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Find the area under the standard normal distribution curve between z = –2.05 and z = 2.05.
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\[ \int_{-2.05}^{2.05} \frac{e^{-\frac{x^2}{2}}}{\sqrt {2 \pi }} \, dx=0.959636 \] You can also find it from a z-table.
thank you
From the tables you get P[ Z< 2.05]= .9798 P[ -2.05 < Z < 2.05]= 2P[ Z< 2.05]-1 = 2(.9798)-1=.9596
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