S2 question

how do you find something like P(10<=X<=13) from the tables and say if it was for poisson, not that it matters.

how would you do it generally?

Reply 1

Joshmeid

Original post by cooldudeman

how do you find something like P(10<=X<=13) from the tables and say if it was for poisson, not that it matters.

how would you do it generally?

P(X \leq 13) - P(X \leq 9)

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Reply 2

metaltron

Original post by cooldudeman

how do you find something like P(10<=X<=13) from the tables and say if it was for poisson, not that it matters.

how would you do it generally?

Firstly you need to be aware of whether the distribution is discrete or continuous. For a discrete distribution such as Poisson:

P(10<=X<=13) = P(X<=13) - P(X<=9). If you don't get it immediately you need to understand why this is. You can post back if you like.

For a continuous distribution, such as the normal distribution, it does not matter if the inequalities are strict or not, since P(X=10) = 0. Therefore you want:

P(10<=X<=13) = P(X<=13) - P(X<=10)

Reply 3

cooldudeman

Original post by Joshmeid

P(X \leq 13) - P(X \leq 9)

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what if wit was P(a<=X<=b)? P(X<=b) - P(X<= a-1)?

(edited 11 years ago)

Reply 4

metaltron

Original post by cooldudeman

what if wit was P(a<=X<=b)? P(X<=b) - P(X<= a-1)?

You should read my other post as well. What you have written is correct for a discrete distribution such as Poisson as it can only take integer values. You'll also want to notice:

P(10<X \leq 13) = P(X \leq 13) - P(X \leq 10)

If you understand why you will be able to negotiate any question requiring use of the tables.

Reply 5

cooldudeman

Original post by metaltron

You should read my other post as well. What you have written is correct for a discrete distribution such as Poisson as it can only take integer values. You'll also want to notice:

P(10<X \leq 13) = P(X \leq 13) - P(X \leq 10)

If you understand why you will be able to negotiate any question requiring use of the tables.

sorry, just read your posts now.
ok so the thing i wrote works for poisson and binomial distribution only since they are discrete?
so for the one you wrote the first part is, 10<X so X>10 so X>=11, 11<=X so P(11<=X<=13) so P(X<=13) - P(X<=11-1)?

if it was one of the ones you wrote for normal distribution, and the inequalities were not strict, would you just turn them to strict, and apply continuity correction and then the same method? like minusing 1 from the first part of the big inequality?

Reply 6

metaltron

Original post by cooldudeman

You are correct for my previous example. For continuous distributions:

P(10 \leq X \leq 13) = P(10 < X < 13) = P(X \leq 13) - P(X \leq 10)

This is because you can include all values not just integers. Also:

P(X=10) = P(X=13) = 0

Reply 7

cooldudeman

Original post by metaltron

You are correct for my previous example. For continuous distributions:

P(10 \leq X \leq 13) = P(10 < X < 13) = P(X \leq 13) - P(X \leq 10)

This is because you can include all values not just integers. Also:

P(X=10) = P(X=13) = 0

ok thanks

Reply 8

cooldudeman

Original post by metaltron

You are correct for my previous example. For continuous distributions:

P(10 \leq X \leq 13) = P(10 < X < 13) = P(X \leq 13) - P(X \leq 10)

This is because you can include all values not just integers. Also:

P(X=10) = P(X=13) = 0

Sorry to bring this topic up again but how do you deal with it for Z so normal distribution. Like
P (A <Z <-B) or if a was negative and b wasnt

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Reply 9

metaltron

Original post by cooldudeman

Sorry to bring this topic up again but how do you deal with it for Z so normal distribution. Like
P (A <Z <-B) or if a was negative and b wasnt

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First you standardise the normal distribution. Then you will treat it as a continuous distribution:

P(a > X > -b) = P(X \leq a) - P(X \leq -b) = P(X \leq a) - (1 - P(X \leq b))

where a and b are positive.

Reply 10

cooldudeman

Original post by metaltron

First you standardise the normal distribution. Then you will treat it as a continuous distribution:

P(a > X > -b) = P(X \leq a) - P(X \leq -b) = P(X \leq a) - (1 - P(X \leq b))

where a and b are positive.

Oh yeah I get tgat. I was being stupid
Can you help with something else since no one is replying to my other thread. With continuous variables like rectangular and the other one thats in S2, how do you deal with conditional probability?
I imagine the standard formulas like P (X> 8 | X> 3)= P (X> 8) ÷ P (X> 3)
and this is when bith inequalities are more than. Is this true for all cases. I really cant do questions like these.

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