The z score that separate the middle 56% of the distribution from the area in the tails of the standard normal distribution is ±0.77.
Given that the z score separates the 56% of the distribution from the area in the tails of the standard normal distribution.
In a normal distribution in with mean μ and standard deviation σ, the z score of a measure X is as under:
Z=(X-μ)/σ
It is used to measure how many standard deviations the measure is from the mean.
After finding the z score we have to look at the z score table and find the p value associated with this z score, which is the percentile of X.
The normal distribution is symmetric which means that the middle 56% is between the 11th and 67th percentile. Looking at the z table the z scores are Z=±0.77.
Hence the z score that separate the middle 56% of the distribution from the area in the tails of the standard normal distribution is ±0.77.
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