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Normal distribution

  • Thread starter TyErd
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Homework Statement


I've attached the question.


Homework Equations


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


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
 

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  • #2
Ray Vickson
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Homework Statement


I've attached the question.


Homework Equations


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


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
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:

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