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The standard normal table tells us the area under a normal curve to the left of a number z.

The tables usually give only the positive value since one can use symmetry to find the corresponding negative values.

The middle 60 percent leaves 20 percent on either side. So we want the z scores that correspond to that 80 percentile which is .804. Therefore the values are are between z scores of -.804 and .804

* * * * *

I make it -0.8416 to 0.8416

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Q: For a normal distribution find the z-scores values that separate the middle 60 percent of the distribution for the 40 percent in the tails?
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