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Standardized variables

Z – score

In order to evaluate the size of the individual deviation of numeric property $x_{i}$ from the arithmetic mean $\mu$ in units of standard deviation we use a z – score (standard score):

$$z_{i}=\frac{x_{i} – \mu}{\sigma}.$$

It is a relative measure of deviation and it measures ”how many standard deviations property $x_{i}$ deviates from $\mu$”. We can place it on a normal distribution curve.

Example 1:  Your test score is $180$, while the test has a mean of $140$ and standard deviation of $15$. If the distribution is normal, what is your z – score? Explain the meaning of the result.


$$z = \frac{180 – 140}{15} = 2.7$$

In conclusion, the score is $2.7$ standard deviations above the mean.


What are standardized variables?

Z – score $z_{i}$ is joined to the element of the population which has a property $x_{i}= \mu + z_{i}\sigma$. Therefore, z – score is a new numeric variable $Z$. This variable operates on the same population as variable $X$. In other words,

$$Z(s) = \frac{X(s)-\mu}{\sigma}.$$

This means that to every element $s$ of the population $S$ deviation of its property $X(s)$ from $\mu$ is joined. Furthermore, it is expressed in the units of $\sigma$. Variable $Z$ is called standardized variable.

Standardized variables are variables that have a common standard. More precisely, they have an arithmetic mean of $0$ and a standard deviation of $1$.

There are plenty of reasons why variables are standardized. For instance, to assure that all variables contribute equally to a scale when items are added together.


More examples

Example 2:  The average test score on the first colloquium is $50$, while average deviation from the mean is $10$. The average test score on the second colloquium is $90$ and standard deviation is $20$.

If a student has a score of $62$ on the first colloquium and $105$ on the second, what can we conclude about his score?


Since the scale ranges are unknown, we can assume that they are different. Therefore, we can make a conclusion about student’s score only by calculating standardized property:

$$z_{1} = \frac{x_{1} – \mu_{1}}{\sigma_{1}} = \frac{62 – 50}{10} = 1.2$$

$$z_{2} = \frac{x_{2} – \mu_{2}}{\sigma_{2}} = \frac{105 – 90}{20} = 0.75$$

In other words, the score on the first colloquium is for $1.2$ deviations better than average. Furthermore, the score on the second colloquium is for $0.75 \sigma$ better than average.


Example 3:  A group of $100$ men competes in running at $100$ metres and in long jump. In the first discipline is $\mu_{1} = 12.8 \ s$ and $\sigma_{1} = 2 \ s$. In the second discipline is $\mu_{2} = 485 \ cm$ and $\sigma_{2} = 50 \ cm$.

If a person A has a result of $12.2 \ s$ in the first discipline and $490 \ cm$ in the second discipline, and person B runs for $13 \ s \ 100 \ m$ and jumps $580 \ cm$ in the distance, who is more successful?


Since the measuring units of the disciplines are different, we have to use a standardized variable. Therefore,

$$z_{A1} = \frac{12.2 – 12.8}{2} = – 0.3,$$

which is in fact $+0.3$ because it is for $0.3 \sigma$ faster than average. Furthermore,

$$z_{B1} = \frac{13 – 12.8}{2} = 0.1,$$

which is in fact $-0 .1$ because it is for $0.1 \sigma$ slower than average.

$$z_{A2} = \frac{490 – 485}{50} = 0.1$$

$$z_{B2} = \frac{580 – 485}{50} = 1.9$$

The average of a person A is $\frac{0.3 + 0.1}{2} = 0.2$ (overall he is approximately average). The average of a person B is $\frac{-0.1 + 1.9}{2} = 0.9$ (overall he is above average).