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The normal approximation to the binomial

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I've attached the question


    2. Relevant equations
    Pr(X<=x)= (x + 0.5 - n*p) / sqrt(n*p*(1-p))


    3. The attempt at a solution
    okay so n=1150, p=0.02 , Pr(X<23)

    =23 + 0.5 - 1150(0.02) / sqrt(1150*0.02*0.98)
    =0.105316

    is that bit right so far. Because it is less than i thought x might've been 22 instead of 23 so when i did that it equaled -0.105316

    im not sure what do after that. Am i suppose to find that value in the normal distribution tables or something?
     

    Attached Files:

  2. jcsd
  3. Apr 16, 2012 #2

    Ray Vickson

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    Your first equation makes no sense. Perhaps you meant to write
    [tex]P(X \leq x) \approx P\left( Z \leq \frac{x + 1/2 - n p)}{\sqrt{n p (1-p)}} \right)? [/tex]
    For n = 1150, p = 0.02 and x = 23, we have [itex] P(Z \leq 0.105316),[/itex] which is similar to what you said---although I am not sure, from what you wrote, that you really understand this. Of course, the question said "less than 23", which means <= 22, so you ought to use x = 22 in the formula. At this point you need to find the probability value from normal tables, or a spreadsheet, or by pressing a button on some brands of hand-held calculators.

    RGV
     
    Last edited: Apr 16, 2012
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