You can put this solution on YOUR website! Find the value of z such that approximately 16.28% of the distribution lies between it and the mean.
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Draw the picture.
Put 16.28% below the mean and above some z-value.
That z-value has a (0.5000-0.1628) = 0.3372 left tail.
Find that z-value:
invNorm(0.3372) = -0.4201
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Cheers,
Stan H.