The variance.
The standardised score decreases.
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.
standard deviation
The sum of the differences between each score and the mean is always zero. This is because the mean is the "center" of the data and any deviation from the mean in one direction is offset by an equal deviation in the opposite direction. This property is essential in understanding the concept of the mean as a measure of central tendency.
The sum of deviations from the mean, for any set of numbers, is always zero. For this reason it is quite useless.
It depends on what the deviation is from. Also, the sum of the deviations from any fixed number will always be zero.
No. The square of a number is always positive, so the sum of several of them must also be positive.
Difference (deviation) from the mean.
the mean %100
IQ is distributed normally, with a mean of 100 and a standard deviation of 15. The z-score of 100 is therefore:(value-mean)*standard deviation= (100-100)*15= 0More generally, a raw score that is equivalent to the mean of a normal distribution will always have a z-score of 0.
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The variance.
score of 92
T-score is used when you don't have the population standard deviation and must use the sample standard deviation as a substitute.
No. The variance of any distribution is the sum of the squares of the deviation from the mean. Since the square of the deviation is essentially the square of the absolute value of the deviation, that means the variance is always positive, be the distribution normal, poisson, or other.