Instruction

1

In order to build the interval variation series, it is first necessary to select the optimal number of intervals and set the length of each of them. In this case, note that the length of the interval should be constant, as in the analysis of variational series compare frequencies from different groups. The optimal number of groups should be selected to reflect the variety of signs together, however, their natural distribution, and also to eliminate distortion together with random fluctuations of frequencies. Note that if the groups is too small, will not be visible distribution pattern, and conversely, if too many random leaps of population units will distort the number distribution.

2

To determine the number of groups in the variational row, use the formula Sterides:

h = 1 + 3,322 x ln(n), where

h – the number of groups in the variational row;

n is the population size.

If the resulting value will be fractional, then the value of the step value of the interval, take the nearest integer.

h = 1 + 3,322 x ln(n), where

h – the number of groups in the variational row;

n is the population size.

If the resulting value will be fractional, then the value of the step value of the interval, take the nearest integer.

3

Then, determine the length of the interval:

i = (Hmag – Xmin)/h, where

HMO – maximum characteristic value in the aggregate;

Xmin – the minimum value of the characteristic in the aggregate.

i = (Hmag – Xmin)/h, where

HMO – maximum characteristic value in the aggregate;

Xmin – the minimum value of the characteristic in the aggregate.

4

Then, complete the boundaries of the interval. They can be specified in different ways: the upper bound of the previous interval and may repeat the lower boundary of the next (5-10, 10-15, 15-20) or not to repeat (5-10, 10,1-15, 15,1-20). For the beginning of the first interval A0 takes the following value:

A0 = GMP – i/2, where

i – length of the interval.

At the end of the j-th interval is assumed A that represents the upper boundary of the j-th interval and the beginning of the (j+1)-th interval:

Aj = A(j-1) + i.

The scale of intervals continues as long as the value of A satisfies the ratio Aj< Hmah + i/2.

A0 = GMP – i/2, where

i – length of the interval.

At the end of the j-th interval is assumed A that represents the upper boundary of the j-th interval and the beginning of the (j+1)-th interval:

Aj = A(j-1) + i.

The scale of intervals continues as long as the value of A satisfies the ratio Aj< Hmah + i/2.

# Advice 2: How to build a variational series

Variation number represented by a certain sequence variants (x(1),...,x(n)), which are arranged in order of descending or-decreasing. The first element of the variational series x(1) is called the minimum: xmin denote it. The last element of this series is called max and is denoted by xmax. On the basis variational series is plotted.

You will need

- - the range;
- - the initial information;
- - notebook;
- - pencil;
- - handle.

Instruction

1

Please note that there are several varieties of variational series: discrete and interval. Each of them has its own peculiarities of construction. Discrete trait variation is variation, the individual values of which differ by a certain value. Continuous variation is considered in the case, if the individual values differ from each other at any value. In the interval variation row of signs do not refer to an individual value but to an entire interval.

2

Before proceeding to the construction of the interval variation series correctly choose the principle on which is based the ranking of individual elements of the interval series. The choice of this or that symptom depends entirely on the homogeneity of the analyzed indicators. For example, if a homogeneous set of indicators, to construct a variational series use the principle of equal intervals.

3

However, before you determine uniform indicators or not, produce a meaningful analysis. Uniformity is determined by constructing a line graph and subsequent analysis to detect abnormal (atypical for the given variational series) of observations. In addition, the principle of equal intervals is used in the construction of variational series with significant irregular, the cause of which is unknown.

4

Correctly determine the amount of space needed to construct the interval variation series: it must be such that, firstly, analyzed the variational series did not seem too bulky, and, secondly, the study clearly discernible features. If the intervals are equal, then the interval size is calculated by the formula: h=R/k where R is the range of variation, and k indicates the number of intervals. Thus R is defined as the difference between xmax and xmin.

5

If you build a discrete variational series, its options are to ascribe the frequency of occurrence of a phenomenon, and the percentages of each option in General the set of analyzed indicators. These shares are calculated as the ratio of specific frequencies to a common indicator, called cystostomy and denote qi. In turn, the relative frequency can be expressed as a percentage and relative numbers.

Note

The correctness of the identification of the relationship between the frequency and the variation of the number depends on how true the schedule will be built.

Useful advice

Sometimes used a method called "splitting intervals": it is necessary to compare the two variational series based on the same symptom, but have different intervals.

# Advice 3: How to build interval series

When

**the number of**distribution already given, you can immediately start his research. But in some problems the initial data is presented just numbers (weight, amount, quantity – any value of the parameter or attribute). In this case, in order to begin your analysis, first you need to build the interval**number**.You will need

- the value of the parameter.

Instruction

1

If the parameter values change over time, use as intervals time intervals, for example, hour, day, month, year. If you select minimum interval, consider the number and scatter of data, try to keep

**the number**distribution was most informative and at the same time compact. For example, if you are given data by months for two years, split for years nothing will be able to speak, and use as a interval a month in some cases will lead to the erosion data. The optimal solution in this case will be broken down by quarters.2

If time for sampling is not matter, form, interval intervals depending on the values. For this rate range, maximum and minimum value, and select the amount of space. You can use this method: subtract from the maximum value to the minimum and the difference, divide by the desired number of intervals. Then set boundaries, of course, better if it will be integers. For example, you are given 32, 33, 35, 38, 45, 47, 48, 50, 58, 59, 63. After settlement you will receive (63-32)/5=6,2. Round the interval size up to 7. Thus, you get the intervals: (32-39), (40-47), (48-55), (56-63).

3

Please note, it is best to make the boundaries of the intervals are not overlapping, i.e. the next interval does not begin with the same number, and more per unit. This will allow you to avoid disagreements and misunderstandings.

4

Once you distribute all the intervals, count the number of values in each of them. Record the results in a table where in one row it will show borders, the other with the number of values lying within the boundaries of this interval. In the above example, the calculation of the number of results will look like this: interval (32-39) includes the values 32, 33, 35, 38 – only 4 values. So, in the first table cell below this interval, enter the number 4. Similarly, calculate values for the following intervals: (40-47) – 2, (48-55) – 2, (56-63) – 3.