Question 857004
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Hi,
using TI function normalcdf(smaller, larger, µ, &#963;):
 mean of 575mm and a standard deviation of 5mm 
P(575 &#8804; x &#8804; 579)= normalcdf(575, 579, 575, 5)   0r using z-value NORMSDIST(.8) - .5
P(x > 570) = normalcdf(570, 9999, 575, 5)    0r using z-value 1 - NORMSDIST(-1.0)
P(x < 580) = normalcdf(-9999, 580, 575, 5)   0r using z-value NORMSDIST(1.0)

2.80% of the tyre will have diameter above what value
z = |NORMSINV(.20/2)| = 1.2816
1.2816 = (X - 575)/5
5*1.2816 + 575 = 581.4mm