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The thickness of a part is to have an upper specification of 0.925 and a lower specification of 0.870 mm. The average of the process is currently 0.917 with a standard deviation of 0.005. Determine the percentage of product above 0.90 mm.

### Solution

Given u = 0.917 and s = 0.005

P(X > 0.9) = P(Z > (0.9 – u) / s)

= P(Z > (0.9 &#████████; █████████████) ██ ████████)

= ███(███ &███; ███████)

= ██ &#███████; █(█ &████; ████████) = ██ &#█████████; ████████████ = ███████████████ = ████████████████