Find the area under the standard normal distribution curve between z=-1 and z=-3
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N(-Z)=1-N(Z), so N(-1)=1-N(1) and N(-3)=1-N(3). So the difference is 1-N(1)-(1-N(3)) = 1-N(1)-1+N(3) = N(3)-N(1).

Use the table to find these values: 0.9987-0.8413=0.1574.

by Top Rated User (1.1m points)
0.8400
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