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AS AQA Mathematics Statistics 1B MS1B June 2014

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Got soooooo much to do and finding it soooooo hard


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Reply 21
how do you go about answering a questions saying - calculate the prob: whether its A or B but NOT both. I know how to do it for when it both and tried the reverse but............
Reply 22
Original post by saad97
Hi guys, does anyone know whether we have to learn any equations for this exam or can they all be found in the formulae booklet? Thanks :smile:


Don't forget the mutually exclusive and independence tests :smile:
Reply 23
Original post by H-SEA
how do you go about answering a questions saying - calculate the prob: whether its A or B but NOT both. I know how to do it for when it both and tried the reverse but............


P(A or B but not both) = P(A only) + P(B only)
or alternatively
P(A or B but not both) = 1 - P(neither A nor B) - P(both A and B)

Assuming only two, independent, variables.

I don't know how to explain it better than this without numbers... do you have an example?

(Also, stats is not my strong point, please correct me if I'm wrong!)
Original post by Cluck
P(A or B but not both) = P(A only) + P(B only)
or alternatively
P(A or B but not both) = 1 - P(neither A nor B) - P(both A and B)

Assuming only two, independent, variables.

I don't know how to explain it better than this without numbers... do you have an example?

(Also, stats is not my strong point, please correct me if I'm wrong!)


I don't think this is quite right, because P(AuB) = P(A) + P(B) - P(AnB), so your formula would be 2P(AnB) too much, so it would be P(A) + P(B) - 2P(AnB), or iff independent P(A) + P(B) - 2P(A)P(B)

Note: Although you may mean P(B') for your P(only A) and P(A') for your P(only B), and I'm still not sure if this would work...

Second note: Stats is my worst module by far as well, so I could be wrong! :L
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(edited 9 years ago)
Reply 25
http://gyazo.com/621d2c29cd4dd955cd26b9b971bc45af
http://gyazo.com/7945fda77e47d20a4007f08cbb8b511b

these may be of some help to anyone, sorry for the terrible photo quality, my phone isnt the best camera.
Reply 26
Original post by TLHroolz
I don't think this is quite right, because P(AuB) = P(A) + P(B) - P(AnB), so your formula would be 2P(AnB) too much, so it would be P(A) + P(B) - 2P(AnB), or iff independent P(A) + P(B) - 2P(A)P(B)

Note: Although you may mean P(B') for your P(only A) and P(A') for your P(only B), and I'm still not sure if this would work...

Second note: Stats is my worst module by far as well, so I could be wrong! :L
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Umm now I'm confued :s-smilie: What you said looks right but I think we may just be using different notation or something, since I've never dropped a probability mark, but at any rate I'm assuming it's too late to worry now! :colondollar:
hey guys

how do you find out how reliable an estimate (one which you have found using the least squares regression line) is when they say that the values of the residuals lie between approximately x and y?
Can someone help me with 7bi please?
Reply 29
Original post by Scarlett44
Can someone help me with 7bi please?


If you look at a normal distribution graph:



you can see that there are likely to be some values at μ - and beyond - however, in this case, μ - is 37 - 25 * 2 = -13, which is negative. As the times in the context in this question cannot be negative, they are unlikely to be normally distributed.
Original post by junaid95
hey guys

how do you find out how reliable an estimate (one which you have found using the least squares regression line) is when they say that the values of the residuals lie between approximately x and y?


The smaller the residuals, the more reliable the estimate. So a residual of +/- 0.5 is very reliable whereas a residual of +/- 50 is likely to be unreliable.
Original post by mike256
If you look at a normal distribution graph:



you can see that there are likely to be some values at μ - and beyond - however, in this case, μ - is 37 - 25 * 2 = -13, which is negative. As the times in the context in this question cannot be negative, they are unlikely to be normally distributed.


Okay thanks, so μ - is something I need to remember for the exam? And if there's a question on normal distribution use that?


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Reply 32
Exam tomorrow guys! Honestly, I'm feeling a lot better about this exam than I did when I first started this thread, and thanks to everyone who's contributed so far :biggrin:

Best advice I can give:

Normal distributions - draw the graph
Binomial distributions - draw a number line
Probability - decide whether or not to represent the probabilities in a tree diagram, Venn diagram or table

If you prefer to work with numbers rather than visual representation, fair enough :smile:

How's everyone feeling???
Original post by PulsarX
z = (x - mu) / sigma

So that's the x value minus the mean, all divided by the standard deviation.


Edit: z = (x μ) / σ


Can you not just use the tables to find z? Or is this equation for something different?

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Reply 34
Original post by winniestar57
Can you not just use the tables to find z? Or is this equation for something different?

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This equation is to find the z value, and then use that value on the normal distribution table to find the probability for that z value.

The tables find the probability USING the z value, not the other way around.
Original post by PulsarX
This equation is to find the z value, and then use that value on the normal distribution table to find the probability for that z value.

The tables find the probability USING the z value, not the other way around.


Oh, I know what you mean now! Haha sorry!

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Reply 36
Original post by Scarlett44
Okay thanks, so μ - is something I need to remember for the exam? And if there's a question on normal distribution use that?


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No, it's not a general thing, it only works in this particular case. I just picked the number of standard deviations that first gave a negative number. It's very rare for a question like this to come up, but you should be familiar with the fact that most of the data in a normal distribution lies within about 3 standard devations either side of the mean.
Original post by mike256
No, it's not a general thing, it only works in this particular case. I just picked the number of standard deviations that first gave a negative number. It's very rare for a question like this to come up, but you should be familiar with the fact that most of the data in a normal distribution lies within about 3 standard devations either side of the mean.


Okay thanks :smile:


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Reply 38
Standard error = (standard deviation of the parent population)/√(sample size).

So in this case it would be √16 / √36 = 2/3 I believe?
Not feeling too confident about this tomorrow :frown: Do you think the core 1 and core 2 exams are a good indicator of the level of difficulty for this paper?


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