You will need

- Calculator, computer

Instruction

1

Let there be a few numbers that characterize any homogeneous magnitudes. For example, the results of izmereniy, weightings, statistical surveys, etc., All represented values must be measured with the same unit. To find the standard deviation, perform the following steps.

Determine the arithmetic mean of all the numbers: add up the numbers and divide the sum by the total number of numbers.

Determine the arithmetic mean of all the numbers: add up the numbers and divide the sum by the total number of numbers.

2

Find the deviation of each number from its mean value: subtract each number from the arithmetic average calculated in the previous paragraph.

3

Determine the variance (dispersion) of numbers: add together the squares of the deviations found previously and divide the resulting amount by the number of numbers.

4

Extract of the variance square root. The resulting number will be the standard deviation of the given set of numbers.

5

Example.

In the house is based on seven patients with fever 34, 35, 36, 37, 38, 39 and 40 degrees Celsius.

It is required to determine the standard deviation from the average temperature.

Solution:

• "the average temperature in the chamber": (34+35+36+37+38+39+40)/7=37 º C;

• deviations of temperature from the average (in this case the normal value): 34-37, 35-37, 36-37, 37-37, 38-37, 39-37, 40-37, it turns out: -3, -2, -1, 0, 1, 2, 3 (º C);

• dispersion: ((-3)2+(-2)2+(-1)2+02+12+22+32)/7=(9+4+1+0+1+4+9)/7=4 (ºС2);

• standard deviation: √4=2 (°C);

Answer: the average chamber temperature – normal: 37 ° C, but the standard deviation of temperature is 2 ° C, which indicates serious problems in patients.

In the house is based on seven patients with fever 34, 35, 36, 37, 38, 39 and 40 degrees Celsius.

It is required to determine the standard deviation from the average temperature.

Solution:

• "the average temperature in the chamber": (34+35+36+37+38+39+40)/7=37 º C;

• deviations of temperature from the average (in this case the normal value): 34-37, 35-37, 36-37, 37-37, 38-37, 39-37, 40-37, it turns out: -3, -2, -1, 0, 1, 2, 3 (º C);

• dispersion: ((-3)2+(-2)2+(-1)2+02+12+22+32)/7=(9+4+1+0+1+4+9)/7=4 (ºС2);

• standard deviation: √4=2 (°C);

Answer: the average chamber temperature – normal: 37 ° C, but the standard deviation of temperature is 2 ° C, which indicates serious problems in patients.

6

If there is a possibility to use the program Excel, the calculation of the variance, respectively, and standard deviation can be significantly simplified.

To do this, place the measurement data in one row (one column) and use an aggregate function VARP. As the function arguments specify the range of cells of the table where you placed the entered number.

To do this, place the measurement data in one row (one column) and use an aggregate function VARP. As the function arguments specify the range of cells of the table where you placed the entered number.