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    (Original post by SeanFM)
    The key thing is understanding the difference between adding multiple distributions (it's usually men and women's weight or stuff like that) and multiplying one distribution.

    Eg if M~N(a,b) and W~N(c,d) then the difference between a man and a woman's weight would be ~N(a-c, b+d) but if you were asked for the probability of a man's height being 1.5 times more than a woman's, you'd be looking for something like (D :=M-1.5W)~N(a-1.5c,b+2.25d) and you'd find P(D>0) and.. that's about it.
    Is it with decimals/fractions that we square?
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    (Original post by L'Evil Wolf)
    Is it with decimals/fractions that we square?
    No it can be integers too. You only need to square when its the variance and its not a combination. The majority fall into two types of questions ill give you an example, sort of from above. Find the probability that the mens height is 2 times greater than women's height. The thing you would consider is P(M>2W) Therefore becomes P(M-2W>0) Let X=M-2W, Then E(X) = E(M)-2E(W) and Var(X)= Var(M) + 2^2Var(W)

    However if the question were to be two women are selected find the probability that the mans height is greater than the total of the two women? you would consider P(M>W1+W2) Which becomes P(M-W1-W2>0) So let Y=M-W1-W2 Then E(Y)= E(M)-E(W1)-E(W2) But because W1 and W2 have the same distribution its E(Y)= E(M) -2E(W) and Var(Y)=Var(M)+Var(W1)+Var(W2) Apply same rule and Var(Y)=Var(M)+2Var(W)

    Hopefully thats all correct and some sort of help
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    (Original post by SeanFM)
    The key thing is understanding the difference between adding multiple distributions (it's usually men and women's weight or stuff like that) and multiplying one distribution.

    Eg if M~N(a,b) and W~N(c,d) then the difference between a man and a woman's weight would be ~N(a-c, b+d) but if you were asked for the probability of a man's height being 1.5 times more than a woman's, you'd be looking for something like (D :=M-1.5W)~N(a-1.5c,b+2.25d) and you'd find P(D>0) and.. that's about it.
    The problem I have is with \displaystyle Var(aX) = a^{2}Var(X)

    If \displaystyle Var(X + Y + Z) = Var(X) + Var(Y) + Var(Z)

    Then why is \displaystyle Var(3X) = Var(X + X + X) \neq 3Var(X)

    I think physicsmaths and Zacken discussed this in the first 10 posts, but I still don't understand
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    Hi guys I have a quick question when doing the Chi-Squared tests is it ok to not do a (Oi-Ei)^2 or Oi^2/Ei Column and work each out individually. Instead just use the formula and plug numbers into there showing what you're done by writing the formula then the a quick example of the numbers your using. E.g Sum(Oi^2/Ei)-N= Sum(11^2/29+12^/33+...+12^/13)= then write the test statistic. Would this still get me full marks provided the test stat is right?

    Also how do we know when two write our hypotheses as H0: A binomial is a suitable model. OR H0: ~B(n,p) is suitable model.
    I cant seem to find when to write which MS are to inconsistent. Binomials just used as an example. Same goes for Poisson, normal ect. How do we know when to write what?
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    (Original post by Euclidean)
    The problem I have is with \displaystyle Var(aX) = a^{2}Var(X)

    If \displaystyle Var(X + Y + Z) = Var(X) + Var(Y) + Var(Z)

    Then why is \displaystyle Var(3X) = Var(X + X + X) \neq 3Var(X)

    I think physicsmaths and Zacken discussed this in the first 10 posts, but I still don't understand
    Are you asking why can't we split Var(X+X) into Var(X)+Var(X) just like we can Var(X+Y)?
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    (Original post by gagafacea1)
    Are you asking why can't we split Var(X+X) into Var(X)+Var(X) just like we can Var(X+Y)?
    Yeah I think so
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    (Original post by Euclidean)
    The problem I have is with \displaystyle Var(aX) = a^{2}Var(X)

    If \displaystyle Var(X + Y + Z) = Var(X) + Var(Y) + Var(Z)

    Then why is \displaystyle Var(3X) = Var(X + X + X) \neq 3Var(X)

    I think physicsmaths and Zacken discussed this in the first 10 posts, but I still don't understand
    Var(3X) is not qual to Var(X + X + X)
    Var(3X) will give you 3^2Var(X)
    BUT Var(X+X+X) Will give you 3Var(X)
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    (Original post by SB0073)
    Var(3X) is not qual to Var(X + X + X)
    Var(3X) will give you 3^2Var(X)
    BUT Var(X+X+X) Will give you 3Var(X)
    That makes sense.

    3X is a scale of the distribution of X which is why the variance is much larger, because we're scaling...

    Thanks
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    (Original post by Euclidean)
    Yeah I think so

    The thing is, X and Y can be different, while X and X cannot. So for example let's say that P(X=a)=p and P(Y=a)=q.
    Now P(X+X=2a)=p, this is only because if X "turned out" to be 'a', then X+X will surely be 2a. While P(X+Y=2a)=pq because to get 2a out of X+Y, we need X to be 'a' AND Y to be 'a', which is not guaranteed.

