**Standard or typified scores are a method of comparing the relative positions of two or more items with respect to the set of observations.**

In other words, the standardized scores return the number of standard deviations that the score x _{i} deviates from the mean.

Mathematically, let x _{i be} the _{i} element of a variable X with mean and standard deviation S. Then, the standardized score of this element i is:

Standardized scores allow you to compare items from different variables and different units of measurement as long as the properties are met.

### Properties

Standard scores do not have units of measure. The units of the numerator cancel out with the units of the denominator. Given this property, the standardized score is also called the standard score.

The absolute value of the score is the number of standard deviations that separate the item from the mean value of the variable where it belongs. Then:

If we consider the sign of the standardized scores, we can establish the position of the element with respect to the mean of the variable.

- Z
_{i}> 0: element*i*is above the mean = element i is to the right of the mean. - Z
_{i}<0: element*i*is below the mean = element i is to the left of the mean.

The standardized scores of all the elements construct a new variable named z _{i} .

This variable z _{i} is obtained from the subtraction (xi – X _{mean} ) and the scale change with the division of the standard deviation (S).

Typification is characterized by having mean 0 and variance 1.

- The mean of all standardized scores is 0.
- The variance of all standardized scores is 1.

### Applications

In statistics and econometrics tables are used probability distribution to find the probability *typified* take an observation given the distribution function follow the variable.

## Practical example

We have two ski resorts A and B in which skiers can do alpine skiing (Alpine) or Nordic skiing (Nordic). We will study which activity is most popular in each ski resort depending on the number of skiers who perform each activity.

Elements | ||||

Seasons | Half | Dev. Standard | Alpine | Nordic |

TO | 96 | 2.6 | 112 | 52 |

B | 22 | 4 | 24 | 41 |

We calculate the standardized scores:

We build the results matrix:

Standardized scores | ||

Seasons | Alpine | Nordic |

TO | 6.1538 | -16,923 |

B | 0.5 | 4.75 |

As a result we have:

Alpine skiing is more popular than Nordic skiing in ski resort A because:

Z _{A, Alpine} > 0, Z _{A, Nordic} <0 and Z _{A, Alpine} > Z _{A, Nordic.}

Nordic skiing is more popular than alpine skiing in ski resort B because

Z _{B, Nordic} > Z _{B, Alpine} with both greater than zero.

Above average:

Z _{A, Alpine} > 0, Z _{B, Alpine} > 0 and Z _{B, Nordic} > 0

Below average:

Z _{A, Nordic} <0