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Normal distribution

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    I've attached the question.


    2. Relevant equations
    μ=np and σ=√(np(1-p))
    Im using the normal distribution function on my calculator to avoid using z tables


    3. The attempt at a solution
    The mean I calculated it to be 160/4 = 40. standard deviation sqrt(160*1/4*3/4)=5.47723
    The question is asking for at least 40(greater or equal to 40) that will choose biscuits but because this is a normal distribution it will be the same as greater than 40 so if I use an X value of 40 the probabilities should just equal 0.5000 right? however the answer is 0.5364
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    You have not used the so-called "continuity correction" or "1/2 correction". If we approximate the binomial B(n,p) by the normal distribution when n is "large" but not really huge, it is better to apply the normal to points half-way between. In this case, it means the following. For the exact Binomial (on 0,1,2,...,160), the event {X ≥ 40} is the same event as {X ≥ 39.5}, because X can take only integer values. If you approximate P{X ≥ 40} by P{Normal ≥ 39.5} you will get a much more accurate answer. Here are the three results:
    (1) Exact binomial P{X ≥ 40} = 0.53031930 ≈ 0.53
    (2) Normal P{X ≥ 40} = 0.50
    (3) Normal P{X ≥ 39.5} = 0.53636776 ≈ 0.54

    RGV
     
    Last edited: Aug 26, 2012
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