3.3: z-score / z-transformation
- Page ID
- 51790
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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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