20 percent
The answer depends on what the standard deviation is.
To find the mean from a raw score, z-score, and standard deviation, you can use the formula: ( \text{Raw Score} = \text{Mean} + (z \times \text{Standard Deviation}) ). Rearranging this gives you the mean: ( \text{Mean} = \text{Raw Score} - (z \times \text{Standard Deviation}) ). Simply substitute the values of the raw score, z-score, and standard deviation into this formula to calculate the mean.
The standardised score decreases.
2 standard deviation's below the mean
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
z-score of a value=(that value minus the mean)/(standard deviation). So a z-score of -1.5 means that a value is 1.5 standard deviations below the mean.
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.
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.