3. Descriptive Statistics: Central Tendency and Dispersion

3.3 z-score / z-transformation

The z-score is the result of transformation of data that converts a dataset of x values, \{ x_{i} \}, that has a mean of \bar{x} and standard deviation s to a set of z values \{z_{i}\} that has a mean of \bar{z} = 0 and a standard deviation of s_{z} = 1. It will be very useful when we need to compute probabilities associated with normal distributions. The z-transformation is defined by

    \[z = \frac{x - \bar{x}}{s} \mbox{\ \ \ \ \ (sample)}\]

    \[z = \frac{x-\mu}{\sigma} \mbox{\ \ \ \ \ (population)}\]

Example 3.12 : Find the z-scores of the data given in the left column of the table below.

Data x_{i} x_{i}^{2} z-score, z_{i}
18 324 (18-9.9)/6.2 = 1.3
15 225 (15-9.9)/6.2 = 0.8
12 144 (12-9.9)/6.2 = 0.3
6 36 (6-9.9)/6.2 = -0.6
8 64 (8-9.9)/6.2 = -0.3
2 4 (2-9.9)/6.2 = -1.3
3 9 (3-9.9)/6.2 = -1.1
5 25 (5-9.5)/6.2 = -0.8
20 400 (20-9.5)/6.2 = -1.7
10 100 (10-9.5)/6.2 = 0.1
\sum x_{i}=99 \sum x_{i}^{2}=1331

The dataset size is n=10. You need to compute the z-score for each data value separately. To do the calculation, both \bar{x} and s are needed. So in addition to the sum of the data, \sum x, we also need the sum of the x^{2} values. The work of getting those sums is shown in the table above. With the x and x^{2} sums we get

    \[\bar{x} = \frac{\sum x_{i}}{n} = \frac{99}{10} = 9.9\]

and

    \[s^{2} &=& \frac{\sum x_{i}^{2} - [\frac{(\sum x_i)^2}{n}]}{n-1} =\frac{1331 - [\frac{99^2}{10}]}{9}=\frac{1331-980.1}{9}=39.0\]

and s = \sqrt{39} = 6.2.

Using these values for \bar{x} and s in the third column of the table above, compute the z-scores as shown. If we had computed the z-scores more accurately, they would add up to zero, \sum z_{i} = 0 (the mean of the z-scores is zero.)

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