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S2 approximations

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...
(edited 11 years ago)
Reply 1
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<43.5)P(x<43.5) as that is the next one down by continuity correction.
However if it was P(x44)P(x\leq 44) then by continuity correction = P(x<44.5)P(x<44.5)
(edited 11 years ago)
Reply 2
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.
Reply 3
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?
Reply 4
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
(edited 11 years ago)
Reply 5
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?
Reply 6
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
Reply 7
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?
Reply 8
Original post by Robbie242
Cheers I'll make a note of that, is my second example wrong though?


Your second example is correct :smile:
Reply 9
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
Original post by Robbie242
Cheers I'll make a note of that, is my second example wrong though?

Yes, that's correct
Reply 11
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
Reply 12
Original post by yaboy
its correct but I dont like the way you wrote it so im gonna go ahead and say its wrong

itscorrectitswrongits correct\not=its wrong
Reply 13
Original post by Robbie242
itscorrectitswrongits 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

itscorrectitswrongits correct\not=its wrong be true?:cool:
Reply 14
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.
Reply 15
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 <
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

itscorrectitswrongits 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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