Question 933223
mean of 1200 and standard deviation of 100. 
{{{z = blue(x - 1200)/blue(100)}}}  
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  Applicants who score in the 90% or higher will be admitted.
z = invNorm(.90) = {{{blue(x - 1200)/blue(100)}}}
invNorm(.90) = 1.2816 (find Using table 0r Calculator 0r Excel function)
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100(1.2816) + 1200 = 1328.16.   1329 score need for admittance
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For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
90% of the Area under the standard normal curve is to the left of z = 1.2816 
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))}}}