Hi,

using TI function normalcdf(smaller, larger, µ, σ):

 mean of 575mm and a standard deviation of 5mm 

P(575 ≤ x ≤ 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 

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