Question 856911
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Hi,
mean 800 miles and standard deviation 90 miles. 
Percentage who preform poorly:  {{{z=(600-800)/90}}} = -2.222, NORMSDIST(-2.222) = .0131
a) 200*.0131 = 2.62  0r ~3
b)|NORMSINV(.01)| = 2.3263 = {{{(X-800)/90}}}, X = 1009.367 = 1009.4
For the normal distribution: 
one  standard deviation from the mean accounts for about 68.2% of the set 
two standard deviations from the mean account for about 95.4%
and three standard deviations from the mean account for about 99.7%.
Important to Understand z -values as they relate to the Standard Normal curve:
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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))}}}