Question 1177196
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Hi

!.  {{{SE = sigma/sqrt(n)}}} = 20
i. {{{n = (sigma/SE)^2}}} = (500/20)^2 = 625

ii. ME ={{{ z*sigma/sqrt(n)}}} =25
   25 = z*20
   z = 25/20 = 1.25  and P (z ≤  1.25) = .8944

2a.  {{{z =blue (x - 14)/blue(4/sqrt(50))}}} 
       1/SE = z =  1.7678  and P(z ≤  1.7678) = .9615
The central limit theorem states that whenever a random sample of size n is taken from any distribution <u>with mean and variance</u>, then the sample mean will be approximately normally distributed with mean and variance
 We are proceeding as if the distribution of this sample(size 50)..its mean will be approximately normally distributed.

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