Question 933215
mean of 1400 and a standard deviation of 200. {{{z = blue(x - 1400)/blue(200)}}}    
 A college will only take applicants in the top 2.5% or higher. 
 What is the lowest score they would need on the entrance exam
 200z + 1400 = x
z = invNorm(.975).
 invNorm(.975) = 1.96  can be found from a table or Using calculator 0r Excel  Function
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 200(invNorm(.975) + 1400 = 200(1.96) + 1400 = 1792, lowest score
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Note: 1 - .025 = .975  (z-value found...97.5% of the Area Under NOrmal curve to the left of that value.
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For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
97.5% of the Area under the standard normal curve is to the left of z = 1.96  
Note: z = 0 (x value: the mean) 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))}}}
stattrek.com a great reference for statistics