mean of 5.82 standard deviation 0.03 millimeters find e two diameters that separate the top 9 and the bottom 9%. Find answer to nearest hundredth if necessary
For the top 9% we need the 91st percentile, i.e. the inverse normal at p = 0.91:
norminv ( p=0.91, mean = 5.82, sd = 0.03 ) = 5.86
And for the bottom 9%:
norminv ( p=0.09, mean = 5.82, sd = 0.03 ) = 5.78
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