Binomial Dist.
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January 30, 2007 at 7:33 am #45977
bbeltusaParticipant@bbeltusa Include @bbeltusa in your post and this person will
be notified via email.Hello
I have a basic doubt regarding the binomial dist. ( when used as an approximation to the normal dist)
The mean is given by ( n*p) and the standard deviation is sqrt(npq).
I always thought the mean and the std. dev. carry the same “units” . In this case , it is different . Am I missing something ?
This doubt cropped up when I was trying to derive the “standard error of the mean” for a proportion sqrt(pq/n)
Thanks in advance to all responses
0January 30, 2007 at 8:01 am #151220
Idiotin math.Participant@Idiotinmath. Include @Idiotinmath. in your post and this person will
be notified via email.Note the four conditions for applying a binomial distribution:2 states (success/failure?),fixed # of trials,trials are independent and probability of success on every trial is the same (use the table for finding the figure for each p,for example p(x=3) equal to .2428 etc,).Don’t mix up ,using sd in the formula as it should only include :sample size,# of defects and the proportion defective (p).
Anyhow the binomial distribution mean and sd (sigma) can be obtained from the following calculations:
the binomial mean=mu=np
the binomial sigma (sd)=Sigma= np(1p) under the square root
ofcourse you have to distinquish between the mean and the sd.
finally the poisson distribution can be an approximation to the binomial when p is equal to or less than 0.1,and the sample size n is fairly large
Good Luck0January 30, 2007 at 9:17 am #151222
bbeltusaParticipant@bbeltusa Include @bbeltusa in your post and this person will
be notified via email.hi idiot!
doesn’t answer my question
bbeltusa
0January 30, 2007 at 1:12 pm #151228
accringtonParticipant@accrington Include @accrington in your post and this person will
be notified via email.You get ‘units’ with continuous data. The binomial disribution is discrete, i.e. there are only two possible outcomes of a single trial, 0 or 1, so there are no units.
When you calculate the mean and the SD of a binomial distribution the ‘units’ are already the same, i.e. there aren’t any, so don’t worry about it.
By the way, the standard error of p is sqrt((p(1 – p)/n)
If you want approximate confidence limits, you can use
p +/ z alpha/2 * s.e.(p)
For 95% C.L.use 1.96*s.e.(p)
Hope this leaves you less doubtful0January 30, 2007 at 1:39 pm #151229
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.Check out your first statement….the binomial is not an approximation of the normal but the other way around. You can use the normal approximation to the binomial. And that only works within a certain range of proportions. Secondly what you are describing as the s.d. isn’t correct. The s.d. for a binomial is sqrt(pq/n) which is what you tried calling std. error of the mean. Please go back and reconfirm all your work.
0January 30, 2007 at 3:31 pm #151244Wow Idiot,I’m impressed. You gave helpful information to someone.Maybe you aren’t as bad as Darth says.
0January 30, 2007 at 3:50 pm #151246
Ken FeldmanParticipant@Darth Include @Darth in your post and this person will
be notified via email.Unfortunately, the original poster didn’t think it was much help.
0January 30, 2007 at 7:23 pm #151269
Idiotinmath.Participant@Idiotinmath. Include @Idiotinmath. in your post and this person will
be notified via email.Thank You Stan.I’m glad to receive your positive comment .
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