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Find the z-scores that separate the middle 78% of the distribution from the area in the tails of the standard normal distrubution

Respuesta :

Using the normal distribution, it is found that the z-scores are [tex]Z = \pm 1.227[/tex].

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we 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 78% is between the 11th and the 89th percentile. Looking at the z-table, the z-scores are [tex]Z = \pm 1.227[/tex].

More can be learned about the normal distribution at https://brainly.com/question/24663213

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