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Find+the+z-scores+that+separate+the+middle+69%+of+the+distribution+from+the+area+in+the+tails+of+the+standard+normal+distribution

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Answer and Explanation:

1 - 0.69/ 2 = 0.155

0.155 = -1.02

(-1.02, 1.02) = z

The z -score separates the middle 69% of the distribution from the area in the tails of the standard normal distribution is  (-1.02, 1.02).

What is a z- score?

Z- score refers to how much a given value differs from the standard deviation.

The z -score separates the middle 69% of the distribution from the area in the tails of the standard normal distribution.

α = 1 - 0.69/ 2

α =  0.155

0.155 = -1.02

z = (-1.02, 1.02)

Learn more about standard deviation:

https://brainly.com/question/11448982

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