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    if you were correcting P(X<44) from a binomial to normal, would it be P(X=<43.5) or P(X<43.5)

    when do you do the equal sign? it never makes a difference to the final answer but i still want to know. might lose marks...
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    (Original post by cooldudeman)
    if you were correcting P(X<44) from a binomial to normal, would it be P(X=<44.5) or P(X<44.5)

    when do you do the equal sign? it never makes a difference to the final answer but i still want to know. might lose marks...
    Think about it for a second, if P(x<44) then P(x&lt;43.5) as that is the next one down by continuity correction.
    However if it was P(x\leq 44) then by continuity correction = P(x&lt;44.5)
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    (Original post by cooldudeman)
    if you were correcting P(X<44) from a binomial to normal, would it be P(X=<44.5) or P(X<44.5)

    when do you do the equal sign? it never makes a difference to the final answer but i still want to know. might lose marks...
    Neither.

    It approximates to P(X<43.5). If X is 43.5 or more, it rounds off to 44, but the inequality clearly states that X cannot be 44.
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    (Original post by justinawe)
    Neither.

    It approximates to P(X<43.5). If X is 43.5 or more, it rounds off to 44, but the inequality clearly states that X cannot be 44.
    Hmm would my less than or equal to be technically wrong or right then, for 43.5, since after all P(x<44) includes 43.5?
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    its p(x less than to 43.5) Learn the continuity rules

    (Original post by Robbie242)
    Hmm would my less than or equal to be technically wrong or right then, for 43.5, since after all P(x<44) includes 43.5?
    its wrong, anything 43.5 or above to automatically round to 44 which is wrong thats why its gotta be <43.5 so that it never manages to make 44. You'll get useto doing it after doing a few normally distribution questions when they give you a Poisson or binomial distribution
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    (Original post by yaboy)
    its p(x less than to 43.5) Learn the continuity rules



    its wrong
    Ok, would my second one also constitute wrong as well?
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    (Original post by Robbie242)
    Hmm would my less than or equal to be technically wrong or right then, for 43.5, since after all P(x<44) includes 43.5?
    Yeah, it's wrong. because if X=43.5, then when rounded off to a whole number, you get X=44. since X<44 it can't be 44, so you can't include 43.5
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    (Original post by justinawe)
    Yeah, it's wrong. because if X=43.5, then when rounded off to a whole number, you get X=44. since X<44 it can't be 44, so you can't include 43.5
    Cheers I'll make a note of that, is my second example wrong though?
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    (Original post by Robbie242)
    Cheers I'll make a note of that, is my second example wrong though?
    Your second example is correct
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    (Original post by Robbie242)
    Ok, would my second one also constitute wrong as well?
    its correct but I dont like the way you wrote it so im gonna go ahead and say its wrong
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    (Original post by Robbie242)
    Cheers I'll make a note of that, is my second example wrong though?
    Yes, that's correct
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    Cheers everyone, glad I didn't make a mistake like this one the exam day, though it'd be likely to only cost a mark, every little helps
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    (Original post by yaboy)
    its correct but I dont like the way you wrote it so im gonna go ahead and say its wrong
    its correct\not=its wrong
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    (Original post by Robbie242)
    its correct\not=its wrong
    0 = 0+0+0...
    0= (1-1)+(1-1)+(1-1).....
    0= 1+(-1+1)+(-1+1)...
    so 0=1

    why cant

    its correct\not=its wrong[/QUOTE] be true?:cool:
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    so it would be P(X<43.5)? so if it was =to, it would round to 44 which would be wrong. i think i get it now.
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    (Original post by cooldudeman)
    so it would be P(X<43.5)? so if it was =to, it would round to 44 which would be wrong. i think i get it now.
    Precisely, if you follow my second example, although it looks similar it is a very different case, think about each problem logically, and which probability range you'd end up with <
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    (Original post by yaboy)
    0 = 0+0+0...
    0= (1-1)+(1-1)+(1-1).....
    0= 1+(-1+1)+(-1+1)...
    so 0=1

    why cant

    its correct\not=its wrong be true?:cool:
    This is actually false because

    0 = (1-1)+(1-1)+(1-1) ...

    means that there will be an equal number of 1's and -1's, as there's a 1 and a -1 in every bracket

    therefore there will always be a -1 at the end of the sequence
 
 
 
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