score of 92
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
1.50
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.
If the Z Score of a test is equal to zero then the raw score of the test is equal to the mean. Z Score = (Raw Score - Mean Score) / Standard Deviation
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 standard deviation.z-score of a value=(that value minus the mean)/(standard deviation)
78
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.
1.50
Bob scored 300 or 700.