    If you substitute X and Y with X1 and X2 respectively, where they both have the same distribution as X, then you'll get P(X_1+X_2=2a)=2p. You see how X1+X2 has a very different distribution from X+X? They both have the same mean, but the variance is different.
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    (Original post by SB0073)
    Var(3X) is not qual to Var(X + X + X)
    Var(3X) will give you 3^2Var(X)
    BUT Var(X+X+X) Will give you 3Var(X)
    I think you meant X1+X2+X3, because 3X does equal X+X+X
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    (Original post by Euclidean)
    The problem I have is with \displaystyle Var(aX) = a^{2}Var(X)

    If \displaystyle Var(X + Y + Z) = Var(X) + Var(Y) + Var(Z)

    Then why is \displaystyle Var(3X) = Var(X + X + X) \neq 3Var(X)

    I think physicsmaths and Zacken discussed this in the first 10 posts, but I still don't understand
    Districution aint X it is X_i. So districution of 3X is different to X_i.


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    (Original post by gagafacea1)
    If you substitute X and Y with X1 and X2 respectively, where they both have the same distribution as X,
    Surely if they have the same distribution as X then they are equal to X? How does that work?

    (Original post by gagafacea1)
    You see how X1+X2 has a very different distribution from X+X? They both have the same mean, but the variance is different.
    I'm slowly getting there with understanding though. So when we have the sample variable which has a multiple like Var(5X_1) = 25Var(X_1) but when there are multiple variables combined like Var(X_1 + 2X_2) = Var(X_1) + 4Var(X_2) we get squares because combinations of X are distinct from combinations of X_i's :beard:

    (Original post by physicsmaths)
    Districution aint X it is X_i. So districution of 3X is different to X_i.Posted from TSR Mobile
    Ah yeah because one's scaled up, Zacken's early comment makes loads of sense now.

    Thanks to both of you
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    (Original post by Euclidean)
    Surely if they have the same distribution as X then they are equal to X? How does that work?



    I'm slowly getting there with understanding though. So when we have the sample variable which has a multiple like Var(5X_1) = 25Var(X_1) but when there are multiple variables combined like Var(X_1 + 2X_2) = Var(X_1) + 4Var(X_2) we get squares because combinations of X are distinct from combinations of X_i's :beard:



    Ah yeah because one's scaled up, Zacken's early comment makes loads of sense now.

    Thanks to both of you
    No but the variance is difference.


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    (Original post by Euclidean)
    Surely if they have the same distribution as X then they are equal to X? How does that work?
    They have the same distribution yes, but they don't have to be equal. They are different random variables, so X1 could be equal to 2 while X2 equal to 7. Just like X and Y are different. This is why I tried to highlight the fact that the probability of getting 2a is different depending on if we had X+X (same variable added to itself) or if we had X1+X2 (different variables added).
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    (Original post by gagafacea1)
    They have the same distribution yes, but they don't have to be equal. They are different random variables, so X1 could be equal to 2 while X2 equal to 7. Just like X and Y are different. This is why I tried to highlight the fact that the probability of getting 2a is different depending on if we had X+X (same variable added to itself) or if we had X1+X2 (different variables added).
    :facepalm:

    That's just clicked for me. Sorry I didn't pick up on it right away, thanks so much
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    (Original post by Euclidean)
    Normal distribution if you're not very confident with it

    Otherwise it took me around 5 hours to study from scratch, the bulk of that being chapter 2 tbh

    Posted from TSR Mobile
    Cheers yeah, just been refreshing my memory on normal distribution.

    And wow, 5 hours!? How did you go through so quickly?
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    (Original post by paradoxequation)
    Cheers yeah, just been refreshing my memory on normal distribution.

    And wow, 5 hours!? How did you go through so quickly?
    https://www.youtube.com/user/jbstatistics

    I didn't practice at all though, straight into the 2000 paper. Going back now and practicing some of the chapters (1 and 2)
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    (Original post by Euclidean)
    https://www.youtube.com/user/jbstatistics

    I didn't practice at all though, straight into the 2000 paper. Going back now and practicing some of the chapters (1 and 2)
    Ah I see, I tend to just do the exercise questions then move on to past papers once content is done. Will check out that channel though.
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    (Original post by Xinchu)
    Hello Sean,
    I have got a S3 question, if you are able to answer it?

    it's a question about goodness of fit( continuous uniform distribution)
    In the S3 text book, there is an example of it,but when I looked at it,I got a little bit confused about how they didn't combine cells with expected frequencies less than 5, my question is :is that always the case just for continuous uniform distribution or was it a mistake in the text book?
    It would be great and helpful if you can answer it for me.
    Thanks.
    :wavey:

    As far as I am aware there is nothing special about continuous uniform distributions so I would say that it is most likely a mistake.
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    (Original post by SeanFM)
    :wavey:

    As far as I am aware there is nothing special about continuous uniform distributions so I would say that it is most likely a mistake.
    xinchu


    I don't think the tag worked so just incase they don't see it
 
 
 
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