Yes.
However, a z-score greater than 3 implies either that the outcome is rare (prob = 0.135%) or that some of the assumptions regarding the variable are incorrect. This could be the distribution itself or its parameters.
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z = 3
T-scores and z-scores measure the deviation from normal. The normal for T-score is 50 with standard deviation of 10. if the score on t-score is more than 50, it means that the person scored above normal (average), and vise versa. The normal for Z-score is 0. If Z-score is above 0, then it means that person scored above normal (average), and vise versa.
The z-score is 0.84. In the related link, look in the body of the table for .3 area (.2995 closest) and it yields the .84 z-score.
It is 0.017864
z = (x - μ) / σ is the formula where x is the raw score and z is the z-score. μ and σ are the mean and standard deviations and must be known numbers. Multiply both sides by σ zσ = x-μ Add μ to both sides μ + zσ = x x = μ + zσ You calculate the raw score x , given the z-score, μ and σ by using the above formula.