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.

# Advice 2: How to build interval variational series

Interval variational series is a table consisting of two columns or rows. In the first indicates the interval characteristic, the variation of which is considered in the second – the number of population units that fall into this interval (frequency).

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.