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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.
Simple! The average deviation for any data set is zero - by definition.
There are no s's in a standard deck of cards, so the probability of selecting any s's, in any sequence of draws, in any strategy of replacement is exactly zero.
Yes, the mean (and median and mode) is the 50th percentile of any normal distribution.
You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).