Solve: Normal curve area between 0 and Z is equal to "\\int_0^Z" "\\frac{1}{\\sqrt{2\\pi}}e^\\frac{-x^2}{2}dx=\\Phi(Z)-\\Phi(0)=\\Phi(Z)," where "\\Phi" is Laplas function.
We have: "\\int_0^Z \\frac{1}{\\sqrt{2\\pi}}e^\\frac{-x^2}{2}dx" "= 0.4332=\\Phi(Z)". Using the table of values of Laplas function we get: Z = 1.5
Answer: Z=1.5
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