Question 675624
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Hi
&#956; = 5.338  ,  &#963; = .068 
vending machine is configured to accept those coins with weights between 5.208 g and 5.468 g.
<font color="red">SAMPLE</font>: 250 different coins
P(5.208 < x < 5.468) = P({{{z <= (5.468-5.338)/(.068/sqrt(250))}}}  <font color="red">-</font>   P({{{z <= (5.208-5.338)/(.068/sqrt(250))}}}
find this difference and multiply times 250 (rounding UP to the nearest integer )
TI normalcdf(z)  gives the portion of the area under the standard normal curve to the LEFT of the z-value entered.
Actually can do in one step for <u>between two z-values</u>: normalcdf(smaller z, larger z)
Important to Understand z -values as they relate to the Standard Normal curve:
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
For ex: normalcdf(2) - normalcdf(-2) would give the portion of the area under the curve between those two z-values 
Note: z = 0 , 50% of the area under the curve is to the left and 50%  to the right
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}