Standard Deviation Dispersion

Introduction to standard deviation dispersion:

Standard Deviation can be defined as the measurement of describing the variability of given data set. Standard deviation is also used for measuring the exact value of the number in the given data set. Dispersion is defined as the measuring that’s also including the average deviation, variance, and then standard deviation. The standard deviation and the variance are most widely used for measuring the dispersion.Please express your views of this topic What is Dispersion by commenting on blog.

Formula for Standard Deviation Dispersion:

The formula for the standard deviation dispersion are given below,

`Variance = (sigma)^2 = (sum(x_r - mu)^2)/N`

The standard deviation, `(sigma)` , is the square root of the variance.

What the formula means:
(1)` x_r -mu` – take the mean value and subtract the mean value with each value given.
(2) `(x_r - mu)^2` – square the each of the results obtained from the step -1.this is to get or remove of any minus signs from the value.
(3) `(sum(x_r - mu)^2)` – this is for adding all the values that is obtained from step-2.
(4) Divide step-3 by N, which is the total number of the given values
(5) For obtaining the standard deviation square root the answer to the step-4.

Example Problems for Standard Deviation Dispersion:

Example problem 1:

Calculate the variance and also standard deviation for the following values: 1, 3, 5, 6, 6, 8, 9, and 10.

Solution:

The mean = `(1+3+5+6+6+8+9+10)/ 8 `
= `48/8`
=6

Therefore (`mu` =6)

Step 1: Subtract the mean value from given values

(1-6)=-5
(3-6)=-3
(5-6)=-1
(6-6)= 0
(6-6)= 0
(8-6)= 2
(9-6)= 3
(10-6)=4

Step 2: In this step we have to calculate the square for the mean values 25, 9, 1, 0,  0, 4, 9, 16.

Step 3: After finding square mean ,sum the square mean 25+9+1+0+0+4+9+16=64.

Step 4: Here , N = 8, therefore variance should be =` 64/ 8` = 8

Step 5: Therefore standard deviation = 2.8284.

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Example problem 2:

Calculate the variance and also standard deviation for the following values: 1, 2, 3, 4, and 5.

Solution:

The mean = `(1+2+3+4+5)/ 5`
= `15/5`
=3

Therefore (`mu` =3)

Step 1: Subtract the mean value from given values

(1-3)=-2
(2-3)=-1
(3-3)=0
(4-3)= 1
(5-3)= 2

Step 2: In this step we have to calculate the square for the mean values 4, 1, 0, 1, 2

Step 3: After finding square mean ,sum the square mean 4+1+0+1+2=8.

Step 4: Here , N = 5, therefore variance should be = `8/5` = 1.6

Step 5: Therefore standard deviation = 1.264.