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Edexcel S2 - 27th June 2016 AM

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Avatar for SSD07
SSD07
3
Original post by Ayman!
X <k/2 means that X < k, so we can just write it as X < k/2.


What is this??? like what topic
Original post by SSD07
What is this??? like what topic


Conditional probability applied to continuous distributions. Quite common as of late in stats papers - came up in my S3
Hi there, im stuck with this question, where for part a)i & a)ii i can't figure out how to hack the alternative method (2), especially for a)ii where for X~B(10,0.4) shouldn't P(X<9) be equivalent to P(x<=1). Plus for part b im lost. :frown: thanks in advance
(edited 7 years ago)
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Original post by taysc
Does anyone have a good way/acronym of remembering skewness in terms of mode median and mean?

Im struggling to remember the pattern


I don't have an acronym to help but for some reason drawing these graphs help. Mode is always the 'peak' or most dense part of the PDF so draw that in, then draw in the median and/or mean depending on it's value being greater or less than the mode. If the PDF you draw is more dense towards the vertical axis (i.e. the mode is closest to the vertical), it is positive skew, and vice-versa. This helped me, maybe I just have an odd way of learning things. :biggrin:

skew2.jpg
S2 June 2011. Need help with question 6b please. Can anyone tell me if/why my working is incorrect?

Unparseable latex formula:

\text{If } P(X $\geqslant$ 31) $\leqslant$ 0.1, Reject H_0.



In the mark scheme they go with working from this approach, but is the opposite true too? I worked from:

Unparseable latex formula:

\text{If } P(X $\leqslant$ 31) $\geqslant$ 0.9, Reject H_0.



My full working and question below:
IMG_3515.jpg513.1KB
Screenshot_1.jpg194.5KB
Original post by yuveethini
Hi there, im stuck with this question, where for part a)i & a)ii i can't figure out how to hack the alternative method (2), especially for a)ii where for X~B(10,0.4) shouldn't P(X<9) be equivalent to P(x<=1). Plus for part b im lost. :frown: thanks in advance


Think of it like this - if you flipped a coin 10 times, and you wanted to model how many times you get a head, say X is the distribution of number of heads and X~B(n,p) then you can also measure it with another distribution, Y, the number of tails. Y~(n,1-p). Why are the probabilities that way? Because if p is the probability of getting a heads, 1-p is the probability of getting a tails.

Now, X = 6 is the same as saying Y = 4. If you read that in English, it's the same as saying getting 6 heads is the same as getting 4 tails. So the distributions might be different, but they're modelling the same outcome (6 heads and 4 tails) and the water example is doing the same thing.

For part aii, the RHS, you may have missed out how they're using X and Y instead of just X. Saying X (the number of customers who order water, and Y is the number who don't) - you say that the probability of strictly less than 9 customers ordering water is the same as the probability of strictly more than 1 customer ordering water, and the probability of that is 1 - p(less than or equal to 1 customer not ordering water).
Original post by MattOGrady
S2 June 2011. Need help with question 6b please. Can anyone tell me if/why my working is incorrect?

Unparseable latex formula:

\text{If } P(X $\geqslant$ 31) $\leqslant$ 0.1, Reject H_0.



In the mark scheme they go with working from this approach, but is the opposite true too? I worked from:

Unparseable latex formula:

\text{If } P(X $\leqslant$ 31) $\geqslant$ 0.9, Reject H_0.



My full working and question below:


The statements are pretty much equivalent (I'm not 100% sure about the first inequality sign in your second line of latex, it may have to be strictly less than 31 but I'm not sure) so I suppose you could work like that, though it'd be a bit strange.
Original post by SSD07
What is this??? like what topic


Look at it this way - if I told you that BOTH X<10 and X<5 (what the intersect is basically), then which inequality implies the other?

X<10 doesn't imply that X<5, because X could be 6,7,8,9 so the first one would be true but the second not.

But X<5 implies that X<10 - if the first one is true, the second one is, so that's why the intersection of the two is X<5.

Similarly, if you had Y> 7 and Y > 11 - the second one implies the first, but not the other way round, so the intersection would be Y > 11.
Original post by SeanFM
The statements are pretty much equivalent (I'm not 100% sure about the first inequality sign in your second line of latex, it may have to be strictly less than 31 but I'm not sure) so I suppose you could work like that, though it'd be a bit strange.


Thank you! It's a normal approximation so I don't think it matters whether it's a strict inequality or not. Doing it this way just means that I can work from tables straight away, without having to do 1 - probability later on. I guess I should just stick with the mark scheme method.
https://57a324a1a586c5508d2813730734691051ac35fd.googledrive.com/host/0B1ZiqBksUHNYZ3M4QzJ0N19IeHc/June%202009%20QP%20-%20S2%20Edexcel.pdf

can someone help me with question 8

why did they e^-2x/15. like i dont get where the exponential came from but i get the 2x/15. :s-smilie:

https://57a324a1a586c5508d2813730734691051ac35fd.googledrive.com/host/0B1ZiqBksUHNYZ3M4QzJ0N19IeHc/June%202009%20MS%20-%20S2%20Edexcel.pdf
Original post by MattOGrady
Thank you! It's a normal approximation so I don't think it matters whether it's a strict inequality or not. Doing it this way just means that I can work from tables straight away, without having to do 1 - probability later on. I guess I should just stick with the mark scheme method.


