literal yes no answer EDIT[probably best not to bother reading it :P]

[Bulk]

My maths book is about 3 steps away, my homework is under my laptop and my laptop is on BFM

rather than leave my comfy chair to find the answer...

when answering a question about probability of an outcome from a binomial distribution of events (poorly worded) ie   X~B(n,P)

when it asks for the mean value, does that simply mean the EXPECTED value, ie np
or the literal mean of outcomes, ie (P(x=0) + P(x=1)+...+P(x=n))/n   ?
or are they both the same thing?

I am then assuming that to find the variance would be the same as a random distribution to work out?

ie e(x^2)-(e(x)^2)

which would simplify into n(p-p^2)   ->     np(1-p)   

and 1-p is the same as q

therefore var = npq, which is good because that's one of the next questions proving that that is the case

just realised i kinda answered my own question writing this question.

I'm going to post it anyway because i am so pleased with myself for actual getting some maths right yay :D

step aside Three60 Pass me that there maths torch

[Goalie]

My maths book is about 3 steps away, my homework is under my laptop and my laptop is on BFM

rather than leave my comfy chair to find the answer...

when answering a question about probability of an outcome from a binomial distribution of events (poorly worded) ie   X~B(n,P)

when it asks for the mean value, does that simply mean the EXPECTED value, ie np
or the literal mean of outcomes, ie (P(x=0) + P(x=1)+...+P(x=n))/n   ?
or are they both the same thing?

I am then assuming that to find the variance would be the same as a random distribution to work out?

ie e(x^2)-(e(x)^2)

which would simplify into n(p-p^2)   ->     np(1-p)   

and 1-p is the same as q

therefore var = npq, which is good because that's one of the next questions proving that that is the case

just realised i kinda answered my own question writing this question.

I'm going to post it anyway because i am so pleased with myself for actual getting some maths right yay :D

step aside Three60 Pass me that there maths torch

I believe that for a binomial distribution the expected value happens to be the same as the mean value.  If you try and work it out, you should find that they are indeed the same.

I believe your variance is correct, either way you write it.  I think most teachers would prefer np(1-p) because it has fewer variables.

[jim360]

One way to look at things is that, in a probability distribution, you talk about the expectation, because you are expecting some result to be the single most likely outcome, but in a statistical distribution you talk about the mean because there is some result that does have more cases than any other. So the mean and expectation are the same result, but in different situations. One is statistics (when you have got the results already) and the other is probability (when you're predicting the results).

Well done on working it out for yourself.

[Bulk]

tty, yeah i worked out the mean and it was the same

and thnx 360, that does clear it up a bit in terms of the language used in questions, i just can't believe i actually did maths correctly, first time this has happened all year