WEBVTT

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Here is the problem: Find P(0 less than Z less than 1.5) for a standard normal variable.

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P(z₁ less than Z less than z₂): area between.

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Find the left areas of both z-scores.

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Subtract.

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So the answer is P(0 less than Z less than 1.5) approximately 0.433193 (43.32 percent).

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43.32 percent of a normal distribution lies between these two z-scores.