Standard deviation is the spread of the data. If each score has 7 added, this would not affect the spread of the data - it would be just as evenly spaced or clumped up, but 7 greater. The only thing that would affect the spread is multiplying every data point by 0.9. This makes distances between the data points 0.9 times as big, and thus makes the standard deviation 0.9 times as big. The standard deviation was 5.6, and so now is 5.6x0.9 = 5.04
The standardised score decreases.
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
score of 92
The absolute value of the standard score becomes smaller.
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.
When you don't have the population standard deviation, but do have the sample standard deviation. The Z score will be better to do as long as it is possible to do it.
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.
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.
The standard deviation.z-score of a value=(that value minus the mean)/(standard deviation)