WEBVTT

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Here is the problem: Find P(negative 1.96 less than Z less than 1.96) 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(negative 1.96 less than Z less than 1.96) approximately 0.950004 (95 percent).

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