONE success.
Therefore we now know for this last question it must be a binomial.

I would suggest you take a quick look on exam solutions dot net just to quickly gather the properties/differences of geometric & binomial equations as these will undoubtably come up in our exam come Wednesday therefore could be the difference.
0
5 years ago
#15
(Original post by musicangel)
Also, just wondering why this was wrong?
Thanks Okay, with the first part it states 'without regard to order' which means we are looking at combinations (how many arrangements) rather than permutations/factorials (how many orders/rearrangements). Therefore we do not need to take into account the repeated 'one' as essentially it is the exact same thing as we don't care for order. So everything was right other than x2.

With the second part it asks you for the total ways to arrange all these numbers (again without regard to repeating). So we have already worked out how many arrangements with 2 'ones' which = 5. Then we worked out how many arrangements with 1 'one' which = 10. Therefore we just need to work out how many arrangements with 0 'ones' and then add! So its 5+10+5C3(=10)=25.

Hopefully this makes sense and helps.
0
#16
(Original post by nathanfield)
It looks like the first few questions are geometric which appears to be the way you have done this which is good!
For the last question, however, it states that there are TWO successes.
This is imperative as Geometric probabilities must only obtain ONE success.
Therefore we now know for this last question it must be a binomial.

I would suggest you take a quick look on exam solutions dot net just to quickly gather the properties/differences of geometric & binomial equations as these will undoubtably come up in our exam come Wednesday therefore could be the difference.
Thank you so much - really appreciate your help.
Just wondering why is has to be binomial/geometric? Why can't you just multiply the probablities together?
0
5 years ago
#17
(Original post by musicangel)
Thank you so much - really appreciate your help.
Just wondering why is has to be binomial/geometric? Why can't you just multiply the probablities together?
I'm not sure I understand, but the way I mentioned is the way you will be able to answer any of these questions
0
#18
(Original post by nathanfield)
Okay, with the first part it states 'without regard to order' which means we are looking at combinations (how many arrangements) rather than permutations/factorials (how many orders/rearrangements). Therefore we do not need to take into account the repeated 'one' as essentially it is the exact same thing as we don't care for order. So everything was right other than x2.

With the second part it asks you for the total ways to arrange all these numbers (again without regard to repeating). So we have already worked out how many arrangements with 2 'ones' which = 5. Then we worked out how many arrangements with 1 'one' which = 10. Therefore we just need to work out how many arrangements with 0 'ones' and then add! So its 5+10+5C3(=10)=25.

Hopefully this makes sense and helps.
Oh okay. I thought that you would treat each '1' as a different number so 1a, 2, 3 and 1b, 2, 3 etc. Why would that not work?

That makes much more sense. Just wondering what I did anything wrong in my method? Is there a way of doing it without working out each case individually?

Sorry for all of the questions! 0
#19
(Original post by nathanfield)
I'm not sure I understand, but the way I mentioned is the way you will be able to answer any of these questions
Ok, so it is always best to use either binomial or geometric. If you had to find out one place in 5 attempts, would you use geometric?

Thank you!
0
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