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In a normal distribution, what z-score is the cut-off score that separates the highest 29.8% of scores from the lowest 70.2% of scores?

User Latox
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Final answer:

The z-score that separates the highest 29.8% of scores from the lowest 70.2% in a normal distribution is approximately 0.55. This value is found using a z-table, which provides the cumulative area under the curve for standard normal distributions.

Step-by-step explanation:

The question asks about the z-score that separates the highest 29.8% of scores from the lowest 70.2% of scores in a normal distribution. Using the empirical rule also known as the 68-95-99.7 rule, we understand that z-scores are standardized ways of describing a value's relationship to the mean in a normal distribution. However, to find a specific z-score that corresponds to an exact cumulative probability, we utilize a z-table.

In the context of the question, we are interested in finding a z-score that corresponds to the cumulative area to the left which in this case would be 70.2%. By consulting a z-table or using statistical software that provides critical values of z, we are looking for the z-score that captures 70.2% of the data to the left. This approximately equates to a z-score of approximately 0.55. This value can be validated by looking up the mean to z area in a z-table for the positive z-score and observing that it closely corresponds with the required cumulative probability.

The question does not require us to translate this z-score to a specific normal distribution with a given mean and standard deviation, but generally, you would use the formula z = (x - µ) / σ where µ is the mean and σ is the standard deviation, and x represents the raw score.

User Diesersamat
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