When comparing data from different distributions, what is the benefit of transforming
data from these distributions to conform to the standard distribution? What role do z-
scores play in this transformation of data from multiple distributions to the standard
normal distribution? What is the relationship between z-scores and percentages? In
your opinion, does one do a better job of representing the proportion of the area under
the standard curve? Give an example that illustrates your answer.
Z-Scores
Standard distribution is very important because most psychological variables have a
normal distribution. In psychology, examples of variables researchers meet include memory,
extroversion, and job satisfaction. Though these variables are not exactly normally
distributed, they are close to that. Again, standard distribution is very essential for
statisticians as it makes their work easier. With a standard distribution, deriving different
statistical tests is straightforward.
One of the key features of standard distribution is Z-scores. This figure is helpful
when a statistician wants to compare between any score and the mean. Sometimes, the
standard distribution is referred to as the z-distribution. (Adeyemi, 2011)
A Z-score is just another way of presenting standard deviation. If a Z-score is 3, this
means 3 standard deviations below and above the mean. This score has a relationship with
percentages when the empirical rule is considered. In a normal distribution, 1, 2 and 3
standard deviations from the mean approximate with 68%, 95%, and 99% of the values,
respectively. (Langley, 2013)
Percentages and Z-scores are effective ways of representing the proportion of area
under a standard curve. The main advantage of z-scores is that they are very effective in
Z-Scores 2
comparison of raw scores. The following example shows how z-scores are useful in
determination of relative performance, something not achievable through percentages.
Consider the performance of math in class A and B.
Class A Class B
Mean 70 84
Raw scores 80 88
Standard Deviation 5.7 3.5
Comparison can be made between student A and B to check on their relative
performance. The Z-score formula gives,
For A,
Z= (80-70) /5.7=1.75
For B,
Z= (88-84)/3.5=1.14
Z-Scores 3
From these results, the relative performance of student A is better than that for student
B.
References
Adeyemi, T. O. (2011). The Effective Use of Standard Scores for Research In Educational
Management. Research Journal of Mathematics and Statistics,3(3), 91-96.