Question

For a standard normal​ curve, find the​ z-score that separates the bottom​ 70% from the top​ 30%.

Answer 1

Using the normal distribution, the z-score that separates the bottom​ 70% from the top​ 30% is z = 0.525.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The z-score that separates the bottom​ 70% from the top​ 30% is z with a p-value of 0.7, hence z = 0.525.

More can be learned about the normal distribution at https://brainly.com/question/24537145

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