I see what you're saying, but both regions include the critical value (which is what matter, you're right with the rest and like 30.9999999 and all of that) and I'm not sure if, by definition, it should be inclusive if you're going to do it that way - maybe I am just making things up :lol:) but fair enough - personally, if you got it right that way I'd give you the marks but.. better safe than sorry, I s'pose :redface:
Original post by SeanFM
I see what you're saying, but both regions include the critical value (which is what matter, you're right with the rest and like 30.9999999 and all of that) and I'm not sure if, by definition, it should be inclusive if you're going to do it that way - maybe I am just making things up :lol:) but fair enough - personally, if you got it right that way I'd give you the marks but.. better safe than sorry, I s'pose :redface:


I get you haha, my teacher used to say "You're arguing about the area of a line" for any probability of an exact value when the variable is continuous, but you are definitely correct. :biggrin: I'll just hope that a normal approximation and hypothesis test don't come up in the same question this year! Cheers dude.
Reply 232
Avatar for imran_
imran_
2
the grade boundaries are so high wtf
Original post by Lilly1234567890


can someone help me with question 8

why did they e^-2x/15. like i dont get where the exponential came from but i get the 2x/15. :s-smilie:



The exponential comes from the Poisson Distribution probability function in the formula booklet:

P(x)=e−λλxx!P(x) = \frac{{e^{ - \lambda } \lambda ^x }}{{x!}}

Where x=0 and λ=2x15x=0 \text{ and } \lambda = \frac{2x}{15} as a result of the question, this becomes:

P(x)=e−2x15P(x) = e^{ - \frac{2x}{15} }

Then just sub in 1.65 and 1.75 giving values either side of 0.8, therefore showing that x=1.7 as required.
Original post by MattOGrady
The exponential comes from the Poisson Distribution probability function in the formula booklet:

P(x)=e−λλxx!P(x) = \frac{{e^{ - \lambda } \lambda ^x }}{{x!}}

Where x=0 and λ=2x15x=0 \text{ and } \lambda = \frac{2x}{15} as a result of the question, this becomes:

P(x)=e−2x15P(x) = e^{ - \frac{2x}{15} }

Then just sub in 1.65 and 1.75 giving values either side of 0.8, therefore showing that x=1.7 as required.


thanks !
Help with this question please








Posted from TSR Mobile
i have always known that if our value is less than 0.05. ten there is sufficient evidence to reject H0. But that doesnt seem to be the case here in q5

https://57a324a1a586c5508d2813730734691051ac35fd.googledrive.com/host/0B1ZiqBksUHNYZ3M4QzJ0N19IeHc/January%202008%20QP%20-%20S2%20Edexcel.pdf

is the mark scheme wrong

https://57a324a1a586c5508d2813730734691051ac35fd.googledrive.com/host/0B1ZiqBksUHNYZ3M4QzJ0N19IeHc/January%202008%20MS%20-%20S2%20Edexcel.pdf
Reply 237
Avatar for taysc
taysc
2
Original post by MattOGrady
I don't have an acronym to help but for some reason drawing these graphs help. Mode is always the 'peak' or most dense part of the PDF so draw that in, then draw in the median and/or mean depending on it's value being greater or less than the mode. If the PDF you draw is more dense towards the vertical axis (i.e. the mode is closest to the vertical), it is positive skew, and vice-versa. This helped me, maybe I just have an odd way of learning things. :biggrin:

skew2.jpg


This really helpful, I'm going to try it today with the last papers that I do.

Thank you
Original post by SeanFM
Think of it like this - if you flipped a coin 10 times, and you wanted to model how many times you get a head, say X is the distribution of number of heads and X~B(n,p) then you can also measure it with another distribution, Y, the number of tails. Y~(n,1-p). Why are the probabilities that way? Because if p is the probability of getting a heads, 1-p is the probability of getting a tails.

Now, X = 6 is the same as saying Y = 4. If you read that in English, it's the same as saying getting 6 heads is the same as getting 4 tails. So the distributions might be different, but they're modelling the same outcome (6 heads and 4 tails) and the water example is doing the same thing.

For part aii, the RHS, you may have missed out how they're using X and Y instead of just X. Saying X (the number of customers who order water, and Y is the number who don't) - you say that the probability of strictly less than 9 customers ordering water is the same as the probability of strictly more than 1 customer ordering water, and the probability of that is 1 - p(less than or equal to 1 customer not ordering water).

Ohhhh i get it now thank you :biggrin:
Also how do i go about doing part b, i just dont understand it :s-smilie:
(edited 7 years ago)
Original post by SeanFM
Number of correct answers = X. number of incorrect answers = 20 - X.

number of points from total correct answers = ...

number of points docked from total incorrect answers = ...

hence...

it's not asking for you to find that it's when you get 4 correct answers and 16 incorrect answers (which doesn't quite make sense, though I see what you are trying to do)

It's asking how many points you get when X answers are right. (implying that 20-X are wrong).


thanks bro

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