Respuesta :

Using the normal distribution, the value that separates the highest 70% of the scores in a normal distribution from the lowest 30% is:

z = -0.52.

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 value that separates the highest 70% of the scores in a normal distribution from the lowest 30% is is the 30th percentile, which is Z with a p-value of -0.3, hence z = -0.52.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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