Find the z-score boundaries that separate a normaldistribution as described in each of the following.
a. The middle 20% from the 80% in the tails
b. The middle 25% from the 75% in the tails
c. The middle 70% from the 30% in the tails
d. The middle 90% from the 10% in the tails
Compute the z – score boundaries for the middle 20% from the 80% in the tails.
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From the given information, the proportion value 0.2000 in the center and 0.8000 divided equally between the two tails. Here, 0.2000 in the center can be divided in half with exactly 0.1000 to the right of the mean and exactly 0.1000 to the left.
From appendix B, table B.1, column D, the 0.1000 is close to 0.987 which corresponds to the z value 0.25 in the column A.
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Thus, the z – score boundaries for the middle 20% from the 80% in the tails is – 0.25 and 0.25.
The boundary values are shown below:
step: 2 of…
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