You can put this solution on YOUR website! find the z scores for which 97% of the distribution's
area lies between -z and z.
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That leaves 1.00-0.97 = 0.03 in the 2 tails.
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-z is the value with a left tail of (1/2)(0.03) = 0.015
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-z = invNorm(0.015) = -2.1701
Then z = 2.1701
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Cheers,
Stan H.
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