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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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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