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Q: Which is better a score of 92 on a test with a mean of 71 and a standard deviation of 15 or a score of 688 on a test with a mean of 493 and a standard deviation of 150?

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No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.

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A large standard deviation means that the data were spread out. It is relative whether or not you consider a standard deviation to be "large" or not, but a larger standard deviation always means that the data is more spread out than a smaller one. For example, if the mean was 60, and the standard deviation was 1, then this is a small standard deviation. The data is not spread out and a score of 74 or 43 would be highly unlikely, almost impossible. However, if the mean was 60 and the standard deviation was 20, then this would be a large standard deviation. The data is spread out more and a score of 74 or 43 wouldn't be odd or unusual at all.

z-score of a value=(that value minus the mean)/(standard deviation). So if a value has a negative z-score, then it is below the mean.

Information is not sufficient to find mean deviation and standard deviation.

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A negative Z-Score corresponds to a negative standard deviation, i.e. an observation that is less than the mean, when the standard deviation is normalized so that the standard deviation is zero when the mean is zero.

A z-score cannot help calculate standard deviation. In fact the very point of z-scores is to remove any contribution from the mean or standard deviation.

No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.

.The test has a mean, or average, standard score of 100 and a standard deviation of 16 (subtests have a mean of 50 and a standard deviation of 8). The standard deviation indicates how far above or below the norm the subject's score is.

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The standard deviation.z-score of a value=(that value minus the mean)/(standard deviation)

mean is 218 with a standard deviation of 16

z-score of a value=(that value minus the mean)/(standard deviation)

z-score of a value=(that value minus the mean)/(standard deviation)

z-score of a value=(that value minus the mean)/(standard deviation)

A z-score requires the mean and standard deviation (or standard error). There is, therefore, not enough information to answer the question.

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