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Binomial Dist.

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  • #45977

    bbeltusa
    Participant

    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
     
     
     
     
     
     

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    #151220

    Idiotin math.
    Participant

    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(1-p) -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  Luck  

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    #151222

    bbeltusa
    Participant

    hi idiot!
    doesn’t answer my question
    bbeltusa
     

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    #151228

    accrington
    Participant

    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 doubtful

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    #151229

    Ken Feldman
    Participant

    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.

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    #151244

    Mikel
    Member

    Wow Idiot,I’m impressed. You gave helpful information to someone.Maybe you aren’t as bad as Darth says.

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    #151246

    Ken Feldman
    Participant

    Unfortunately, the original poster didn’t think it was much help.

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    #151269

    Idiotinmath.
    Participant

    Thank You Stan.I’m  glad  to  receive  your  positive  comment .

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