You can put this solution on YOUR website! Find the value of z such that approximately 10.26% of the distribution lies between it and the mean.
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Find the z-value with a left tail of 0.50+0.1026 = 0.6026
invNorm(0.6026) =0.2601
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Find the z-value with a left tail of 0.50-0.1026 = 0.3974
invNorm(0.3974) = -0.2601
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Cheers,
Stan H.